{"id":1412,"date":"2018-04-11T22:51:42","date_gmt":"2018-04-12T02:51:42","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/chapter\/4-1-writing-and-balancing-chemical-equations\/"},"modified":"2019-05-14T13:40:38","modified_gmt":"2019-05-14T17:40:38","slug":"4-1-writing-and-balancing-chemical-equations","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/chapter\/4-1-writing-and-balancing-chemical-equations\/","title":{"raw":"6.1 Writing and Balancing Chemical Equations","rendered":"6.1 Writing and Balancing Chemical Equations"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Objectives<\/h3>\r\nBy the end of this section, you will be able to:\r\n<ul>\r\n \t<li>Derive chemical equations from narrative descriptions of chemical reactions.<\/li>\r\n \t<li>Write and balance chemical equations in molecular, total ionic, and net ionic formats.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-idp47467664\">The preceding chapter introduced the use of element symbols to represent individual atoms. When atoms gain or lose electrons to yield ions, or combine with other atoms to form molecules, their symbols are modified or combined to generate chemical formulas that appropriately represent these species. Extending this symbolism to represent both the identities and the relative quantities of substances undergoing a chemical (or physical) change involves writing and balancing a <strong>chemical equation<\/strong>. Consider as an example the reaction between one methane molecule (CH<sub>4<\/sub>) and two diatomic oxygen molecules (O<sub>2<\/sub>) to produce one carbon dioxide molecule (CO<sub>2<\/sub>) and two water molecules (H<sub>2<\/sub>O). The chemical equation representing this process is provided in the upper half of <a href=\"#CNX_Chem_04_01_rxn2\" class=\"autogenerated-content\">Figure 1<\/a>, with space-filling molecular models shown in the lower half of the figure.<\/p>\r\n\r\n<figure id=\"CNX_Chem_04_01_rxn2\"><figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]<a href=\"https:\/\/opentextbc.ca\/chemistry\/wp-content\/uploads\/sites\/150\/2016\/05\/CNX_Chem_04_01_rxn2.jpg\"><img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_01_rxn2-2.jpg\" alt=\"This figure shows a balanced chemical equation followed below by a representation of the equation using space-filling models. The equation reads C H subscript 4 plus 2 O subscript 2 arrow C O subscript 2 plus 2 H subscript 2 O. Under the C H subscript 4, the molecule is shown with a central black sphere, representing a C atom, to which 4 smaller white spheres, representing H atoms, are distributed evenly around. All four H atoms are bonded to the central black C atom. This is followed by a plus sign. Under the 2 O subscript 2, two molecules are shown. The molecules are each composed of two red spheres bonded together. The red spheres represent O atoms. To the right of an arrow and under the C O subscript 2, appears a single molecule with a black central sphere with two red spheres bonded to the left and right. Following a plus sign and under the 2 H subscript 2 O, are two molecules, each with a central red sphere and two smaller white spheres attached to the lower right and lower left sides of the central red sphere. Note that in space filling models of molecules, spheres appear slightly compressed in regions where there is a bond between two atoms.\" width=\"975\" height=\"444\" \/><\/a> <strong>Figure 1.<\/strong> The reaction between methane and oxygen to yield carbon dioxide and water (shown at bottom) may be represented by a chemical equation using formulas (top).[\/caption]\r\n\r\n<\/figcaption><\/figure>\r\n<p id=\"fs-idp80311968\">This example illustrates the fundamental aspects of any chemical equation:<\/p>\r\n\r\n<ol id=\"fs-idm9329568\">\r\n \t<li>The substances undergoing reaction are called <strong>reactants<\/strong>, and their formulas are placed on the left side of the equation.<\/li>\r\n \t<li>The substances generated by the reaction are called <strong>products<\/strong>, and their formulas are placed on the right sight of the equation.<\/li>\r\n \t<li>Plus signs (+) separate individual reactant and product formulas, and an arrow ($latex \\longrightarrow$) separates the reactant and product (left and right) sides of the equation.<\/li>\r\n \t<li>The relative numbers of reactant and product species are represented by <strong>coefficients<\/strong> (numbers placed immediately to the left of each formula). A coefficient of 1 is typically omitted.<\/li>\r\n<\/ol>\r\n<p id=\"fs-idm33251968\">It is common practice to use the smallest possible whole-number coefficients in a chemical equation, as is done in this example. Realize, however, that these coefficients represent the <em>relative<\/em> numbers of reactants and products, and, therefore, they may be correctly interpreted as ratios. Methane and oxygen react to yield carbon dioxide and water in a 1:2:1:2 ratio. This ratio is satisfied if the numbers of these molecules are, respectively, 1-2-1-2, or 2-4-2-4, or 3-6-3-6, and so on (<a href=\"#CNX_Chem_04_01_rxn3\" class=\"autogenerated-content\">Figure 2<\/a>). Likewise, these coefficients may be interpreted with regard to any amount (number) unit, and so this equation may be correctly read in many ways, including:<\/p>\r\n\r\n<ul id=\"fs-idp42981680\">\r\n \t<li><em>One<\/em> methane molecule and <em>two<\/em> oxygen molecules react to yield <em>one<\/em> carbon dioxide molecule and <em>two<\/em> water molecules.<\/li>\r\n \t<li><em>One dozen<\/em> methane molecules and <em>two dozen<\/em> oxygen molecules react to yield <em>one dozen<\/em> carbon dioxide molecules and <em>two dozen<\/em> water molecules.<\/li>\r\n \t<li><em>One mole<\/em> of methane molecules and <em>2 moles<\/em> of oxygen molecules react to yield <em>1 mole<\/em> of carbon dioxide molecules and <em>2 moles<\/em> of water molecules.<\/li>\r\n<\/ul>\r\n<figure id=\"CNX_Chem_04_01_rxn3\"><figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"1300\"]<a href=\"https:\/\/opentextbc.ca\/chemistry\/wp-content\/uploads\/sites\/150\/2016\/05\/CNX_Chem_04_01_rxn3.jpg\"><img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_01_rxn3-2.jpg\" alt=\"This image has a left side, labeled, \u201cMixture before reaction\u201d separated by a vertical dashed line from right side labeled, \u201cMixture after reaction.\u201d On the left side of the figure, two types of molecules are illustrated with space-filling models. Six of the molecules have only two red spheres bonded together. Three of the molecules have four small white spheres evenly distributed about and bonded to a central, larger black sphere. On the right side of the dashed vertical line, two types of molecules which are different from those on the left side are shown. Six of the molecules have a central red sphere to which smaller white spheres are bonded. The white spheres are not opposite each other on the red atoms, giving the molecule a bent shape or appearance. The second molecule type has a central black sphere to which two red spheres are attached on opposite sides, resulting in a linear shape or appearance. Note that in space filling models of molecules, spheres appear slightly compressed in regions where there is a bond between two atoms. On each side of the dashed line, twelve red, three black, and twelve white spheres are present.\" width=\"1300\" height=\"504\" \/><\/a> <strong>Figure 2.<\/strong> Regardless of the absolute numbers of molecules involved, the ratios between numbers of molecules of each species that react (the reactants) and molecules of each species that form (the products) are the same and are given by the chemical reaction equation.[\/caption]\r\n\r\n<\/figcaption><\/figure>\r\n<section id=\"fs-idp23923488\">\r\n<h2>Balancing Equations<\/h2>\r\n<p id=\"fs-idp13839872\">The chemical equation described in Figure 1 is <strong>balanced<\/strong>, meaning that equal numbers of atoms for each element involved in the reaction are represented on the reactant and product sides. This is a requirement the equation must satisfy to be consistent with the law of conservation of matter. It may be confirmed by simply summing the numbers of atoms on either side of the arrow and comparing these sums to ensure they are equal. Note that the number of atoms for a given element is calculated by multiplying the coefficient of any formula containing that element by the element\u2019s subscript in the formula. If an element appears in more than one formula on a given side of the equation, the number of atoms represented in each must be computed and then added together. For example, both product species in the example reaction, CO<sub>2<\/sub> and H<sub>2<\/sub>O, contain the element oxygen, and so the number of oxygen atoms on the product side of the equation is<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp219991408\" style=\"text-align: center\">$latex (1 \\;\\text{CO}_2 \\;\\text{molecule} \\times \\frac{2 \\;\\text{O atoms}}{\\text{CO}_2 \\;\\text{molecule}}) + (2\\;\\text{H}_2\\text{O molecule} \\times \\frac{1 \\;\\text{O atom}}{\\text{H}_2\\text{O molecule}}) = 4 \\;\\text{O atoms}$<\/div>\r\n<p id=\"fs-idp52518864\">The equation for the reaction between methane and oxygen to yield carbon dioxide and water is confirmed to be balanced per this approach, as shown here:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp38334880\" style=\"text-align: center\">$latex \\text{CH}_4 + 2\\text{O}_2 \\longrightarrow \\text{CO}_2 + 2\\text{H}_2\\text{O}$<\/div>\r\n<table id=\"fs-idp140513680\" class=\"medium unnumbered\" summary=\"This is a table with four columns and four rows. The columns are labeled, \u201cElement,\u201d \u201cReactants,\u201d \u201cProducts,\u201d and \u201cBalanced?\u201d. Under the \u201cElement\u201d column are the letters C, H, and O. Under the \u201cReactants\u201d column are the equations \u201c1 times 1 equals 1,\u201d \u201c4 times 1 equals 4,\u201d and \u201c2 times 2 equals 4.\u201d Under the \u201cProducts\u201d column are the equations, \u201c1 times 1 equals 1,\u201d \u201c2 times 2 equals 4,\u201d and \u201c( 1 times 2 ) plus ( 2 times 1 ) equals 4.\u201d Under the \u201cBalanced?\u201d column are, \u201c1 equals 1, yes,\u201d \u201c4 equals 4, yes,\u201d \u201c4 equals 4, yes.\u201d\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Element<\/th>\r\n<th>Reactants<\/th>\r\n<th>Products<\/th>\r\n<th>Balanced?<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>C<\/td>\r\n<td>1 \u00d7 1 = 1<\/td>\r\n<td>1 \u00d7 1 = 1<\/td>\r\n<td>1 = 1, yes<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>H<\/td>\r\n<td>4 \u00d7 1 = 4<\/td>\r\n<td>2 \u00d7 2 = 4<\/td>\r\n<td>4 = 4, yes<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>O<\/td>\r\n<td>2 \u00d7 2 = 4<\/td>\r\n<td>(1 \u00d7 2) + (2 \u00d7 1) = 4<\/td>\r\n<td>4 = 4, yes<\/td>\r\n<\/tr>\r\n<tr>\r\n<td colspan=\"4\"><strong>Table 1.<\/strong><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-idp8426240\">A balanced chemical equation often may be derived from a qualitative description of some chemical reaction by a fairly simple approach known as balancing by inspection. Consider as an example the decomposition of water to yield molecular hydrogen and oxygen. This process is represented qualitatively by an <em>unbalanced<\/em> chemical equation:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idm60111376\" style=\"text-align: center\">$latex \\text{H}_2\\text{O} \\longrightarrow \\text{H}_2 + \\text{O}_2 \\;(\\text{unbalanced})$<\/div>\r\n<p id=\"fs-idp20832160\">Comparing the number of H and O atoms on either side of this equation confirms its imbalance:<\/p>\r\n\r\n<table id=\"fs-idp104786160\" class=\"medium unnumbered\" summary=\"This is a table with four columns and three rows. The columns are labeled, \u201cElement,\u201d \u201cReactants,\u201d \u201cProducts,\u201d and \u201cBalanced?\u201d. Under the \u201cElement\u201d column are the letters H and O. Under the \u201cReactants\u201d column are the equations \u201c1 times 2 equals 2,\u201d and \u201c1 times 1 equals 1.\u201d Under the \u201cProducts\u201d column are the equations, \u201c1 times 2 equals 2,\u201d and \u201c1 times 2 equals 2.\u201d Under the \u201cBalanced?\u201d column are, \u201c2 equals 2, yes,\u201d and \u201c1 does not equal 2, no.\u201d\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Element<\/th>\r\n<th>Reactants<\/th>\r\n<th>Products<\/th>\r\n<th>Balanced?<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>H<\/td>\r\n<td>1 \u00d7 2 = 2<\/td>\r\n<td>1 \u00d7 2 = 2<\/td>\r\n<td>2 = 2, yes<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>O<\/td>\r\n<td>1 \u00d7 1 = 1<\/td>\r\n<td>1 \u00d7 2 = 2<\/td>\r\n<td>1 \u2260 2, no<\/td>\r\n<\/tr>\r\n<tr>\r\n<td colspan=\"4\"><strong>Table 2.<\/strong><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-idp116453584\">The numbers of H atoms on the reactant and product sides of the equation are equal, but the numbers of O atoms are not. To achieve balance, the <em>coefficients<\/em> of the equation may be changed as needed. Keep in mind, of course, that the <em>formula subscripts<\/em> define, in part, the identity of the substance, and so these cannot be changed without altering the qualitative meaning of the equation. For example, changing the reactant formula from H<sub>2<\/sub>O to H<sub>2<\/sub>O<sub>2<\/sub> would yield balance in the number of atoms, but doing so also changes the reactant\u2019s identity (it\u2019s now hydrogen peroxide and not water). The O atom balance may be achieved by changing the coefficient for H<sub>2<\/sub>O to 2.<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idm1006576\" style=\"text-align: center\">$latex 2\\text{H}_2\\text{O} \\longrightarrow \\text{H}_2 + \\text{O}_2 \\;(\\text{unbalanced})$<\/div>\r\n<table id=\"fs-idm15543696\" class=\"medium unnumbered\" summary=\"This is a table with four columns and three rows. The columns are labeled, \u201cElement,\u201d \u201cReactants,\u201d \u201cProducts,\u201d and \u201cBalanced?\u201d. Under the \u201cElement\u201d column are the letters H and O. Under the \u201cReactants\u201d column are the equations \u201c2 times 2 equals 4,\u201d and \u201c2 times 1 equals 2.\u201d The first 2 in the \u201c2 times 2 equals 4\u201d equation is bold. Under the \u201cProducts\u201d column are the equations, \u201c1 times 2 equals 2,\u201d and \u201c1 times 2 equals 2.\u201d Under the \u201cBalanced?\u201d column are, \u201c4 does not equal 2, no,\u201d and \u201c2 equals 2, yes.\u201d\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Element<\/th>\r\n<th>Reactants<\/th>\r\n<th>Products<\/th>\r\n<th>Balanced?<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>H<\/td>\r\n<td><strong>2<\/strong>\u00a0\u00d7 2 = 4<\/td>\r\n<td>1 \u00d7 2 = 2<\/td>\r\n<td>4 \u2260 2, no<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>O<\/td>\r\n<td>2 \u00d7 1 = 2<\/td>\r\n<td>1 \u00d7 2 = 2<\/td>\r\n<td>2 = 2, yes<\/td>\r\n<\/tr>\r\n<tr>\r\n<td colspan=\"4\"><strong>Table 3.<\/strong><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-idp53902400\">The H atom balance was upset by this change, but it is easily reestablished by changing the coefficient for the H<sub>2<\/sub> product to 2.<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idm94539648\" style=\"text-align: center\">$latex 2\\text{H}_2\\text{O} \\longrightarrow 2\\text{H}_2 + \\text{O}_2 \\;(\\text{balanced})$<\/div>\r\n<table id=\"fs-idp151419504\" class=\"medium unnumbered\" summary=\"This is a table with four columns and three rows. The columns are labeled, \u201cElement,\u201d \u201cReactants,\u201d \u201cProducts,\u201d and \u201cBalanced?\u201d. Under the \u201cElement\u201d column are the letters H and O. Under the \u201cReactants\u201d column are the equations \u201c2 times 2 equals 4,\u201d and \u201c2 times 1 equals 2.\u201d Under the \u201cProducts\u201d column are the equations, \u201c2 times 2 equals 2,\u201d and \u201c1 times 2 equals 2.\u201d The first 2 in the \u201c2 times 2 equals 2\u201d equation is bold. Under the \u201cBalanced?\u201d column are, \u201c4 equals 4, yes,\u201d and \u201c2 equals 2, yes.\u201d\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Element<\/th>\r\n<th>Reactants<\/th>\r\n<th>Products<\/th>\r\n<th>Balanced?<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>H<\/td>\r\n<td>2 \u00d7 2 = 4<\/td>\r\n<td><strong>2<\/strong> \u00d7 2 = 4<\/td>\r\n<td>4 = 4, yes<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>O<\/td>\r\n<td>2 \u00d7 1 = 2<\/td>\r\n<td>1 \u00d7 2 = 2<\/td>\r\n<td>2 = 2, yes<\/td>\r\n<\/tr>\r\n<tr>\r\n<td colspan=\"4\"><strong>Table 4.<\/strong><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-idp147875376\">These coefficients yield equal numbers of both H and O atoms on the reactant and product sides, and the balanced equation is, therefore:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp105215584\" style=\"text-align: center\">$latex 2\\text{H}_2\\text{O} \\longrightarrow 2\\text{H}_2 + \\text{O}_2 $<\/div>\r\n<div class=\"textbox shaded\" id=\"fs-idp1783792\">\r\n<h3>Example 1<\/h3>\r\n<p id=\"fs-idp22283888\">Write a balanced equation for the reaction of molecular nitrogen (N<sub>2<\/sub>) and oxygen (O<sub>2<\/sub>) to form dinitrogen pentoxide.<\/p>\r\n&nbsp;\r\n<p id=\"fs-idp154895936\"><strong>Solution<\/strong>\r\nFirst, write the unbalanced equation.<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp147872544\" style=\"text-align: center\">$latex \\text{N}_2 + \\text{O}_2 \\longrightarrow \\text{N}_2 \\text{O}_5 \\;(\\text{unbalanced})$<\/div>\r\n<p id=\"fs-idp239172416\">Next, count the number of each type of atom present in the unbalanced equation.<\/p>\r\n\r\n<table id=\"fs-idp107503280\" class=\"medium unnumbered\" summary=\"This is a table with four columns and three rows. The columns are labeled, \u201cElement,\u201d \u201cReactants,\u201d \u201cProducts,\u201d and \u201cBalanced?\u201d. Under the \u201cElement\u201d column are the letters N and O. Under the \u201cReactants\u201d column are the equations \u201c1 times 2 equals 2,\u201d and \u201c1 times 2 equals 2.\u201d Under the \u201cProducts\u201d column are the equations, \u201c1 times 2 equals 2,\u201d and \u201c1 times 5 equals 5.\u201d Under the \u201cBalanced?\u201d column are, \u201c2 equals 2, yes,\u201d and \u201c2 does not equal 5, no.\u201d\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Element<\/th>\r\n<th>Reactants<\/th>\r\n<th>Products<\/th>\r\n<th>Balanced?<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>N<\/td>\r\n<td>1 \u00d7 2 = 2<\/td>\r\n<td>1 \u00d7 2 = 2<\/td>\r\n<td>2 = 2, yes<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>O<\/td>\r\n<td>1 \u00d7 2 = 2<\/td>\r\n<td>1 \u00d7 5 = 5<\/td>\r\n<td>2 \u2260 5, no<\/td>\r\n<\/tr>\r\n<tr>\r\n<td colspan=\"4\"><strong>Table 5.<\/strong><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-idp266947552\">Though nitrogen is balanced, changes in coefficients are needed to balance the number of oxygen atoms. To balance the number of oxygen atoms, a reasonable first attempt would be to change the coefficients for the O<sub>2<\/sub> and N<sub>2<\/sub>O<sub>5<\/sub> to integers that will yield 10 O atoms (the least common multiple for the O atom subscripts in these two formulas).<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp80836160\" style=\"text-align: center\">$latex \\text{N}_2 + 5\\text{O}_2 \\longrightarrow 2\\text{N}_2\\text{O}_5 \\;(\\text{unbalanced})$<\/div>\r\n<table id=\"fs-idp7305424\" class=\"medium unnumbered\" summary=\"This is a table with four columns and three rows. The columns are labeled, \u201cElement,\u201d \u201cReactants,\u201d \u201cProducts,\u201d and \u201cBalanced?\u201d. Under the \u201cElement\u201d column are the letters N and O. Under the \u201cReactants\u201d column are the equations \u201c1 times 2 equals 2,\u201d and \u201c5 times 2 equals 10.\u201d The 5 in the second equation is bold. Under the \u201cProducts\u201d column are the equations, \u201c2 times 2 equals 4,\u201d and \u201c2 times 5 equals 10.\u201d The initial 2 in each equation is bold. Under the \u201cBalanced?\u201d column are, \u201c2 does not equal 4, no,\u201d and \u201c10 equals 10, yes.\u201d\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Element<\/th>\r\n<th>Reactants<\/th>\r\n<th>Products<\/th>\r\n<th>Balanced?<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>N<\/td>\r\n<td>1 \u00d7\u00d7 2 = 2<\/td>\r\n<td><strong>2<\/strong> \u00d7 2 = 4<\/td>\r\n<td>2 \u2260 4, no<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>O<\/td>\r\n<td><strong>5<\/strong> \u00d7 2 = 10<\/td>\r\n<td><strong>2<\/strong> \u00d7 5 = 10<\/td>\r\n<td>10 = 10, yes<\/td>\r\n<\/tr>\r\n<tr>\r\n<td colspan=\"4\"><strong>Table 6.<\/strong><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-idp27772864\">The N atom balance has been upset by this change; it is restored by changing the coefficient for the reactant N<sub>2<\/sub> to 2.<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp107722208\" style=\"text-align: center\">$latex 2\\text{N}_2 + 5\\text{O}_2 \\longrightarrow 2\\text{N}_2 \\text{O}_5 $<\/div>\r\n<table id=\"fs-idm9607408\" class=\"medium unnumbered\" summary=\"This is a table with four columns and three rows. The columns are labeled, \u201cElement,\u201d \u201cReactants,\u201d \u201cProducts,\u201d and \u201cBalanced?\u201d. Under the \u201cElement\u201d column are the letters N and O. Under the \u201cReactants\u201d column are the equations \u201c2 times 2 equals 4,\u201d and \u201c5 times 2 equals 10.\u201d The initial 2 in the first equation is bold. Under the \u201cProducts\u201d column are the equations, \u201c2 times 2 equals 4,\u201d and \u201c2 times 5 equals 10.\u201d Under the \u201cBalanced?\u201d column are, \u201c4 equals 4, yes,\u201d and \u201c10 equals 10, yes.\u201d\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Element<\/th>\r\n<th>Reactants<\/th>\r\n<th>Products<\/th>\r\n<th>Balanced?<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>N<\/td>\r\n<td><strong>2<\/strong> \u00d7 2 = 4<\/td>\r\n<td>2 \u00d7 2 = 4<\/td>\r\n<td>4 = 4, yes<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>O<\/td>\r\n<td>5 \u00d7 2 = 10<\/td>\r\n<td>2 \u00d7 5 = 10<\/td>\r\n<td>10 = 10, yes<\/td>\r\n<\/tr>\r\n<tr>\r\n<td colspan=\"4\"><strong>Table 7.<\/strong><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-idp17727056\">The numbers of N and O atoms on either side of the equation are now equal, and so the equation is balanced.<\/p>\r\n&nbsp;\r\n<p id=\"fs-idp79125936\"><em><strong>Test Yourself<\/strong><\/em>\r\nWrite a balanced equation for the decomposition of ammonium nitrate to form molecular nitrogen, molecular oxygen, and water. (Hint: Balance oxygen last, since it is present in more than one molecule on the right side of the equation.)<\/p>\r\n&nbsp;\r\n\r\n<em><strong>Answer<\/strong><\/em>\r\n\r\n$latex 2\\text{NH}_4 \\text{NO}_3 \\longrightarrow 2\\text{N}_2 + \\text{O}_2 + 4\\text{H}_2\\text{O} $\r\n\r\n<\/div>\r\n<p id=\"fs-idp261408544\">It is sometimes convenient to use fractions instead of integers as intermediate coefficients in the process of balancing a chemical equation. When balance is achieved, all the equation\u2019s coefficients may then be multiplied by a whole number to convert the fractional coefficients to integers without upsetting the atom balance. For example, consider the reaction of ethane (C<sub>2<\/sub>H<sub>6<\/sub>) with oxygen to yield H<sub>2<\/sub>O and CO<sub>2<\/sub>, represented by the unbalanced equation:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp25579440\" style=\"text-align: center\">$latex \\text{C}_2 \\text{H}_6 + \\text{O}_2 \\longrightarrow \\text{H}_2 \\text{O} + \\text{C} \\text{O}_2 \\;(\\text{unbalanced}) $<\/div>\r\n<p id=\"fs-idp115010864\">Following the usual inspection approach, one might first balance C and H atoms by changing the coefficients for the two product species, as shown:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp71981056\" style=\"text-align: center\">$latex \\text{C}_2 \\text{H}_6 + \\text{O}_2 \\longrightarrow 3\\text{H}_2 \\text{O} + 2\\text{C} \\text{O}_2 \\;(\\text{unbalanced}) $<\/div>\r\n<p id=\"fs-idp16121888\">This results in seven O atoms on the product side of the equation, an odd number\u2014no integer coefficient can be used with the O<sub>2<\/sub> reactant to yield an odd number, so a fractional coefficient, $latex \\frac{7}{2}$, is used instead to yield a provisional balanced equation:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp37794464\" style=\"text-align: center\">$latex \\text{C}_2 \\text{H}_6 + \\frac{7}{2}\\text{O}_2 \\longrightarrow 3\\text{H}_2 \\text{O} + 2\\text{C} \\text{O}_2 \\; $<\/div>\r\n<p id=\"fs-idp54942992\">A conventional balanced equation with integer-only coefficients is derived by multiplying each coefficient by 2:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idm26016768\" style=\"text-align: center\">$latex 2\\text{C}_2 \\text{H}_6 + 7\\text{O}_2 \\longrightarrow 6\\text{H}_2 \\text{O} + 4\\text{C} \\text{O}_2 \\; $<\/div>\r\n<p id=\"fs-idp57635360\">Finally with regard to balanced equations, recall that convention dictates use of the <em>smallest whole-number coefficients<\/em>. Although the equation for the reaction between molecular nitrogen and molecular hydrogen to produce ammonia is, indeed, balanced,<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp93360688\" style=\"text-align: center\">$latex 3\\text{N}_2 + 9\\text{H}_2 \\longrightarrow 6\\text{N} \\text{H}_3 $<\/div>\r\n<p id=\"fs-idp117890176\">the coefficients are not the smallest possible integers representing the relative numbers of reactant and product molecules. Dividing each coefficient by the greatest common factor, 3, gives the preferred equation:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idm15551488\" style=\"text-align: center\">$latex \\text{N}_2 + 3\\text{H}_2 \\longrightarrow 2\\text{N} \\text{H}_3 $<\/div>\r\n<div id=\"fs-idp54281248\" class=\"textbox shaded\">\r\n\r\n<img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Interactive_200DPI-3-2.png\" alt=\"\u00a0\" width=\"140\" height=\"87\" class=\"alignleft\" \/>\r\n\r\n&nbsp;\r\n<p id=\"fs-idp94095504\">Use this interactive <a href=\"http:\/\/openstaxcollege.org\/l\/16BalanceEq\">tutorial<\/a> for additional practice balancing equations.<\/p>\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/section><section id=\"fs-idp50900304\">\r\n<h2>Additional Information in Chemical Equations<\/h2>\r\n<p id=\"fs-idp46664848\">The physical states of reactants and products in chemical equations very often are indicated with a parenthetical abbreviation following the formulas. Common abbreviations include <em>s<\/em> for solids, <em>l<\/em> for liquids, <em>g<\/em> for gases, and <em>aq<\/em> for substances dissolved in water (<em>aqueous solutions<\/em>, as introduced in the preceding chapter). These notations are illustrated in the example equation here:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp55940384\" style=\"text-align: center\">$latex 2\\text{Na}(s) + 2\\text{H}_2 \\text{O}(l) \\longrightarrow 2\\text{NaOH}(aq) + \\text{H}_2(g) $<\/div>\r\n<p id=\"fs-idp38373104\">This equation represents the reaction that takes place when sodium metal is placed in water. The solid sodium reacts with liquid water to produce molecular hydrogen gas and the ionic compound sodium hydroxide (a solid in pure form, but readily dissolved in water).<\/p>\r\n<p id=\"fs-idp53095760\">Special conditions necessary for a reaction are sometimes designated by writing a word or symbol above or below the equation\u2019s arrow. For example, a reaction carried out by heating may be indicated by the uppercase Greek letter delta (\u0394) over the arrow.<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp144225920\" style=\"text-align: center\">$latex \\text{CaCO}_3(s) \\;\\xrightarrow{\\Delta} \\; \\text{CaO}(s) + \\text{CO}_2(g) $<\/div>\r\n<p id=\"fs-idp104838720\">Other examples of these special conditions will be encountered in more depth in later chapters.<\/p>\r\n\r\n<\/section><section id=\"fs-idp44459328\">\r\n<h2>Ionic Compounds in Solution<\/h2>\r\n<p id=\"ball-ch04_s03_p02\" class=\"para editable block\">One important aspect about ionic compounds that differs from molecular compounds has to do with dissolving in a liquid, such as water. When molecular compounds, such as sugar, dissolve in water, the individual molecules drift apart from each other. When ionic compounds dissolve, <em class=\"emphasis\">the ions physically separate from each other<\/em>. We can use a chemical equation to represent this process\u2014for example, with NaCl:<\/p>\r\n\r\n<\/section>&nbsp;\r\n\r\n<section id=\"fs-idp44459328\"><a href=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/Equation-1.png\"><img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/Equation-1-300x72.png\" alt=\"\" width=\"300\" height=\"72\" class=\"wp-image-4870 size-medium alignleft\" \/><\/a>\r\n<p style=\"text-align: center\"><a href=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/Ionic-Compounds-1.png\" style=\"font-weight: bold;font-size: 14pt\"><img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/Ionic-Compounds-1-300x300.png\" alt=\"\" width=\"200\" height=\"201\" class=\"wp-image-2169\" \/><\/a><\/p>\r\n<p style=\"text-align: center\"><strong>Figure 3.<\/strong>\u00a0The dissolution of sodium chloride.<\/p>\r\n\r\n<div class=\"figure large medium-height editable block\" id=\"ball-ch04_s03_f01\">\r\n\r\nWhen NaCl dissolves in water, the ions separate and go their own way in solution; the ions are now written with their respective charges, and the (aq) phase label emphasizes that they are dissolved (<a class=\"xref\" href=\"#ball-ch04_s03_f01\">Figure 3 \"Ionic Solutions\"<\/a>). This process is called <span class=\"margin_term\"><a class=\"glossterm\">dissociation<\/a><\/span>; we say that the ions <em class=\"emphasis\">dissociate<\/em>.\r\n<p class=\"para\">When an ionic compound dissociates in water, water molecules surround each ion and separate it from the rest of the solid. Each ion goes its own way in solution.<\/p>\r\n\r\n<\/div>\r\n<p id=\"ball-ch04_s03_p04\" class=\"para editable block\">All ionic compounds that dissolve behave this way. (This behaviour was first suggested by the Swedish chemist Svante August Arrhenius [1859\u20131927] as part of his PhD dissertation in 1884. Interestingly, his PhD examination team had a hard time believing that ionic compounds would behave like this, so they gave Arrhenius a barely passing grade. Later, this work was cited when Arrhenius was awarded the Nobel Prize in Chemistry.)<\/p>\r\n<p class=\"para editable block\">Keep in mind that when the ions separate, <em class=\"emphasis\">all<\/em>\u00a0of the ions separate. Thus, when CaCl<sub class=\"subscript\">2<\/sub> dissolves, the one Ca<sup class=\"superscript\">2+<\/sup> ion and the two Cl<sup class=\"superscript\">\u2212<\/sup> ions separate from each other:<\/p>\r\n<p style=\"text-align: center\"><span class=\"informalequation block\">CaCl<sub>2<\/sub>(s) $latex \\longrightarrow$ Ca<sup>2+<\/sup>(aq) + Cl<sup>-<\/sup>(aq) + Cl<sup>-<\/sup>(aq) <\/span><\/p>\r\n<p style=\"text-align: center\"><span class=\"informalequation block\">or<\/span><\/p>\r\n<p style=\"text-align: center\"><span class=\"informalequation block\">CaCl<sub>2<\/sub>(s) $latex \\longrightarrow$ Ca<sup>2+<\/sup>(aq) + 2Cl<sup>-<\/sup>(aq)<\/span><\/p>\r\n<p id=\"ball-ch04_s03_p05\" class=\"para editable block\">That is, the two chloride ions go off on their own. They do not remain as Cl<sub class=\"subscript\">2<\/sub> (that would be elemental chlorine; these are chloride ions); they do not stick together to make Cl<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">\u2212<\/sup> or Cl<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">2\u2212<\/sup>. They become dissociated ions in their own right. Polyatomic ions also retain their overall identity when they are dissolved.<\/p>\r\n\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Example 2<\/h3>\r\n<p id=\"ball-ch04_s03_p06\" class=\"para\">Write the chemical equation that represents the dissociation of each ionic compound.<\/p>\r\n<p class=\"para\">a) KBr \u00a0 \u00a0 \u00a0b)\u00a0Na<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub><\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\r\n<p class=\"simpara\">a) KBr(s)\u00a0$latex \\longrightarrow$ K<sup class=\"superscript\">+<\/sup>(aq) +\u00a0Br<sup class=\"superscript\">\u2212<\/sup>(aq)<\/p>\r\n<p class=\"simpara\">b) Not only do the two sodium ions go their own way, but the sulfate ion stays together as the sulfate ion. The dissolving equation is<\/p>\r\n<span class=\"informalequation\"><span class=\"mathphrase\">Na<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub>(s) $latex \\longrightarrow$ 2Na<sup class=\"superscript\">+<\/sup>(aq) +\u00a0SO<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">2\u2212<\/sup>(aq)<\/span><\/span>\r\n\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\r\n<p id=\"ball-ch04_s03_p07\" class=\"para\">Write the chemical equation that represents the dissociation of (NH<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">2<\/sub>S.<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\r\n<p id=\"ball-ch04_s03_p08\" class=\"para\">(NH<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">2<\/sub>S(s) $latex \\longrightarrow$ 2NH<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup>(aq) +\u00a0S<sup class=\"superscript\">2\u2212<\/sup>(aq)<\/p>\r\n\r\n<\/div>\r\n<h2>Equations for Ionic Reactions<\/h2>\r\n<p id=\"fs-idm1065424\">Given the abundance of water on earth, it stands to reason that a great many chemical reactions take place in aqueous media. When ions are involved in these reactions, the chemical equations may be written with various levels of detail appropriate to their intended use. To illustrate this, consider a reaction between ionic compounds taking place in an aqueous solution. When aqueous solutions of CaCl<sub>2<\/sub> and AgNO<sub>3<\/sub> are mixed, a reaction takes place producing aqueous Ca(NO<sub>3<\/sub>)<sub>2<\/sub> and solid AgCl:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp15979216\" style=\"text-align: center\">$latex \\text{CaCl}_2(aq) + 2\\text{AgNO}_3(aq) \\longrightarrow \\text{Ca(NO}_3)_2(aq) + 2\\text{AgCl}(s)$<\/div>\r\n<p id=\"fs-idp204481712\">This balanced equation, derived in the usual fashion, is called a <strong>molecular equation<\/strong> because it doesn\u2019t explicitly represent the ionic species that are present in solution. When ionic compounds dissolve in water, they may <em>dissociate<\/em> into their constituent ions, which are subsequently dispersed homogenously throughout the resulting solution (a thorough discussion of this important process is provided in the chapter on solutions). Ionic compounds dissolved in water are, therefore, more realistically represented as dissociated ions, in this case:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idm1980352\" style=\"text-align: center\">$latex \\begin{array}{r @{{}\\longrightarrow{}} l} \\text{CaCl}_2(aq) &amp; \\text{Ca}^{2+}(aq) + 2 \\text{Cl}^{-}(aq) \\\\[0.5em] 2 \\text{AgNO}_3(aq) &amp; 2\\text{Ag}^{+}(aq) + 2 {\\text{NO}_3}^{-}(aq) \\\\[0.5em] \\text{Ca(NO}_3)_2(aq) &amp; \\text{Ca}^{2+}(aq) + 2 {\\text{NO}_3}^{-}(aq) \\end{array}$<\/div>\r\n<p id=\"fs-idp24388656\">Unlike these three ionic compounds, AgCl does not dissolve in water to a significant extent, as signified by its physical state notation, <em>s<\/em>.<\/p>\r\n<p id=\"fs-idp6577008\">Explicitly representing all dissolved ions results in a <strong>complete ionic equation<\/strong>. In this particular case, the formulas for the dissolved ionic compounds are replaced by formulas for their dissociated ions:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp116409536\" style=\"text-align: center\">$latex \\text{Ca}^{2+}(aq) + 2\\text{Cl}^{-}(aq) + 2\\text{Ag}^{+}(aq) + 2{\\text{NO}_3}^{-}(aq) \\longrightarrow \\text{Ca}^{2+}(aq) + 2{\\text{NO}_3}^{-}(aq) + 2\\text{AgCl}(s) $<\/div>\r\n<p id=\"fs-idp146097648\">Examining this equation shows that two chemical species are present in identical form on both sides of the arrow, Ca<sup>2+<\/sup>(<em>aq<\/em>) and NO<sub>3<\/sub><sup>\u2212<\/sup>(aq).NO<sub>3<\/sub><sup>\u2212<\/sup>(aq). These <strong>spectator ions<\/strong>\u2014ions whose presence is required to maintain charge neutrality\u2014are neither chemically nor physically changed by the process, and so they may be eliminated from the equation to yield a more succinct representation called a <strong>net ionic equation<\/strong>:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp235470368\" style=\"text-align: center\">$latex \\rule[0.5ex]{4em}{0.1ex}\\hspace{-4em} \\text{Ca}^{2+}(aq) + 2\\text{Cl}^{-}(aq) + 2\\text{Ag}^{+}(aq) + \\rule[0.5ex]{4.5em}{0.1ex}\\hspace{-4.5em} 2\\text{NO}_3^{-}(aq) \\longrightarrow \\rule[0.5ex]{4em}{0.1ex}\\hspace{-4em} \\text{Ca}^{2+}(aq) + \\rule[0.5ex]{4.5em}{0.1ex}\\hspace{-4.5em} 2{\\text{NO}_3}^{-}(aq) + 2\\text{AgCl}(s) $<\/div>\r\n<div class=\"equation\"><\/div>\r\n<div class=\"equation\" style=\"text-align: center\">$latex 2\\text{Cl}^{-}(aq) + 2\\text{Ag}^{+}(aq) \\longrightarrow 2\\text{AgCl}(s) $<\/div>\r\n<p id=\"fs-idp8534256\">Following the convention of using the smallest possible integers as coefficients, this equation is then written:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp266868512\" style=\"text-align: center\">$latex \\text{Cl}^{-}(aq) + \\text{Ag}^{+}(aq) \\longrightarrow \\text{AgCl}(s) $<\/div>\r\n<p id=\"fs-idp42466432\">This net ionic equation indicates that solid silver chloride may be produced from dissolved chloride and silver(I) ions, regardless of the source of these ions. These molecular and complete ionic equations provide additional information, namely, the ionic compounds used as sources of Cl<sup>\u2212<\/sup> and Ag<sup>+<\/sup>.<\/p>\r\n\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Example 3<\/h3>\r\n<p id=\"ball-ch04_s03_p11\" class=\"para\">Write the complete ionic equation for each chemical reaction.<\/p>\r\n<p class=\"para\">a) KBr(aq) +\u00a0AgC<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub>(aq) $latex \\longrightarrow$ KC<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub>(aq) +\u00a0AgBr(s)<\/p>\r\n<p class=\"para\">b) MgSO<sub class=\"subscript\">4<\/sub>(aq) +\u00a0Ba(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub>(aq) $latex \\longrightarrow$ Mg(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub>(aq) +\u00a0BaSO<sub class=\"subscript\">4<\/sub>(s)<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\r\n<p id=\"ball-ch04_s03_p12\" class=\"para\">For any ionic compound that is aqueous, we will write the compound as separated ions.<\/p>\r\n<p class=\"para\">a) The complete ionic equation is<\/p>\r\n<span class=\"informalequation\"><span class=\"mathphrase\">K<sup class=\"superscript\">+<\/sup>(aq) +\u00a0Br<sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a0Ag<sup class=\"superscript\">+<\/sup>(aq) +\u00a0C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq) $latex \\longrightarrow$ K<sup class=\"superscript\">+<\/sup>(aq) +\u00a0C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a0AgBr(s)<\/span><\/span>\r\n\r\nb) The complete ionic equation is\r\n\r\n<span class=\"informalequation\"><span class=\"mathphrase\">Mg<sup class=\"superscript\">2+<\/sup>(aq) +\u00a0SO<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">2\u2212<\/sup>(aq) +\u00a0Ba<sup class=\"superscript\">2+<\/sup>(aq) +\u00a02NO<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq) $latex \\longrightarrow$ Mg<sup class=\"superscript\">2+<\/sup>(aq) +\u00a02NO<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a0BaSO<sub class=\"subscript\">4<\/sub>(s)<\/span><\/span>\r\n\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\r\n<p id=\"ball-ch04_s03_p13\" class=\"para\">Write the complete ionic equation for<\/p>\r\n<span class=\"informalequation\"><span class=\"mathphrase\">CaCl<sub class=\"subscript\">2<\/sub>(aq) +\u00a0Pb(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub>(aq) $latex \\longrightarrow$ Ca(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub>(aq) +\u00a0PbCl<sub class=\"subscript\">2<\/sub>(s)<\/span><\/span>\r\n\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\r\n<p id=\"ball-ch04_s03_p14\" class=\"para\">Ca<sup class=\"superscript\">2+<\/sup>(aq) +\u00a02Cl<sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a0Pb<sup class=\"superscript\">2+<\/sup>(aq) +\u00a02NO<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq) $latex \\longrightarrow$ Ca<sup class=\"superscript\">2+<\/sup>(aq) +\u00a02NO<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a0PbCl<sub class=\"subscript\">2<\/sub>(s)<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Example 4<\/h3>\r\n<p id=\"ball-ch04_s03_p19\" class=\"para\">Write the net ionic equation for each chemical reaction.<\/p>\r\n<p class=\"para\">a) K<sup class=\"superscript\">+<\/sup>(aq) +\u00a0Br<sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a0Ag<sup class=\"superscript\">+<\/sup>(aq) +\u00a0C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq) $latex \\longrightarrow$ K<sup class=\"superscript\">+<\/sup>(aq) +\u00a0C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a0AgBr(s)<\/p>\r\n<p class=\"para\">b) Mg<sup class=\"superscript\">2+<\/sup>(aq) +\u00a0SO<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">2\u2212<\/sup>(aq) +\u00a0Ba<sup class=\"superscript\">2+<\/sup>(aq) +\u00a02 NO<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq) $latex \\longrightarrow$ Mg<sup class=\"superscript\">2+<\/sup>(aq) +\u00a02 NO<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a0BaSO<sub class=\"subscript\">4<\/sub>(s)<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\r\n<p class=\"simpara\">a) In the first equation, the K<sup class=\"superscript\">+<\/sup>(aq) and C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq) ions are spectator ions, so they are canceled:<\/p>\r\n<span class=\"informalequation\">\u00a0K<sup>+<\/sup>(aq) + Br<sup>\u2212<\/sup>(aq) + Ag<sup>+<\/sup>(aq) + C<sub>2<\/sub>H<sub>3<\/sub>O<sub>2<\/sub><sup>\u2212<\/sup>(aq) $latex \\longrightarrow$ K<sup>+<\/sup>(aq) + C<sub>2<\/sub>H<sub>3<\/sub>O<sub>2<\/sub><sup>\u2212<\/sup>(aq) + AgBr(s)<\/span>\r\n<p id=\"ball-ch04_s03_p20\" class=\"para\">The net ionic equation is<\/p>\r\n<span class=\"informalequation\"><span class=\"mathphrase\">Br<sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a0Ag<sup class=\"superscript\">+<\/sup>(aq) $latex \\longrightarrow$ AgBr(s)<\/span><\/span>\r\n\r\nb) In the second equation, the Mg<sup class=\"superscript\">2+<\/sup>(aq) and NO<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq) ions are spectator ions, so they are canceled:\r\n\r\n<span class=\"informalequation\">Mg<sup>2+<\/sup>(aq) + SO<sub>4<\/sub><sup>2\u2212<\/sup>(aq) + Ba<sup>2+<\/sup>(aq) +\u00a02 NO<sub>3<\/sub><sup>\u2212<\/sup>(aq) $latex \\longrightarrow$ Mg<sup>2+<\/sup>(aq) + 2 NO<sub>3<\/sub><sup>\u2212<\/sup>(aq) + BaSO<sub>4<\/sub>(s)<\/span>\r\n<p id=\"ball-ch04_s03_p21\" class=\"para\">The net ionic equation is<\/p>\r\n<span class=\"informalequation\"><span class=\"mathphrase\">SO<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">2\u2212<\/sup>(aq) +\u00a0Ba<sup class=\"superscript\">2+<\/sup>(aq) $latex \\longrightarrow$ BaSO<sub class=\"subscript\">4<\/sub>(s)<\/span><\/span>\r\n\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\r\n<p id=\"ball-ch04_s03_p22\" class=\"para\">Write the net ionic equation for<\/p>\r\n<span class=\"informalequation\"><span class=\"mathphrase\">CaCl<sub class=\"subscript\">2<\/sub>(aq) +\u00a0Pb(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub>(aq) $latex \\longrightarrow$ Ca(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub>(aq) +\u00a0PbCl<sub class=\"subscript\">2<\/sub>(s)<\/span><\/span>\r\n\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\r\n<p id=\"ball-ch04_s03_p23\" class=\"para\">Pb<sup class=\"superscript\">2+<\/sup>(aq) +\u00a02Cl<sup class=\"superscript\">\u2212<\/sup>(aq) $latex \\longrightarrow$ PbCl<sub class=\"subscript\">2<\/sub>(s)<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\" id=\"fs-idp44418832\">\r\n<h3>Example 5<\/h3>\r\n<p id=\"fs-idm27450736\">When carbon dioxide is dissolved in an aqueous solution of sodium hydroxide, the mixture reacts to yield aqueous sodium carbonate and liquid water. Write balanced molecular, complete ionic, and net ionic equations for this process.<\/p>\r\n&nbsp;\r\n<p id=\"fs-idp41941568\"><strong>Solution<\/strong>\r\nBegin by identifying formulas for the reactants and products and arranging them properly in chemical equation form:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp114521376\" style=\"text-align: center\">$latex \\text{CO}_2(aq) + \\text{NaOH}(aq) \\longrightarrow \\text{Na}_2 \\text{CO}_3(aq) + \\text{H}_2 \\text{O}(l) \\;(\\text{unbalanced}) $<\/div>\r\n<p id=\"fs-idp277452576\">Balance is achieved easily in this case by changing the coefficient for NaOH to 2, resulting in the molecular equation for this reaction:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp54247424\" style=\"text-align: center\">$latex \\text{CO}_2(aq) + 2\\text{NaOH}(aq) \\longrightarrow \\text{Na}_2 \\text{CO}_3(aq) + \\text{H}_2 \\text{O}(l) $<\/div>\r\n<p id=\"fs-idp71749568\">The two dissolved ionic compounds, NaOH and Na<sub>2<\/sub>CO<sub>3<\/sub>, can be represented as dissociated ions to yield the complete ionic equation:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp41721264\" style=\"text-align: center\">$latex \\text{CO}_2(aq) + 2\\text{Na}^{+}(aq) + 2\\text{OH}^{-}(aq) \\longrightarrow 2\\text{Na}^{+}(aq) + {\\text{CO}_3}^{2-}(aq) + \\text{H}_2 \\text{O}(l) $<\/div>\r\n<p id=\"fs-idp111986480\">Finally, identify the spectator ion(s), in this case Na<sup>+<\/sup>(<em>aq<\/em>), and remove it from each side of the equation to generate the net ionic equation:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp56132768\">\r\n<p style=\"text-align: center\">$latex \\text{CO}_2(aq) + \\rule[0.5ex]{4.25em}{0.1ex}\\hspace{-4.25em} 2\\text{Na}^{+}(aq) + 2\\text{OH}^{-}(aq) \\longrightarrow \\rule[0.5ex]{4.25em}{0.1ex}\\hspace{-4.25em} 2\\text{Na}^{+}(aq) + {\\text{CO}_3}^{2-}(aq) + \\text{H}_2 \\text{O}(l) $$latex \\text{CO}_2(aq) + 2\\text{OH}^{-}(aq) \\longrightarrow {\\text{CO}_3}^{2-}(aq) + \\text{H}_2 \\text{O}(l) $<\/p>\r\n\r\n<\/div>\r\n&nbsp;\r\n<p id=\"fs-idp71485312\"><em><strong>Test Yourself<\/strong><\/em>\r\nDiatomic chlorine and sodium hydroxide (lye) are commodity chemicals produced in large quantities, along with diatomic hydrogen, via the electrolysis of brine, according to the following unbalanced equation:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp47302160\" style=\"text-align: center\">$latex \\text{NaCl}(aq) + \\text{H}_2 \\text{O}(l) \\;\\;\\xrightarrow{\\text{electricity}}\\;\\; \\text{NaOH}(aq) + \\text{H}_2(g) + \\text{Cl}_2(g) $<\/div>\r\n<p id=\"fs-idp124894848\">Write balanced molecular, complete ionic, and net ionic equations for this process.<\/p>\r\n&nbsp;\r\n\r\n<em><strong>Answers<\/strong><\/em>\r\n\r\n$latex 2\\text{NaCl}(aq) + 2\\text{H}_2 \\text{O} \\longrightarrow 2 \\text{NaOH}(aq) + \\text{H}_2(g) + \\text{Cl}_2(g) (\\text{molecular}) $\r\n\r\n$latex 2\\text{Na}^{+}(aq) + 2\\text{Cl}^{-}(aq) + 2\\text{H}_2 \\text{O} \\longrightarrow 2\\text{Na}^{+}(aq) + 2\\text{OH}^{-}(aq) + \\text{H}_2(g) + \\text{Cl}_2(g) (\\text{complete ionic}) $\r\n\r\n$latex 2\\text{Cl}^{-}(aq) + 2\\text{H}_2 \\text{O} \\longrightarrow 2\\text{OH}^{-}(aq) + 2\\text{H}_2(g) + \\text{Cl}_2(g) (\\text{net ionic}) $\r\n\r\n<\/div>\r\n<\/section><section id=\"fs-idp114420064\" class=\"summary\">\r\n<div class=\"callout block\" id=\"ball-ch04_s03_n05\">\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Chemistry Is Everywhere: Soluble and Insoluble Ionic Compounds<\/h3>\r\n<p id=\"ball-ch04_s03_p52\" class=\"para\">The concept of solubility versus insolubility in ionic compounds is a matter of degree. Some ionic compounds are very soluble, some are only moderately soluble, and some are soluble so little that they are considered insoluble. For most ionic compounds, there is also a limit to the amount of compound can be dissolved in a sample of water. For example, you can dissolve a maximum of 36.0 g of NaCl in 100 g of water at room temperature, but you can dissolve only 0.00019 g of AgCl in 100 g of water. We consider NaCl soluble but AgCl insoluble.<\/p>\r\n\r\n\r\n[caption id=\"attachment_3210\" align=\"aligncenter\" width=\"385\"]<a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/800px-Grand_canyon_yavapal_point_2010.jpg\"><img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/800px-Grand_canyon_yavapal_point_2010-1.jpg\" alt=\"The Grand Canyon was formed by water running through rock for billions of years, very slowly dissolving it. Note the Colorado River is still present in the lower part of the photo. \u201cGrand canyon yavapal point 2010\u2032\u2032 by chensiyuan is licensed under Creative Commons\" class=\"wp-image-3210\" height=\"251\" width=\"385\" \/><\/a> <strong>Figure 4.<\/strong>\u00a0The Grand Canyon was formed by water running through rock for billions of years, very slowly dissolving it. Note the Colorado River is still present in the lower part of the photo. \u201cGrand canyon yavapal point 2010\u2032\u2032 by chensiyuan is licensed under Creative Commons[\/caption]\r\n<p class=\"para\">One place where solubility is important is in the tank-type water heater found in many homes in the United States. Domestic water frequently contains small amounts of dissolved ionic compounds, including calcium carbonate (CaCO<sub class=\"subscript\">3<\/sub>). However, CaCO<sub class=\"subscript\">3<\/sub> has the relatively unusual property of being less soluble in hot water than in cold water. So as the water heater operates by heating water, CaCO<sub class=\"subscript\">3<\/sub> can precipitate if there is enough of it in the water. This precipitate, called <em class=\"emphasis\">limescale<\/em>, can also contain magnesium compounds, hydrogen carbonate compounds, and phosphate compounds. The problem is that too much limescale can impede the function of a water heater, requiring more energy to heat water to a specific temperature or even blocking water pipes into or out of the water heater, causing dysfunction.<\/p>\r\n<p id=\"ball-ch04_s03_p54\" class=\"para\">Another place where solubility versus insolubility is an issue is the Grand Canyon. We usually think of rock as insoluble. But it is actually ever so slightly soluble. This means that over a period of about two billion years, the Colorado River carved rock from the surface by slowly dissolving it, eventually generating a spectacular series of gorges and canyons. And all because of solubility!<\/p>\r\n\r\n<\/div>\r\n<h2><span style=\"font-family: Roboto, Helvetica, Arial, sans-serif\">Key Concepts and Summary<\/span><\/h2>\r\n<\/div>\r\n<p id=\"fs-idp43480096\">Chemical equations are symbolic representations of chemical and physical changes. Formulas for the substances undergoing the change (reactants) and substances generated by the change (products) are separated by an arrow and preceded by integer coefficients indicating their relative numbers. Balanced equations are those whose coefficients result in equal numbers of atoms for each element in the reactants and products. Chemical reactions in aqueous solution that involve ionic reactants or products may be represented more realistically by complete ionic equations and, more succinctly, by net ionic equations.<\/p>\r\n\r\n<\/section><section id=\"fs-idm23851744\" class=\"exercises\">\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Exercises<\/h3>\r\n1. What does it mean to say an equation is balanced? Why is it important for an equation to be balanced?\r\n\r\n2. Balance the following equations:\r\n<p id=\"fs-idp156664864\">a) $latex \\text{PCl}_5(s) + \\text{H}_2 \\text{O}(l) \\longrightarrow \\text{POCl}_3(l) + \\text{HCl}(aq) $<\/p>\r\n<p id=\"fs-idm5703840\">b) $latex \\text{Cu}(s) + \\text{HNO}_3(aq) \\longrightarrow \\text{Cu(NO}_3)_2(aq) + \\text{H}_2 \\text{O}(l) + \\text{NO}(g) $<\/p>\r\n<p id=\"fs-idp26705376\">c) $latex \\text{H}_2(g) + \\text{I}_2(s) \\longrightarrow \\text{HI}(s) $<\/p>\r\n<p id=\"fs-idm2545856\">d) $latex \\text{Fe}(s) + \\text{O}_2(g) \\longrightarrow \\text{Fe}_2 \\text{O}_3(s) $<\/p>\r\n<p id=\"fs-idp38879216\">e) $latex \\text{Na}(s) + \\text{H}_2 \\text{O}(l) \\longrightarrow \\text{NaOH}(aq) + \\text{H}_2(g) $<\/p>\r\n<p id=\"fs-idp261699472\">f) $latex \\text{(NH}_4)_2 \\text{Cr}_2\\text{O}_7(s) \\longrightarrow \\text{Cr}_2\\text{O}_3(s) + \\text{N}_2(g) + \\text{H}_2 \\text{O}(g) $<\/p>\r\n<p id=\"fs-idp78017472\">g) $latex \\text{P}_4(s) + \\text{Cl}_2(g) \\longrightarrow \\text{PCl}_3(l) $<\/p>\r\n<p id=\"fs-idp44039696\">h) $latex \\text{PtCl}_4(s) \\longrightarrow \\text{Pt}(s) + \\text{Cl}_2(g) $<\/p>\r\n3. Balance the following equations:\r\n<p id=\"fs-idp14691792\">a) $latex \\text{Ag}(s) + \\text{H}_2 \\text{S}(g) + \\text{O}_2(g) \\longrightarrow \\text{Ag}_2 \\text{S}(s) + \\text{H}_2 \\text{O}(l)$<\/p>\r\n<p id=\"fs-idp41951632\">b) $latex \\text{P}_4(s) + \\text{O}_2(g) \\longrightarrow \\text{P}_4 \\text{O}_{10}(s)$<\/p>\r\n<p id=\"fs-idp215876720\">c) $latex \\text{Pb}(s) + \\text{H}_2 \\text{O}(l) + \\text{O}_2(g) \\longrightarrow \\text{Pb(OH)}_2(s) $<\/p>\r\n<p id=\"fs-idp7611408\">d) $latex \\text{Fe}(s) + \\text{H}_2 \\text{O}(l) \\longrightarrow \\text{Fe}_3 \\text{O}_4(s) + \\text{H}_2(g) $<\/p>\r\n<p id=\"fs-idp51158528\">e) $latex \\text{Sc}_2 \\text{O}_3(s) + \\text{SO}_3(l) \\longrightarrow \\text{Sc}_2 \\text{(SO}_4)_3(s) $<\/p>\r\n<p id=\"fs-idp77422352\">f) $latex \\text{Ca}_3 \\text{(PO}_4)_2(aq) + \\text{H}_3 \\text{PO}_4(aq) \\longrightarrow \\text{Ca(H}_2 \\text{PO}_4)_2(aq) $<\/p>\r\n<p id=\"fs-idp48923840\">g) $latex \\text{Al}(s) + \\text{H}_2 \\text{SO}_4(aq) \\longrightarrow \\text{Al}_2 \\text{(SO}_4)_3(s) + \\text{H}_2(g) $<\/p>\r\n<p id=\"fs-idp73552\">h) $latex \\text{TiCl}_4(s) + \\text{H}_2 \\text{O}(g) \\longrightarrow \\text{TiO}_2(s) + \\text{HCl}(g) $<\/p>\r\n4. Write a balanced molecular equation describing each of the following chemical reactions.\r\n<p id=\"fs-idp105395536\">a) Solid calcium carbonate is heated and decomposes to solid calcium oxide and carbon dioxide gas.<\/p>\r\n<p id=\"fs-idp105396032\">b) Gaseous butane, C<sub>4<\/sub>H<sub>10<\/sub>, reacts with diatomic oxygen gas to yield gaseous carbon dioxide and water vapor.<\/p>\r\n<p id=\"fs-idp45919792\">c) Aqueous solutions of magnesium chloride and sodium hydroxide react to produce solid magnesium hydroxide and aqueous sodium chloride.<\/p>\r\n<p id=\"fs-idp45920336\">d) Water vapor reacts with sodium metal to produce solid sodium hydroxide and hydrogen gas.<\/p>\r\n5. Colorful fireworks often involve the decomposition of barium nitrate and potassium chlorate and the reaction of the metals magnesium, aluminum, and iron with oxygen.\r\n<p id=\"fs-idm22159824\">a) Write the formulas of barium nitrate and potassium chlorate.<\/p>\r\n<p id=\"fs-idm22159440\">b) The decomposition of solid potassium chlorate leads to the formation of solid potassium chloride and diatomic oxygen gas. Write an equation for the reaction.<\/p>\r\n<p id=\"fs-idm22158880\">c) The decomposition of solid barium nitrate leads to the formation of solid barium oxide, diatomic nitrogen gas, and diatomic oxygen gas. Write an equation for the reaction.<\/p>\r\n<p id=\"fs-idp208977456\">d) Write separate equations for the reactions of the solid metals magnesium, aluminum, and iron with diatomic oxygen gas to yield the corresponding metal oxides. (Assume the iron oxide contains Fe<sup>3+<\/sup> ions.)<\/p>\r\n6. Fill in the blank with a single chemical formula for a covalent compound that will balance the equation:<img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_01_basehyd_img-2.jpg\" alt=\"This figure shows a chemical reaction. On the left side of the reaction arrow, a structural formula for a molecule is shown on the far left. It has a C atom on the left to which H atoms are bonded above, below, and to the left. To the right, another C atom is bonded which has H atoms bonded above and below. To the right, another C atom is bonded, which has a double bonded O atom above and another O atom singly bonded to the right. To the right of the singly bonded O atom, an H atom is bonded. This is followed by a plus sign and N a O H. A reaction arrow appears to the right, which is followed by another structural formula. It has a C atom on the left to which H atoms are bonded above, below, and to the left. To the right, another C atom is bonded which has H atoms bonded above and below. To the right, another C atom is bonded, which has a double bonded O atom above and another O atom singly bonded to the right. The singly bonded O atom is followed by a superscript negative sign. This is followed to the right by a plus sign, N a superscript positive sign, another plus sign, and a horizontal line segment, indicating a space for an answer to be written.\" \/>\r\n\r\n7. Aqueous hydrogen fluoride (hydrofluoric acid) is used to etch glass and to analyze minerals for their silicon content. Hydrogen fluoride will also react with sand (silicon dioxide).\r\n<p id=\"fs-idm98641232\">a) Write an equation for the reaction of solid silicon dioxide with hydrofluoric acid to yield gaseous silicon tetrafluoride and liquid water.<\/p>\r\n<p id=\"fs-idp126128736\">b) The mineral fluorite (calcium fluoride) occurs extensively in Illinois. Solid calcium fluoride can also be prepared by the reaction of aqueous solutions of calcium chloride and sodium fluoride, yielding aqueous sodium chloride as the other product. Write complete and net ionic equations for this reaction.<\/p>\r\n8. From the balanced molecular equations, write the complete ionic and net ionic equations for the following:\r\n<p id=\"fs-idp41636720\">a) $latex \\text{K}_2 \\text{C}_2 \\text{O}_4(aq) + \\text{Ba(OH)}_2(aq) \\longrightarrow 2\\text{KOH}(aq) + \\text{BaC}_2 \\text{O}_2(s) $<\/p>\r\n<p id=\"fs-idp46970864\">b) $latex {\\text{Pb(NO}_3)}_2(aq) + \\text{H}_2 \\text{SO}_4(aq) \\longrightarrow \\text{PbSO}_4(s) + 2\\text{HNO}_3(aq) $<\/p>\r\n<p id=\"fs-idm2481664\">c) $latex \\text{CaCO}_3(s) + \\text{H}_2 \\text{SO}_4(aq) \\longrightarrow \\text{CaSO}_4(s) + \\text{CO}_2(g) + \\text{H}_2\\text{O}(l) $<\/p>\r\n<p id=\"ball-ch04_s01_qs01_p1\" class=\"para\">9. From the statement \u201cnitrogen gas and hydrogen gas react to produce ammonia gas,\u201d identify the reactants and the products.<\/p>\r\n<p class=\"para\"><span style=\"font-size: 1em\">10. From the statement \u201ca solution of magnesium hydroxide reacts with a solution of nitric acid to produce a solution of magnesium nitrate and water,\u201d identify the reactants and the products.<\/span><\/p>\r\n<p class=\"para\"><span style=\"font-size: 1em\">11. Write and balance the chemical equation described by Exercise 1.<\/span><\/p>\r\n<p class=\"para\"><span style=\"font-size: 1em\">12. Write and balance the chemical equation described by Exercise 2.<\/span><\/p>\r\n<p class=\"para\"><span style=\"font-size: 1em\">13. Balance: ___NaClO<\/span><sub class=\"subscript\">3<\/sub><span style=\"font-size: 1em\"> $latex \\longrightarrow$ ___NaCl +\u00a0___O<\/span><sub class=\"subscript\">2<\/sub><\/p>\r\n<p class=\"para\"><span style=\"font-size: 1em\">14. Balance: ___N<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\"> +\u00a0___H<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\"> $latex \\longrightarrow$ ___N<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">H<\/span><sub class=\"subscript\">4<\/sub><\/p>\r\n<p class=\"para\"><span style=\"font-size: 1em\">15. Balance: ___Al +\u00a0___O<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\"> $latex \\longrightarrow$ ___Al<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">O<\/span><sub class=\"subscript\">3<\/sub><\/p>\r\n<p class=\"para\"><span style=\"font-size: 1em\">16. Balance: ___C<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">H<\/span><sub class=\"subscript\">4<\/sub><span style=\"font-size: 1em\"> +\u00a0___O<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\"> $latex \\longrightarrow$ ___CO<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\"> +\u00a0___H<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">O<\/span><\/p>\r\n<p class=\"para\"><span style=\"font-size: 1em\">17. Balance:\u00a0___N<sub class=\"subscript\">2<\/sub><sub>(g)<\/sub> +\u00a0___H<sub class=\"subscript\">2<\/sub><sub>(g)<\/sub> $latex \\longrightarrow$ ___NH<sub class=\"subscript\">3<\/sub><sub>(g)<\/sub><\/span><\/p>\r\n<p id=\"ball-ch04_s03_qs01_p1\" class=\"para\">18. Write a chemical equation that represents NaBr(s) dissociating in water.<\/p>\r\n<p class=\"para\"><span style=\"font-size: 1em\">19. Write a chemical equation that represents (NH<\/span><sub class=\"subscript\">4<\/sub><span style=\"font-size: 1em\">)<\/span><sub class=\"subscript\">3<\/sub><span style=\"font-size: 1em\">PO<\/span><sub class=\"subscript\">4<\/sub><span style=\"font-size: 1em\">(s) dissociating in water.<\/span><\/p>\r\n<p class=\"para\"><span style=\"font-size: 1em\">20. Write the complete ionic equation for the reaction of FeCl<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">(aq) and AgNO<\/span><sub class=\"subscript\">3<\/sub><span style=\"font-size: 1em\">(aq). You may have to consult the solubility rules.<\/span><\/p>\r\n<p class=\"para\"><span style=\"font-size: 1em\">21. Write the complete ionic equation for the reaction of KCl(aq) and NaC<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">H<\/span><sub class=\"subscript\">3<\/sub><span style=\"font-size: 1em\">O<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">(aq). You may have to consult the solubility rules.<\/span><\/p>\r\n<p class=\"para\"><span style=\"font-size: 1em\">22. Write the net ionic equation for the reaction of FeCl<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">(aq) and AgNO<\/span><sub class=\"subscript\">3<\/sub><span style=\"font-size: 1em\">(aq). You may have to consult the solubility rules.<\/span><\/p>\r\n<p class=\"para\"><span style=\"font-size: 1em\">23. Write the net ionic equation for the reaction of KCl(aq) and NaC<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">H<\/span><sub class=\"subscript\">3<\/sub><span style=\"font-size: 1em\">O<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">(aq). You may have to consult the solubility rules.<\/span><\/p>\r\n<p class=\"para\"><span style=\"font-size: 1em\">24. Identify the spectator ions in Exercises 20 and 21.<\/span><\/p>\r\n&nbsp;\r\n\r\n<strong>Answers<\/strong>\r\n<p id=\"fs-idm10510096\">1. An equation is balanced when the same number of each element is represented on the reactant and product sides. Equations must be balanced to accurately reflect the law of conservation of matter.<\/p>\r\n<p id=\"fs-idm90785536\">2.\u00a0a) $latex \\text{PCl}_5(s) + \\text{H}_2 \\text{O}(l) \\longrightarrow \\text{POCl}_3(l) + 2\\text{HCl}(aq) $<\/p>\r\nb) $latex 3\\text{Cu}(s) + 8\\text{HNO}_3(aq) \\longrightarrow 3\\text{Cu(NO}_3)_2(aq) + 4\\text{H}_2 \\text{O}(l) + 2\\text{NO}(g) $\r\n\r\nc) $latex \\text{H}_2(g) + \\text{I}_2(s) \\longrightarrow 2\\text{HI}(s) $\r\n\r\nd) $latex 4\\text{Fe}(s) + 3\\text{O}_2(g) \\longrightarrow 2\\text{Fe}_2 \\text{O}_3(s) $\r\n\r\ne) $latex 2\\text{Na}(s) + 2\\text{H}_2 \\text{O}(l) \\longrightarrow 2\\text{NaOH}(aq) + \\text{H}_2(g) $\r\n\r\nf) $latex \\text{(NH}_4)_2 \\text{Cr}_2\\text{2O}_7(s) \\longrightarrow \\text{Cr}_2\\text{O}_3(s) + \\text{N}_2(g) + 4\\text{H}_2 \\text{O}(l) $\r\n\r\ng) $latex \\text{P}_4(s) + 6\\text{Cl}_2(g) \\longrightarrow 4\\text{PCl}_3(l) $\r\n\r\nh) $latex \\text{PtCl}_4(s) \\longrightarrow \\text{Pt}(s) + 2\\text{Cl}_2(g) $\r\n\r\n3.\u00a0a) $latex 4\\text{Ag}(s) + 2\\text{H}_2 \\text{S}(g) + \\text{O}_2(g) \\longrightarrow 2\\text{Ag}_2 \\text{S}(s) + 2\\text{H}_2 \\text{O}(l)$\r\n<p id=\"fs-idp41951632\">b) $latex \\text{P}_4(s) + 5\\text{O}_2(g) \\longrightarrow \\text{P}_4 \\text{O}_{10}(s)$<\/p>\r\n<p id=\"fs-idp215876720\">c) $latex 2\\text{Pb}(s) + 2\\text{H}_2 \\text{O}(l) + \\text{O}_2(g) \\longrightarrow 2\\text{Pb(OH)}_2(s) $<\/p>\r\n<p id=\"fs-idp7611408\">d) $latex 3\\text{Fe}(s) + 4\\text{H}_2 \\text{O}(l) \\longrightarrow \\text{Fe}_3 \\text{O}_4(s) + 4\\text{H}_2(g) $<\/p>\r\n<p id=\"fs-idp51158528\">e) $latex \\text{Sc}_2 \\text{O}_3(s) + 3\\text{SO}_3(l) \\longrightarrow \\text{Sc}_2 \\text{(SO}_4)_3(s) $<\/p>\r\n<p id=\"fs-idp77422352\">f) $latex \\text{Ca}_3 \\text{(PO}_4)_2(aq) + 4\\text{H}_3 \\text{PO}_4(aq) \\longrightarrow 3\\text{Ca(H}_2 \\text{PO}_4)_2(aq) $<\/p>\r\n<p id=\"fs-idp48923840\">g) $latex 2\\text{Al}(s) + 3\\text{H}_2 \\text{SO}_4(aq) \\longrightarrow \\text{Al}_2 \\text{(SO}_4)_3(s) + 3\\text{H}_2(g) $<\/p>\r\n<p id=\"fs-idp73552\">h) $latex \\text{TiCl}_4(s) + 2\\text{H}_2 \\text{O}(g) \\longrightarrow \\text{TiO}_2(s) + 4\\text{HCl}(g) $<\/p>\r\n4.a) $latex \\text{CaCO}_3(s) \\longrightarrow \\text{CaO}(s) + \\text{CO}_2(g) $\r\nb) $latex 2\\text{C}_4 \\text{H}_{10}(g) + 13 \\text{O}_2(g) \\longrightarrow 8\\text{CO}_2(g) + 10\\text{H}_2\\text{O}(g) $\r\nc) $latex \\text{MgCl}_{2}(aq) + 2 \\text{NaOH}(aq) \\longrightarrow \\text{Mg(OH)}_2(s) + 2 \\text{NaCl}(aq) $\r\nd) $latex 2\\text{H}_2 \\text{O}(g) + 2 \\text{Na}(s) \\longrightarrow 2\\text{NaOH}(s) + \\text{H}_2(g) $\r\n<p id=\"fs-idp208978816\">5.\u00a0a) $latex \\text{Ba(NO}_3)_2$ , $latex \\text{KClO}_3 $\r\nb) $latex 2 \\text{KClO}_3(s) \\longrightarrow 2 \\text{KCl}(s) + 3\\text{O}_2(g)$\r\nc) $latex 2 \\text{Ba(NO}_3)_2(s) \\longrightarrow 2\\text{BaO}(s) + 2\\text{N}_2(g) + 5\\text{O}_2(g)$\r\nd) $latex 2 \\text{Mg}(s) + \\text{O}_2(g) \\longrightarrow 2 \\text{MgO}(s) $;\u00a0$latex 4\\text{Al}(s) + 3\\text{O}_2(g) \\longrightarrow 2\\text{Al}_2 \\text{O}_3(g) $;\u00a0<span style=\"line-height: 1.5\">$latex 4\\text{Fe}(s) + 3\\text{O}_2(g) \\longrightarrow 2\\text{Fe}_2 \\text{O}_3(s)$<\/span><\/p>\r\n6. H<sub>2<\/sub>O\r\n\r\n7.\u00a0a) $latex 4\\text{HF}(aq) + \\text{SiO}_2(s) \\longrightarrow \\text{SiF}_4(g) + 2\\text{H}_2 \\text{O}(l) $\r\nb) complete: $latex 2\\text{Na}^{+}(aq) + 2\\text{F}^{-}(aq) + \\text{Ca}^{2+}(aq) + 2\\text{Cl}^{-}(aq) \\longrightarrow \\text{CaF}_2(s) + 2\\text{Na}^{+}(aq) + 2\\text{Cl}^{-}(aq) $\r\n\r\nnet:\u00a0$latex 2\\text{F}^{-}(aq) + \\text{Ca}^{2+}(aq) \\longrightarrow \\text{CaF}_2(s)$\r\n<p id=\"fs-idp198600416\">8.\u00a0a) complete:\u00a0$latex 2\\text{K}^{+}(aq) + {\\text{C}_2 \\text{O}_4}^{2-}(aq) + \\text{Ba}^{2+}(aq) + 2\\text{OH}^{-}(aq) \\longrightarrow 2\\text{K}^{+}(aq) + 2\\text{OH}^{-}(aq) + \\text{BaC}_2 \\text{O}_4(s) $\r\nnet: $latex \\text{Ba}^{2+}(aq) + {\\text{C}_2 {\\text{O}_4}}^{2-}(aq) \\longrightarrow \\text{BaC}_2 \\text{O}_4(s) $<\/p>\r\nb) complete: $latex \\text{Pb}^{2+}(aq) + 2{\\text{NO}_3}^{-}(aq) + 2\\text{H}^{+}(aq) + {\\text{SO}_4}^{2-}(aq) \\longrightarrow \\text{PbSO}_4(s) + 2\\text{H}^{+}(aq) + 2{\\text{NO}_3}^{-}(aq) $\r\n\r\nnet: $latex \\text{Pb}^{2+}(aq) + {\\text{SO}_4}^{2-}(aq) \\longrightarrow \\text{PbSO}_4(s) $\r\n\r\nc) complete:\u00a0$latex \\text{CaCO}_3(s) + 2\\text{H}^{+}(aq) + {\\text{SO}_4}^{2-}(aq) \\longrightarrow \\text{CaSO}_4(s) + \\text{CO}_2(g) + \\text{H}_2 \\text{O}(l) $\r\n\r\nnet: $latex \\text{CaCO}_3(s) + 2\\text{H}^{+}(aq) + {\\text{SO}_4}^{2-}(aq) \\longrightarrow \\text{CaSO}_4(s) + \\text{CO}_2(g) + \\text{H}_2 \\text{O}(l) $\r\n<div class=\"qandaset block\" id=\"ball-ch04_s01_qs01_ans\">\r\n\r\n9.\u00a0<span style=\"font-size: 1em\">reactants: nitrogen and hydrogen; product: ammonia<\/span>\r\n\r\n<\/div>\r\n<div class=\"qandaset block\">\r\n\r\n10.\u00a0reactants: magnesium hydroxide and nitric acid; products: magnesium nitrate and water\r\n\r\n11.\u00a0N<sub class=\"subscript\">2<\/sub><sub>(g)<\/sub> +\u00a03 H<sub class=\"subscript\">2<\/sub><sub>(g)<\/sub> $latex \\longrightarrow$ 2 NH<sub class=\"subscript\">3<\/sub><sub>(g)<\/sub>\r\n\r\n12.\u00a0Mg(OH)<sub class=\"subscript\">2<\/sub><sub>(aq)<\/sub> +\u00a02 HNO<sub class=\"subscript\">3<\/sub><sub>(aq)<\/sub> $latex \\longrightarrow$ Mg(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub><sub>(aq)<\/sub> +\u00a02 H<sub class=\"subscript\">2<\/sub>O<span class=\"informalequation block\"><sub>(\u2113)<\/sub><\/span>\r\n\r\n13.\u00a02 NaClO<sub class=\"subscript\">3<\/sub> $latex \\longrightarrow$ 2 NaCl +\u00a03 O<sub class=\"subscript\">2<\/sub>\r\n\r\n14.\u00a0<span style=\"font-size: 1em\">N<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\"> +\u00a02 H<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\"> $latex \\longrightarrow$ N<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">H<\/span><sub class=\"subscript\">4<\/sub>\r\n\r\n15.\u00a04Al +\u00a03O<sub class=\"subscript\">2<\/sub> $latex \\longrightarrow$ 2Al<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub>\r\n\r\n16.\u00a0<span style=\"font-size: 1em\">C<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">H<\/span><sub class=\"subscript\">4<\/sub><span style=\"font-size: 1em\"> +\u00a03 O<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\"> $latex \\longrightarrow$ 2 CO<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\"> +\u00a02 H<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">O<\/span>\r\n\r\n17.\u00a0N<sub class=\"subscript\">2<\/sub><sub>(g)<\/sub> +\u00a03 H<sub class=\"subscript\">2<\/sub><sub>(g)<\/sub> $latex \\longrightarrow$ 2 NH<sub class=\"subscript\">3<\/sub><sub>(g)<\/sub>\r\n\r\n18.\u00a0NaBr(s) $latex \\longrightarrow$\u00a0Na<sup class=\"superscript\">+<\/sup>(aq) +\u00a0Br<sup class=\"superscript\">\u2212<\/sup>(aq)\r\n\r\n19.\u00a0(NH<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">3<\/sub>PO<sub class=\"subscript\">4<\/sub>(s) $latex \\longrightarrow$\u00a03 NH<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup>(aq) +\u00a0PO<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">3\u2212<\/sup>(aq)\r\n\r\n20.\u00a0Fe<sup class=\"superscript\">2+<\/sup>(aq) +\u00a02 Cl<sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a02 Ag<sup class=\"superscript\">+<\/sup>(aq) +\u00a02 NO<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq) $latex \\longrightarrow$ Fe<sup class=\"superscript\">2+<\/sup>(aq) +\u00a02 NO<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a02 AgCl(s)\r\n\r\n21.\u00a0K<sup class=\"superscript\">+<\/sup>(aq) +\u00a0Cl<sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a0Na<sup class=\"superscript\">+<\/sup>(aq) +\u00a0C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq) $latex \\longrightarrow$ Na<sup class=\"superscript\">+<\/sup>(aq) +\u00a0Cl<sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a0K<sup class=\"superscript\">+<\/sup>(aq) +\u00a0C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq)\r\n\r\n22.\u00a02 Cl<sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a02 Ag<sup class=\"superscript\">+<\/sup>(aq) $latex \\longrightarrow$ 2 AgCl(s)\r\n\r\n23.\u00a0There is no overall reaction.\r\n\r\n24.\u00a0In Exercise 20, Fe<sup class=\"superscript\">2+<\/sup>(aq) and NO<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq) are spectator ions; in Exercise 21, Na<sup class=\"superscript\">+<\/sup>(aq) and Cl<sup class=\"superscript\">\u2212<\/sup>(aq) are spectator ions.\r\n\r\n<\/div>\r\n<\/div>\r\n<\/section>\r\n<div>\r\n<dl id=\"fs-idp25650896\" class=\"definition\"><\/dl>\r\n<h2>Glossary<\/h2>\r\n<strong>balanced equation:<\/strong>\u00a0chemical equation with equal numbers of atoms for each element in the reactant and product\r\n\r\n<strong>chemical equation:<\/strong>\u00a0symbolic representation of a chemical reaction\r\n\r\n<strong>coefficient:<\/strong>\u00a0number placed in front of symbols or formulas in a chemical equation to indicate their relative amount\r\n\r\n<strong>complete ionic equation:<\/strong>\u00a0chemical equation in which all dissolved ionic reactants and products, including spectator ions, are explicitly represented by formulas for their dissociated ions\r\n\r\n<strong>molecular equation:<\/strong>\u00a0chemical equation in which all reactants and products are represented as neutral substances\r\n\r\n<strong>net ionic equation:<\/strong>\u00a0chemical equation in which only those dissolved ionic reactants and products that undergo a chemical or physical change are represented (excludes spectator ions)\r\n\r\n<strong>product:<\/strong>\u00a0substance formed by a chemical or physical change; shown on the right side of the arrow in a chemical equation\r\n\r\n<strong>reactant:<\/strong>\u00a0substance undergoing a chemical or physical change; shown on the left side of the arrow in a chemical equation\r\n\r\n<strong>spectator ion:<\/strong>\u00a0ion that does not undergo a chemical or physical change during a reaction, but its presence is required to maintain charge neutrality\r\n\r\n<\/div>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Objectives<\/h3>\n<p>By the end of this section, you will be able to:<\/p>\n<ul>\n<li>Derive chemical equations from narrative descriptions of chemical reactions.<\/li>\n<li>Write and balance chemical equations in molecular, total ionic, and net ionic formats.<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-idp47467664\">The preceding chapter introduced the use of element symbols to represent individual atoms. When atoms gain or lose electrons to yield ions, or combine with other atoms to form molecules, their symbols are modified or combined to generate chemical formulas that appropriately represent these species. Extending this symbolism to represent both the identities and the relative quantities of substances undergoing a chemical (or physical) change involves writing and balancing a <strong>chemical equation<\/strong>. Consider as an example the reaction between one methane molecule (CH<sub>4<\/sub>) and two diatomic oxygen molecules (O<sub>2<\/sub>) to produce one carbon dioxide molecule (CO<sub>2<\/sub>) and two water molecules (H<sub>2<\/sub>O). The chemical equation representing this process is provided in the upper half of <a href=\"#CNX_Chem_04_01_rxn2\" class=\"autogenerated-content\">Figure 1<\/a>, with space-filling molecular models shown in the lower half of the figure.<\/p>\n<figure id=\"CNX_Chem_04_01_rxn2\"><figcaption>\n<figure style=\"width: 975px\" class=\"wp-caption aligncenter\"><a href=\"https:\/\/opentextbc.ca\/chemistry\/wp-content\/uploads\/sites\/150\/2016\/05\/CNX_Chem_04_01_rxn2.jpg\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_01_rxn2-2.jpg\" alt=\"This figure shows a balanced chemical equation followed below by a representation of the equation using space-filling models. The equation reads C H subscript 4 plus 2 O subscript 2 arrow C O subscript 2 plus 2 H subscript 2 O. Under the C H subscript 4, the molecule is shown with a central black sphere, representing a C atom, to which 4 smaller white spheres, representing H atoms, are distributed evenly around. All four H atoms are bonded to the central black C atom. This is followed by a plus sign. Under the 2 O subscript 2, two molecules are shown. The molecules are each composed of two red spheres bonded together. The red spheres represent O atoms. To the right of an arrow and under the C O subscript 2, appears a single molecule with a black central sphere with two red spheres bonded to the left and right. Following a plus sign and under the 2 H subscript 2 O, are two molecules, each with a central red sphere and two smaller white spheres attached to the lower right and lower left sides of the central red sphere. Note that in space filling models of molecules, spheres appear slightly compressed in regions where there is a bond between two atoms.\" width=\"975\" height=\"444\" \/><\/a><figcaption class=\"wp-caption-text\"><strong>Figure 1.<\/strong> The reaction between methane and oxygen to yield carbon dioxide and water (shown at bottom) may be represented by a chemical equation using formulas (top).<\/figcaption><\/figure>\n<\/figcaption><\/figure>\n<p id=\"fs-idp80311968\">This example illustrates the fundamental aspects of any chemical equation:<\/p>\n<ol id=\"fs-idm9329568\">\n<li>The substances undergoing reaction are called <strong>reactants<\/strong>, and their formulas are placed on the left side of the equation.<\/li>\n<li>The substances generated by the reaction are called <strong>products<\/strong>, and their formulas are placed on the right sight of the equation.<\/li>\n<li>Plus signs (+) separate individual reactant and product formulas, and an arrow ([latex]\\longrightarrow[\/latex]) separates the reactant and product (left and right) sides of the equation.<\/li>\n<li>The relative numbers of reactant and product species are represented by <strong>coefficients<\/strong> (numbers placed immediately to the left of each formula). A coefficient of 1 is typically omitted.<\/li>\n<\/ol>\n<p id=\"fs-idm33251968\">It is common practice to use the smallest possible whole-number coefficients in a chemical equation, as is done in this example. Realize, however, that these coefficients represent the <em>relative<\/em> numbers of reactants and products, and, therefore, they may be correctly interpreted as ratios. Methane and oxygen react to yield carbon dioxide and water in a 1:2:1:2 ratio. This ratio is satisfied if the numbers of these molecules are, respectively, 1-2-1-2, or 2-4-2-4, or 3-6-3-6, and so on (<a href=\"#CNX_Chem_04_01_rxn3\" class=\"autogenerated-content\">Figure 2<\/a>). Likewise, these coefficients may be interpreted with regard to any amount (number) unit, and so this equation may be correctly read in many ways, including:<\/p>\n<ul id=\"fs-idp42981680\">\n<li><em>One<\/em> methane molecule and <em>two<\/em> oxygen molecules react to yield <em>one<\/em> carbon dioxide molecule and <em>two<\/em> water molecules.<\/li>\n<li><em>One dozen<\/em> methane molecules and <em>two dozen<\/em> oxygen molecules react to yield <em>one dozen<\/em> carbon dioxide molecules and <em>two dozen<\/em> water molecules.<\/li>\n<li><em>One mole<\/em> of methane molecules and <em>2 moles<\/em> of oxygen molecules react to yield <em>1 mole<\/em> of carbon dioxide molecules and <em>2 moles<\/em> of water molecules.<\/li>\n<\/ul>\n<figure id=\"CNX_Chem_04_01_rxn3\"><figcaption>\n<figure style=\"width: 1300px\" class=\"wp-caption aligncenter\"><a href=\"https:\/\/opentextbc.ca\/chemistry\/wp-content\/uploads\/sites\/150\/2016\/05\/CNX_Chem_04_01_rxn3.jpg\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_01_rxn3-2.jpg\" alt=\"This image has a left side, labeled, \u201cMixture before reaction\u201d separated by a vertical dashed line from right side labeled, \u201cMixture after reaction.\u201d On the left side of the figure, two types of molecules are illustrated with space-filling models. Six of the molecules have only two red spheres bonded together. Three of the molecules have four small white spheres evenly distributed about and bonded to a central, larger black sphere. On the right side of the dashed vertical line, two types of molecules which are different from those on the left side are shown. Six of the molecules have a central red sphere to which smaller white spheres are bonded. The white spheres are not opposite each other on the red atoms, giving the molecule a bent shape or appearance. The second molecule type has a central black sphere to which two red spheres are attached on opposite sides, resulting in a linear shape or appearance. Note that in space filling models of molecules, spheres appear slightly compressed in regions where there is a bond between two atoms. On each side of the dashed line, twelve red, three black, and twelve white spheres are present.\" width=\"1300\" height=\"504\" \/><\/a><figcaption class=\"wp-caption-text\"><strong>Figure 2.<\/strong> Regardless of the absolute numbers of molecules involved, the ratios between numbers of molecules of each species that react (the reactants) and molecules of each species that form (the products) are the same and are given by the chemical reaction equation.<\/figcaption><\/figure>\n<\/figcaption><\/figure>\n<section id=\"fs-idp23923488\">\n<h2>Balancing Equations<\/h2>\n<p id=\"fs-idp13839872\">The chemical equation described in Figure 1 is <strong>balanced<\/strong>, meaning that equal numbers of atoms for each element involved in the reaction are represented on the reactant and product sides. This is a requirement the equation must satisfy to be consistent with the law of conservation of matter. It may be confirmed by simply summing the numbers of atoms on either side of the arrow and comparing these sums to ensure they are equal. Note that the number of atoms for a given element is calculated by multiplying the coefficient of any formula containing that element by the element\u2019s subscript in the formula. If an element appears in more than one formula on a given side of the equation, the number of atoms represented in each must be computed and then added together. For example, both product species in the example reaction, CO<sub>2<\/sub> and H<sub>2<\/sub>O, contain the element oxygen, and so the number of oxygen atoms on the product side of the equation is<\/p>\n<div class=\"equation\" id=\"fs-idp219991408\" style=\"text-align: center\">[latex](1 \\;\\text{CO}_2 \\;\\text{molecule} \\times \\frac{2 \\;\\text{O atoms}}{\\text{CO}_2 \\;\\text{molecule}}) + (2\\;\\text{H}_2\\text{O molecule} \\times \\frac{1 \\;\\text{O atom}}{\\text{H}_2\\text{O molecule}}) = 4 \\;\\text{O atoms}[\/latex]<\/div>\n<p id=\"fs-idp52518864\">The equation for the reaction between methane and oxygen to yield carbon dioxide and water is confirmed to be balanced per this approach, as shown here:<\/p>\n<div class=\"equation\" id=\"fs-idp38334880\" style=\"text-align: center\">[latex]\\text{CH}_4 + 2\\text{O}_2 \\longrightarrow \\text{CO}_2 + 2\\text{H}_2\\text{O}[\/latex]<\/div>\n<table id=\"fs-idp140513680\" class=\"medium unnumbered\" summary=\"This is a table with four columns and four rows. The columns are labeled, \u201cElement,\u201d \u201cReactants,\u201d \u201cProducts,\u201d and \u201cBalanced?\u201d. Under the \u201cElement\u201d column are the letters C, H, and O. Under the \u201cReactants\u201d column are the equations \u201c1 times 1 equals 1,\u201d \u201c4 times 1 equals 4,\u201d and \u201c2 times 2 equals 4.\u201d Under the \u201cProducts\u201d column are the equations, \u201c1 times 1 equals 1,\u201d \u201c2 times 2 equals 4,\u201d and \u201c( 1 times 2 ) plus ( 2 times 1 ) equals 4.\u201d Under the \u201cBalanced?\u201d column are, \u201c1 equals 1, yes,\u201d \u201c4 equals 4, yes,\u201d \u201c4 equals 4, yes.\u201d\">\n<thead>\n<tr valign=\"top\">\n<th>Element<\/th>\n<th>Reactants<\/th>\n<th>Products<\/th>\n<th>Balanced?<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>C<\/td>\n<td>1 \u00d7 1 = 1<\/td>\n<td>1 \u00d7 1 = 1<\/td>\n<td>1 = 1, yes<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>H<\/td>\n<td>4 \u00d7 1 = 4<\/td>\n<td>2 \u00d7 2 = 4<\/td>\n<td>4 = 4, yes<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>O<\/td>\n<td>2 \u00d7 2 = 4<\/td>\n<td>(1 \u00d7 2) + (2 \u00d7 1) = 4<\/td>\n<td>4 = 4, yes<\/td>\n<\/tr>\n<tr>\n<td colspan=\"4\"><strong>Table 1.<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-idp8426240\">A balanced chemical equation often may be derived from a qualitative description of some chemical reaction by a fairly simple approach known as balancing by inspection. Consider as an example the decomposition of water to yield molecular hydrogen and oxygen. This process is represented qualitatively by an <em>unbalanced<\/em> chemical equation:<\/p>\n<div class=\"equation\" id=\"fs-idm60111376\" style=\"text-align: center\">[latex]\\text{H}_2\\text{O} \\longrightarrow \\text{H}_2 + \\text{O}_2 \\;(\\text{unbalanced})[\/latex]<\/div>\n<p id=\"fs-idp20832160\">Comparing the number of H and O atoms on either side of this equation confirms its imbalance:<\/p>\n<table id=\"fs-idp104786160\" class=\"medium unnumbered\" summary=\"This is a table with four columns and three rows. The columns are labeled, \u201cElement,\u201d \u201cReactants,\u201d \u201cProducts,\u201d and \u201cBalanced?\u201d. Under the \u201cElement\u201d column are the letters H and O. Under the \u201cReactants\u201d column are the equations \u201c1 times 2 equals 2,\u201d and \u201c1 times 1 equals 1.\u201d Under the \u201cProducts\u201d column are the equations, \u201c1 times 2 equals 2,\u201d and \u201c1 times 2 equals 2.\u201d Under the \u201cBalanced?\u201d column are, \u201c2 equals 2, yes,\u201d and \u201c1 does not equal 2, no.\u201d\">\n<thead>\n<tr valign=\"top\">\n<th>Element<\/th>\n<th>Reactants<\/th>\n<th>Products<\/th>\n<th>Balanced?<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>H<\/td>\n<td>1 \u00d7 2 = 2<\/td>\n<td>1 \u00d7 2 = 2<\/td>\n<td>2 = 2, yes<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>O<\/td>\n<td>1 \u00d7 1 = 1<\/td>\n<td>1 \u00d7 2 = 2<\/td>\n<td>1 \u2260 2, no<\/td>\n<\/tr>\n<tr>\n<td colspan=\"4\"><strong>Table 2.<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-idp116453584\">The numbers of H atoms on the reactant and product sides of the equation are equal, but the numbers of O atoms are not. To achieve balance, the <em>coefficients<\/em> of the equation may be changed as needed. Keep in mind, of course, that the <em>formula subscripts<\/em> define, in part, the identity of the substance, and so these cannot be changed without altering the qualitative meaning of the equation. For example, changing the reactant formula from H<sub>2<\/sub>O to H<sub>2<\/sub>O<sub>2<\/sub> would yield balance in the number of atoms, but doing so also changes the reactant\u2019s identity (it\u2019s now hydrogen peroxide and not water). The O atom balance may be achieved by changing the coefficient for H<sub>2<\/sub>O to 2.<\/p>\n<div class=\"equation\" id=\"fs-idm1006576\" style=\"text-align: center\">[latex]2\\text{H}_2\\text{O} \\longrightarrow \\text{H}_2 + \\text{O}_2 \\;(\\text{unbalanced})[\/latex]<\/div>\n<table id=\"fs-idm15543696\" class=\"medium unnumbered\" summary=\"This is a table with four columns and three rows. The columns are labeled, \u201cElement,\u201d \u201cReactants,\u201d \u201cProducts,\u201d and \u201cBalanced?\u201d. Under the \u201cElement\u201d column are the letters H and O. Under the \u201cReactants\u201d column are the equations \u201c2 times 2 equals 4,\u201d and \u201c2 times 1 equals 2.\u201d The first 2 in the \u201c2 times 2 equals 4\u201d equation is bold. Under the \u201cProducts\u201d column are the equations, \u201c1 times 2 equals 2,\u201d and \u201c1 times 2 equals 2.\u201d Under the \u201cBalanced?\u201d column are, \u201c4 does not equal 2, no,\u201d and \u201c2 equals 2, yes.\u201d\">\n<thead>\n<tr valign=\"top\">\n<th>Element<\/th>\n<th>Reactants<\/th>\n<th>Products<\/th>\n<th>Balanced?<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>H<\/td>\n<td><strong>2<\/strong>\u00a0\u00d7 2 = 4<\/td>\n<td>1 \u00d7 2 = 2<\/td>\n<td>4 \u2260 2, no<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>O<\/td>\n<td>2 \u00d7 1 = 2<\/td>\n<td>1 \u00d7 2 = 2<\/td>\n<td>2 = 2, yes<\/td>\n<\/tr>\n<tr>\n<td colspan=\"4\"><strong>Table 3.<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-idp53902400\">The H atom balance was upset by this change, but it is easily reestablished by changing the coefficient for the H<sub>2<\/sub> product to 2.<\/p>\n<div class=\"equation\" id=\"fs-idm94539648\" style=\"text-align: center\">[latex]2\\text{H}_2\\text{O} \\longrightarrow 2\\text{H}_2 + \\text{O}_2 \\;(\\text{balanced})[\/latex]<\/div>\n<table id=\"fs-idp151419504\" class=\"medium unnumbered\" summary=\"This is a table with four columns and three rows. The columns are labeled, \u201cElement,\u201d \u201cReactants,\u201d \u201cProducts,\u201d and \u201cBalanced?\u201d. Under the \u201cElement\u201d column are the letters H and O. Under the \u201cReactants\u201d column are the equations \u201c2 times 2 equals 4,\u201d and \u201c2 times 1 equals 2.\u201d Under the \u201cProducts\u201d column are the equations, \u201c2 times 2 equals 2,\u201d and \u201c1 times 2 equals 2.\u201d The first 2 in the \u201c2 times 2 equals 2\u201d equation is bold. Under the \u201cBalanced?\u201d column are, \u201c4 equals 4, yes,\u201d and \u201c2 equals 2, yes.\u201d\">\n<thead>\n<tr valign=\"top\">\n<th>Element<\/th>\n<th>Reactants<\/th>\n<th>Products<\/th>\n<th>Balanced?<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>H<\/td>\n<td>2 \u00d7 2 = 4<\/td>\n<td><strong>2<\/strong> \u00d7 2 = 4<\/td>\n<td>4 = 4, yes<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>O<\/td>\n<td>2 \u00d7 1 = 2<\/td>\n<td>1 \u00d7 2 = 2<\/td>\n<td>2 = 2, yes<\/td>\n<\/tr>\n<tr>\n<td colspan=\"4\"><strong>Table 4.<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-idp147875376\">These coefficients yield equal numbers of both H and O atoms on the reactant and product sides, and the balanced equation is, therefore:<\/p>\n<div class=\"equation\" id=\"fs-idp105215584\" style=\"text-align: center\">[latex]2\\text{H}_2\\text{O} \\longrightarrow 2\\text{H}_2 + \\text{O}_2[\/latex]<\/div>\n<div class=\"textbox shaded\" id=\"fs-idp1783792\">\n<h3>Example 1<\/h3>\n<p id=\"fs-idp22283888\">Write a balanced equation for the reaction of molecular nitrogen (N<sub>2<\/sub>) and oxygen (O<sub>2<\/sub>) to form dinitrogen pentoxide.<\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-idp154895936\"><strong>Solution<\/strong><br \/>\nFirst, write the unbalanced equation.<\/p>\n<div class=\"equation\" id=\"fs-idp147872544\" style=\"text-align: center\">[latex]\\text{N}_2 + \\text{O}_2 \\longrightarrow \\text{N}_2 \\text{O}_5 \\;(\\text{unbalanced})[\/latex]<\/div>\n<p id=\"fs-idp239172416\">Next, count the number of each type of atom present in the unbalanced equation.<\/p>\n<table id=\"fs-idp107503280\" class=\"medium unnumbered\" summary=\"This is a table with four columns and three rows. The columns are labeled, \u201cElement,\u201d \u201cReactants,\u201d \u201cProducts,\u201d and \u201cBalanced?\u201d. Under the \u201cElement\u201d column are the letters N and O. Under the \u201cReactants\u201d column are the equations \u201c1 times 2 equals 2,\u201d and \u201c1 times 2 equals 2.\u201d Under the \u201cProducts\u201d column are the equations, \u201c1 times 2 equals 2,\u201d and \u201c1 times 5 equals 5.\u201d Under the \u201cBalanced?\u201d column are, \u201c2 equals 2, yes,\u201d and \u201c2 does not equal 5, no.\u201d\">\n<thead>\n<tr valign=\"top\">\n<th>Element<\/th>\n<th>Reactants<\/th>\n<th>Products<\/th>\n<th>Balanced?<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>N<\/td>\n<td>1 \u00d7 2 = 2<\/td>\n<td>1 \u00d7 2 = 2<\/td>\n<td>2 = 2, yes<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>O<\/td>\n<td>1 \u00d7 2 = 2<\/td>\n<td>1 \u00d7 5 = 5<\/td>\n<td>2 \u2260 5, no<\/td>\n<\/tr>\n<tr>\n<td colspan=\"4\"><strong>Table 5.<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-idp266947552\">Though nitrogen is balanced, changes in coefficients are needed to balance the number of oxygen atoms. To balance the number of oxygen atoms, a reasonable first attempt would be to change the coefficients for the O<sub>2<\/sub> and N<sub>2<\/sub>O<sub>5<\/sub> to integers that will yield 10 O atoms (the least common multiple for the O atom subscripts in these two formulas).<\/p>\n<div class=\"equation\" id=\"fs-idp80836160\" style=\"text-align: center\">[latex]\\text{N}_2 + 5\\text{O}_2 \\longrightarrow 2\\text{N}_2\\text{O}_5 \\;(\\text{unbalanced})[\/latex]<\/div>\n<table id=\"fs-idp7305424\" class=\"medium unnumbered\" summary=\"This is a table with four columns and three rows. The columns are labeled, \u201cElement,\u201d \u201cReactants,\u201d \u201cProducts,\u201d and \u201cBalanced?\u201d. Under the \u201cElement\u201d column are the letters N and O. Under the \u201cReactants\u201d column are the equations \u201c1 times 2 equals 2,\u201d and \u201c5 times 2 equals 10.\u201d The 5 in the second equation is bold. Under the \u201cProducts\u201d column are the equations, \u201c2 times 2 equals 4,\u201d and \u201c2 times 5 equals 10.\u201d The initial 2 in each equation is bold. Under the \u201cBalanced?\u201d column are, \u201c2 does not equal 4, no,\u201d and \u201c10 equals 10, yes.\u201d\">\n<thead>\n<tr valign=\"top\">\n<th>Element<\/th>\n<th>Reactants<\/th>\n<th>Products<\/th>\n<th>Balanced?<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>N<\/td>\n<td>1 \u00d7\u00d7 2 = 2<\/td>\n<td><strong>2<\/strong> \u00d7 2 = 4<\/td>\n<td>2 \u2260 4, no<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>O<\/td>\n<td><strong>5<\/strong> \u00d7 2 = 10<\/td>\n<td><strong>2<\/strong> \u00d7 5 = 10<\/td>\n<td>10 = 10, yes<\/td>\n<\/tr>\n<tr>\n<td colspan=\"4\"><strong>Table 6.<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-idp27772864\">The N atom balance has been upset by this change; it is restored by changing the coefficient for the reactant N<sub>2<\/sub> to 2.<\/p>\n<div class=\"equation\" id=\"fs-idp107722208\" style=\"text-align: center\">[latex]2\\text{N}_2 + 5\\text{O}_2 \\longrightarrow 2\\text{N}_2 \\text{O}_5[\/latex]<\/div>\n<table id=\"fs-idm9607408\" class=\"medium unnumbered\" summary=\"This is a table with four columns and three rows. The columns are labeled, \u201cElement,\u201d \u201cReactants,\u201d \u201cProducts,\u201d and \u201cBalanced?\u201d. Under the \u201cElement\u201d column are the letters N and O. Under the \u201cReactants\u201d column are the equations \u201c2 times 2 equals 4,\u201d and \u201c5 times 2 equals 10.\u201d The initial 2 in the first equation is bold. Under the \u201cProducts\u201d column are the equations, \u201c2 times 2 equals 4,\u201d and \u201c2 times 5 equals 10.\u201d Under the \u201cBalanced?\u201d column are, \u201c4 equals 4, yes,\u201d and \u201c10 equals 10, yes.\u201d\">\n<thead>\n<tr valign=\"top\">\n<th>Element<\/th>\n<th>Reactants<\/th>\n<th>Products<\/th>\n<th>Balanced?<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>N<\/td>\n<td><strong>2<\/strong> \u00d7 2 = 4<\/td>\n<td>2 \u00d7 2 = 4<\/td>\n<td>4 = 4, yes<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>O<\/td>\n<td>5 \u00d7 2 = 10<\/td>\n<td>2 \u00d7 5 = 10<\/td>\n<td>10 = 10, yes<\/td>\n<\/tr>\n<tr>\n<td colspan=\"4\"><strong>Table 7.<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-idp17727056\">The numbers of N and O atoms on either side of the equation are now equal, and so the equation is balanced.<\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-idp79125936\"><em><strong>Test Yourself<\/strong><\/em><br \/>\nWrite a balanced equation for the decomposition of ammonium nitrate to form molecular nitrogen, molecular oxygen, and water. (Hint: Balance oxygen last, since it is present in more than one molecule on the right side of the equation.)<\/p>\n<p>&nbsp;<\/p>\n<p><em><strong>Answer<\/strong><\/em><\/p>\n<p>[latex]2\\text{NH}_4 \\text{NO}_3 \\longrightarrow 2\\text{N}_2 + \\text{O}_2 + 4\\text{H}_2\\text{O}[\/latex]<\/p>\n<\/div>\n<p id=\"fs-idp261408544\">It is sometimes convenient to use fractions instead of integers as intermediate coefficients in the process of balancing a chemical equation. When balance is achieved, all the equation\u2019s coefficients may then be multiplied by a whole number to convert the fractional coefficients to integers without upsetting the atom balance. For example, consider the reaction of ethane (C<sub>2<\/sub>H<sub>6<\/sub>) with oxygen to yield H<sub>2<\/sub>O and CO<sub>2<\/sub>, represented by the unbalanced equation:<\/p>\n<div class=\"equation\" id=\"fs-idp25579440\" style=\"text-align: center\">[latex]\\text{C}_2 \\text{H}_6 + \\text{O}_2 \\longrightarrow \\text{H}_2 \\text{O} + \\text{C} \\text{O}_2 \\;(\\text{unbalanced})[\/latex]<\/div>\n<p id=\"fs-idp115010864\">Following the usual inspection approach, one might first balance C and H atoms by changing the coefficients for the two product species, as shown:<\/p>\n<div class=\"equation\" id=\"fs-idp71981056\" style=\"text-align: center\">[latex]\\text{C}_2 \\text{H}_6 + \\text{O}_2 \\longrightarrow 3\\text{H}_2 \\text{O} + 2\\text{C} \\text{O}_2 \\;(\\text{unbalanced})[\/latex]<\/div>\n<p id=\"fs-idp16121888\">This results in seven O atoms on the product side of the equation, an odd number\u2014no integer coefficient can be used with the O<sub>2<\/sub> reactant to yield an odd number, so a fractional coefficient, [latex]\\frac{7}{2}[\/latex], is used instead to yield a provisional balanced equation:<\/p>\n<div class=\"equation\" id=\"fs-idp37794464\" style=\"text-align: center\">[latex]\\text{C}_2 \\text{H}_6 + \\frac{7}{2}\\text{O}_2 \\longrightarrow 3\\text{H}_2 \\text{O} + 2\\text{C} \\text{O}_2 \\;[\/latex]<\/div>\n<p id=\"fs-idp54942992\">A conventional balanced equation with integer-only coefficients is derived by multiplying each coefficient by 2:<\/p>\n<div class=\"equation\" id=\"fs-idm26016768\" style=\"text-align: center\">[latex]2\\text{C}_2 \\text{H}_6 + 7\\text{O}_2 \\longrightarrow 6\\text{H}_2 \\text{O} + 4\\text{C} \\text{O}_2 \\;[\/latex]<\/div>\n<p id=\"fs-idp57635360\">Finally with regard to balanced equations, recall that convention dictates use of the <em>smallest whole-number coefficients<\/em>. Although the equation for the reaction between molecular nitrogen and molecular hydrogen to produce ammonia is, indeed, balanced,<\/p>\n<div class=\"equation\" id=\"fs-idp93360688\" style=\"text-align: center\">[latex]3\\text{N}_2 + 9\\text{H}_2 \\longrightarrow 6\\text{N} \\text{H}_3[\/latex]<\/div>\n<p id=\"fs-idp117890176\">the coefficients are not the smallest possible integers representing the relative numbers of reactant and product molecules. Dividing each coefficient by the greatest common factor, 3, gives the preferred equation:<\/p>\n<div class=\"equation\" id=\"fs-idm15551488\" style=\"text-align: center\">[latex]\\text{N}_2 + 3\\text{H}_2 \\longrightarrow 2\\text{N} \\text{H}_3[\/latex]<\/div>\n<div id=\"fs-idp54281248\" class=\"textbox shaded\">\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Interactive_200DPI-3-2.png\" alt=\"\u00a0\" width=\"140\" height=\"87\" class=\"alignleft\" \/><\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-idp94095504\">Use this interactive <a href=\"http:\/\/openstaxcollege.org\/l\/16BalanceEq\">tutorial<\/a> for additional practice balancing equations.<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<\/section>\n<section id=\"fs-idp50900304\">\n<h2>Additional Information in Chemical Equations<\/h2>\n<p id=\"fs-idp46664848\">The physical states of reactants and products in chemical equations very often are indicated with a parenthetical abbreviation following the formulas. Common abbreviations include <em>s<\/em> for solids, <em>l<\/em> for liquids, <em>g<\/em> for gases, and <em>aq<\/em> for substances dissolved in water (<em>aqueous solutions<\/em>, as introduced in the preceding chapter). These notations are illustrated in the example equation here:<\/p>\n<div class=\"equation\" id=\"fs-idp55940384\" style=\"text-align: center\">[latex]2\\text{Na}(s) + 2\\text{H}_2 \\text{O}(l) \\longrightarrow 2\\text{NaOH}(aq) + \\text{H}_2(g)[\/latex]<\/div>\n<p id=\"fs-idp38373104\">This equation represents the reaction that takes place when sodium metal is placed in water. The solid sodium reacts with liquid water to produce molecular hydrogen gas and the ionic compound sodium hydroxide (a solid in pure form, but readily dissolved in water).<\/p>\n<p id=\"fs-idp53095760\">Special conditions necessary for a reaction are sometimes designated by writing a word or symbol above or below the equation\u2019s arrow. For example, a reaction carried out by heating may be indicated by the uppercase Greek letter delta (\u0394) over the arrow.<\/p>\n<div class=\"equation\" id=\"fs-idp144225920\" style=\"text-align: center\">[latex]\\text{CaCO}_3(s) \\;\\xrightarrow{\\Delta} \\; \\text{CaO}(s) + \\text{CO}_2(g)[\/latex]<\/div>\n<p id=\"fs-idp104838720\">Other examples of these special conditions will be encountered in more depth in later chapters.<\/p>\n<\/section>\n<section id=\"fs-idp44459328\">\n<h2>Ionic Compounds in Solution<\/h2>\n<p id=\"ball-ch04_s03_p02\" class=\"para editable block\">One important aspect about ionic compounds that differs from molecular compounds has to do with dissolving in a liquid, such as water. When molecular compounds, such as sugar, dissolve in water, the individual molecules drift apart from each other. When ionic compounds dissolve, <em class=\"emphasis\">the ions physically separate from each other<\/em>. We can use a chemical equation to represent this process\u2014for example, with NaCl:<\/p>\n<\/section>\n<p>&nbsp;<\/p>\n<section id=\"fs-idp44459328\"><a href=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/Equation-1.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/Equation-1-300x72.png\" alt=\"\" width=\"300\" height=\"72\" class=\"wp-image-4870 size-medium alignleft\" srcset=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/Equation-1-300x72.png 300w, https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/Equation-1-65x16.png 65w, https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/Equation-1-225x54.png 225w, https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/Equation-1-350x84.png 350w, https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/Equation-1.png 482w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/a><\/p>\n<p style=\"text-align: center\"><a href=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/Ionic-Compounds-1.png\" style=\"font-weight: bold;font-size: 14pt\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/Ionic-Compounds-1-300x300.png\" alt=\"\" width=\"200\" height=\"201\" class=\"wp-image-2169\" srcset=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/Ionic-Compounds-1-300x300.png 300w, https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/Ionic-Compounds-1-150x150.png 150w, https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/Ionic-Compounds-1-65x65.png 65w, https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/Ionic-Compounds-1-225x226.png 225w, https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/Ionic-Compounds-1-350x351.png 350w, https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/Ionic-Compounds-1.png 600w\" sizes=\"auto, (max-width: 200px) 100vw, 200px\" \/><\/a><\/p>\n<p style=\"text-align: center\"><strong>Figure 3.<\/strong>\u00a0The dissolution of sodium chloride.<\/p>\n<div class=\"figure large medium-height editable block\" id=\"ball-ch04_s03_f01\">\n<p>When NaCl dissolves in water, the ions separate and go their own way in solution; the ions are now written with their respective charges, and the (aq) phase label emphasizes that they are dissolved (<a class=\"xref\" href=\"#ball-ch04_s03_f01\">Figure 3 &#8220;Ionic Solutions&#8221;<\/a>). This process is called <span class=\"margin_term\"><a class=\"glossterm\">dissociation<\/a><\/span>; we say that the ions <em class=\"emphasis\">dissociate<\/em>.<\/p>\n<p class=\"para\">When an ionic compound dissociates in water, water molecules surround each ion and separate it from the rest of the solid. Each ion goes its own way in solution.<\/p>\n<\/div>\n<p id=\"ball-ch04_s03_p04\" class=\"para editable block\">All ionic compounds that dissolve behave this way. (This behaviour was first suggested by the Swedish chemist Svante August Arrhenius [1859\u20131927] as part of his PhD dissertation in 1884. Interestingly, his PhD examination team had a hard time believing that ionic compounds would behave like this, so they gave Arrhenius a barely passing grade. Later, this work was cited when Arrhenius was awarded the Nobel Prize in Chemistry.)<\/p>\n<p class=\"para editable block\">Keep in mind that when the ions separate, <em class=\"emphasis\">all<\/em>\u00a0of the ions separate. Thus, when CaCl<sub class=\"subscript\">2<\/sub> dissolves, the one Ca<sup class=\"superscript\">2+<\/sup> ion and the two Cl<sup class=\"superscript\">\u2212<\/sup> ions separate from each other:<\/p>\n<p style=\"text-align: center\"><span class=\"informalequation block\">CaCl<sub>2<\/sub>(s) [latex]\\longrightarrow[\/latex] Ca<sup>2+<\/sup>(aq) + Cl<sup>&#8211;<\/sup>(aq) + Cl<sup>&#8211;<\/sup>(aq) <\/span><\/p>\n<p style=\"text-align: center\"><span class=\"informalequation block\">or<\/span><\/p>\n<p style=\"text-align: center\"><span class=\"informalequation block\">CaCl<sub>2<\/sub>(s) [latex]\\longrightarrow[\/latex] Ca<sup>2+<\/sup>(aq) + 2Cl<sup>&#8211;<\/sup>(aq)<\/span><\/p>\n<p id=\"ball-ch04_s03_p05\" class=\"para editable block\">That is, the two chloride ions go off on their own. They do not remain as Cl<sub class=\"subscript\">2<\/sub> (that would be elemental chlorine; these are chloride ions); they do not stick together to make Cl<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">\u2212<\/sup> or Cl<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">2\u2212<\/sup>. They become dissociated ions in their own right. Polyatomic ions also retain their overall identity when they are dissolved.<\/p>\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 2<\/h3>\n<p id=\"ball-ch04_s03_p06\" class=\"para\">Write the chemical equation that represents the dissociation of each ionic compound.<\/p>\n<p class=\"para\">a) KBr \u00a0 \u00a0 \u00a0b)\u00a0Na<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub><\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\n<p class=\"simpara\">a) KBr(s)\u00a0[latex]\\longrightarrow[\/latex] K<sup class=\"superscript\">+<\/sup>(aq) +\u00a0Br<sup class=\"superscript\">\u2212<\/sup>(aq)<\/p>\n<p class=\"simpara\">b) Not only do the two sodium ions go their own way, but the sulfate ion stays together as the sulfate ion. The dissolving equation is<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">Na<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub>(s) [latex]\\longrightarrow[\/latex] 2Na<sup class=\"superscript\">+<\/sup>(aq) +\u00a0SO<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">2\u2212<\/sup>(aq)<\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\n<p id=\"ball-ch04_s03_p07\" class=\"para\">Write the chemical equation that represents the dissociation of (NH<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">2<\/sub>S.<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\n<p id=\"ball-ch04_s03_p08\" class=\"para\">(NH<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">2<\/sub>S(s) [latex]\\longrightarrow[\/latex] 2NH<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup>(aq) +\u00a0S<sup class=\"superscript\">2\u2212<\/sup>(aq)<\/p>\n<\/div>\n<h2>Equations for Ionic Reactions<\/h2>\n<p id=\"fs-idm1065424\">Given the abundance of water on earth, it stands to reason that a great many chemical reactions take place in aqueous media. When ions are involved in these reactions, the chemical equations may be written with various levels of detail appropriate to their intended use. To illustrate this, consider a reaction between ionic compounds taking place in an aqueous solution. When aqueous solutions of CaCl<sub>2<\/sub> and AgNO<sub>3<\/sub> are mixed, a reaction takes place producing aqueous Ca(NO<sub>3<\/sub>)<sub>2<\/sub> and solid AgCl:<\/p>\n<div class=\"equation\" id=\"fs-idp15979216\" style=\"text-align: center\">[latex]\\text{CaCl}_2(aq) + 2\\text{AgNO}_3(aq) \\longrightarrow \\text{Ca(NO}_3)_2(aq) + 2\\text{AgCl}(s)[\/latex]<\/div>\n<p id=\"fs-idp204481712\">This balanced equation, derived in the usual fashion, is called a <strong>molecular equation<\/strong> because it doesn\u2019t explicitly represent the ionic species that are present in solution. When ionic compounds dissolve in water, they may <em>dissociate<\/em> into their constituent ions, which are subsequently dispersed homogenously throughout the resulting solution (a thorough discussion of this important process is provided in the chapter on solutions). Ionic compounds dissolved in water are, therefore, more realistically represented as dissociated ions, in this case:<\/p>\n<div class=\"equation\" id=\"fs-idm1980352\" style=\"text-align: center\">[latex]\\begin{array}{r @{{}\\longrightarrow{}} l} \\text{CaCl}_2(aq) & \\text{Ca}^{2+}(aq) + 2 \\text{Cl}^{-}(aq) \\\\[0.5em] 2 \\text{AgNO}_3(aq) & 2\\text{Ag}^{+}(aq) + 2 {\\text{NO}_3}^{-}(aq) \\\\[0.5em] \\text{Ca(NO}_3)_2(aq) & \\text{Ca}^{2+}(aq) + 2 {\\text{NO}_3}^{-}(aq) \\end{array}[\/latex]<\/div>\n<p id=\"fs-idp24388656\">Unlike these three ionic compounds, AgCl does not dissolve in water to a significant extent, as signified by its physical state notation, <em>s<\/em>.<\/p>\n<p id=\"fs-idp6577008\">Explicitly representing all dissolved ions results in a <strong>complete ionic equation<\/strong>. In this particular case, the formulas for the dissolved ionic compounds are replaced by formulas for their dissociated ions:<\/p>\n<div class=\"equation\" id=\"fs-idp116409536\" style=\"text-align: center\">[latex]\\text{Ca}^{2+}(aq) + 2\\text{Cl}^{-}(aq) + 2\\text{Ag}^{+}(aq) + 2{\\text{NO}_3}^{-}(aq) \\longrightarrow \\text{Ca}^{2+}(aq) + 2{\\text{NO}_3}^{-}(aq) + 2\\text{AgCl}(s)[\/latex]<\/div>\n<p id=\"fs-idp146097648\">Examining this equation shows that two chemical species are present in identical form on both sides of the arrow, Ca<sup>2+<\/sup>(<em>aq<\/em>) and NO<sub>3<\/sub><sup>\u2212<\/sup>(aq).NO<sub>3<\/sub><sup>\u2212<\/sup>(aq). These <strong>spectator ions<\/strong>\u2014ions whose presence is required to maintain charge neutrality\u2014are neither chemically nor physically changed by the process, and so they may be eliminated from the equation to yield a more succinct representation called a <strong>net ionic equation<\/strong>:<\/p>\n<div class=\"equation\" id=\"fs-idp235470368\" style=\"text-align: center\">[latex]\\rule[0.5ex]{4em}{0.1ex}\\hspace{-4em} \\text{Ca}^{2+}(aq) + 2\\text{Cl}^{-}(aq) + 2\\text{Ag}^{+}(aq) + \\rule[0.5ex]{4.5em}{0.1ex}\\hspace{-4.5em} 2\\text{NO}_3^{-}(aq) \\longrightarrow \\rule[0.5ex]{4em}{0.1ex}\\hspace{-4em} \\text{Ca}^{2+}(aq) + \\rule[0.5ex]{4.5em}{0.1ex}\\hspace{-4.5em} 2{\\text{NO}_3}^{-}(aq) + 2\\text{AgCl}(s)[\/latex]<\/div>\n<div class=\"equation\"><\/div>\n<div class=\"equation\" style=\"text-align: center\">[latex]2\\text{Cl}^{-}(aq) + 2\\text{Ag}^{+}(aq) \\longrightarrow 2\\text{AgCl}(s)[\/latex]<\/div>\n<p id=\"fs-idp8534256\">Following the convention of using the smallest possible integers as coefficients, this equation is then written:<\/p>\n<div class=\"equation\" id=\"fs-idp266868512\" style=\"text-align: center\">[latex]\\text{Cl}^{-}(aq) + \\text{Ag}^{+}(aq) \\longrightarrow \\text{AgCl}(s)[\/latex]<\/div>\n<p id=\"fs-idp42466432\">This net ionic equation indicates that solid silver chloride may be produced from dissolved chloride and silver(I) ions, regardless of the source of these ions. These molecular and complete ionic equations provide additional information, namely, the ionic compounds used as sources of Cl<sup>\u2212<\/sup> and Ag<sup>+<\/sup>.<\/p>\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 3<\/h3>\n<p id=\"ball-ch04_s03_p11\" class=\"para\">Write the complete ionic equation for each chemical reaction.<\/p>\n<p class=\"para\">a) KBr(aq) +\u00a0AgC<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub>(aq) [latex]\\longrightarrow[\/latex] KC<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub>(aq) +\u00a0AgBr(s)<\/p>\n<p class=\"para\">b) MgSO<sub class=\"subscript\">4<\/sub>(aq) +\u00a0Ba(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub>(aq) [latex]\\longrightarrow[\/latex] Mg(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub>(aq) +\u00a0BaSO<sub class=\"subscript\">4<\/sub>(s)<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\n<p id=\"ball-ch04_s03_p12\" class=\"para\">For any ionic compound that is aqueous, we will write the compound as separated ions.<\/p>\n<p class=\"para\">a) The complete ionic equation is<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">K<sup class=\"superscript\">+<\/sup>(aq) +\u00a0Br<sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a0Ag<sup class=\"superscript\">+<\/sup>(aq) +\u00a0C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq) [latex]\\longrightarrow[\/latex] K<sup class=\"superscript\">+<\/sup>(aq) +\u00a0C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a0AgBr(s)<\/span><\/span><\/p>\n<p>b) The complete ionic equation is<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">Mg<sup class=\"superscript\">2+<\/sup>(aq) +\u00a0SO<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">2\u2212<\/sup>(aq) +\u00a0Ba<sup class=\"superscript\">2+<\/sup>(aq) +\u00a02NO<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq) [latex]\\longrightarrow[\/latex] Mg<sup class=\"superscript\">2+<\/sup>(aq) +\u00a02NO<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a0BaSO<sub class=\"subscript\">4<\/sub>(s)<\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\n<p id=\"ball-ch04_s03_p13\" class=\"para\">Write the complete ionic equation for<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">CaCl<sub class=\"subscript\">2<\/sub>(aq) +\u00a0Pb(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub>(aq) [latex]\\longrightarrow[\/latex] Ca(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub>(aq) +\u00a0PbCl<sub class=\"subscript\">2<\/sub>(s)<\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\n<p id=\"ball-ch04_s03_p14\" class=\"para\">Ca<sup class=\"superscript\">2+<\/sup>(aq) +\u00a02Cl<sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a0Pb<sup class=\"superscript\">2+<\/sup>(aq) +\u00a02NO<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq) [latex]\\longrightarrow[\/latex] Ca<sup class=\"superscript\">2+<\/sup>(aq) +\u00a02NO<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a0PbCl<sub class=\"subscript\">2<\/sub>(s)<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 4<\/h3>\n<p id=\"ball-ch04_s03_p19\" class=\"para\">Write the net ionic equation for each chemical reaction.<\/p>\n<p class=\"para\">a) K<sup class=\"superscript\">+<\/sup>(aq) +\u00a0Br<sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a0Ag<sup class=\"superscript\">+<\/sup>(aq) +\u00a0C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq) [latex]\\longrightarrow[\/latex] K<sup class=\"superscript\">+<\/sup>(aq) +\u00a0C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a0AgBr(s)<\/p>\n<p class=\"para\">b) Mg<sup class=\"superscript\">2+<\/sup>(aq) +\u00a0SO<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">2\u2212<\/sup>(aq) +\u00a0Ba<sup class=\"superscript\">2+<\/sup>(aq) +\u00a02 NO<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq) [latex]\\longrightarrow[\/latex] Mg<sup class=\"superscript\">2+<\/sup>(aq) +\u00a02 NO<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a0BaSO<sub class=\"subscript\">4<\/sub>(s)<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\n<p class=\"simpara\">a) In the first equation, the K<sup class=\"superscript\">+<\/sup>(aq) and C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq) ions are spectator ions, so they are canceled:<\/p>\n<p><span class=\"informalequation\">\u00a0K<sup>+<\/sup>(aq) + Br<sup>\u2212<\/sup>(aq) + Ag<sup>+<\/sup>(aq) + C<sub>2<\/sub>H<sub>3<\/sub>O<sub>2<\/sub><sup>\u2212<\/sup>(aq) [latex]\\longrightarrow[\/latex] K<sup>+<\/sup>(aq) + C<sub>2<\/sub>H<sub>3<\/sub>O<sub>2<\/sub><sup>\u2212<\/sup>(aq) + AgBr(s)<\/span><\/p>\n<p id=\"ball-ch04_s03_p20\" class=\"para\">The net ionic equation is<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">Br<sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a0Ag<sup class=\"superscript\">+<\/sup>(aq) [latex]\\longrightarrow[\/latex] AgBr(s)<\/span><\/span><\/p>\n<p>b) In the second equation, the Mg<sup class=\"superscript\">2+<\/sup>(aq) and NO<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq) ions are spectator ions, so they are canceled:<\/p>\n<p><span class=\"informalequation\">Mg<sup>2+<\/sup>(aq) + SO<sub>4<\/sub><sup>2\u2212<\/sup>(aq) + Ba<sup>2+<\/sup>(aq) +\u00a02 NO<sub>3<\/sub><sup>\u2212<\/sup>(aq) [latex]\\longrightarrow[\/latex] Mg<sup>2+<\/sup>(aq) + 2 NO<sub>3<\/sub><sup>\u2212<\/sup>(aq) + BaSO<sub>4<\/sub>(s)<\/span><\/p>\n<p id=\"ball-ch04_s03_p21\" class=\"para\">The net ionic equation is<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">SO<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">2\u2212<\/sup>(aq) +\u00a0Ba<sup class=\"superscript\">2+<\/sup>(aq) [latex]\\longrightarrow[\/latex] BaSO<sub class=\"subscript\">4<\/sub>(s)<\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\n<p id=\"ball-ch04_s03_p22\" class=\"para\">Write the net ionic equation for<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">CaCl<sub class=\"subscript\">2<\/sub>(aq) +\u00a0Pb(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub>(aq) [latex]\\longrightarrow[\/latex] Ca(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub>(aq) +\u00a0PbCl<sub class=\"subscript\">2<\/sub>(s)<\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\n<p id=\"ball-ch04_s03_p23\" class=\"para\">Pb<sup class=\"superscript\">2+<\/sup>(aq) +\u00a02Cl<sup class=\"superscript\">\u2212<\/sup>(aq) [latex]\\longrightarrow[\/latex] PbCl<sub class=\"subscript\">2<\/sub>(s)<\/p>\n<\/div>\n<div class=\"textbox shaded\" id=\"fs-idp44418832\">\n<h3>Example 5<\/h3>\n<p id=\"fs-idm27450736\">When carbon dioxide is dissolved in an aqueous solution of sodium hydroxide, the mixture reacts to yield aqueous sodium carbonate and liquid water. Write balanced molecular, complete ionic, and net ionic equations for this process.<\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-idp41941568\"><strong>Solution<\/strong><br \/>\nBegin by identifying formulas for the reactants and products and arranging them properly in chemical equation form:<\/p>\n<div class=\"equation\" id=\"fs-idp114521376\" style=\"text-align: center\">[latex]\\text{CO}_2(aq) + \\text{NaOH}(aq) \\longrightarrow \\text{Na}_2 \\text{CO}_3(aq) + \\text{H}_2 \\text{O}(l) \\;(\\text{unbalanced})[\/latex]<\/div>\n<p id=\"fs-idp277452576\">Balance is achieved easily in this case by changing the coefficient for NaOH to 2, resulting in the molecular equation for this reaction:<\/p>\n<div class=\"equation\" id=\"fs-idp54247424\" style=\"text-align: center\">[latex]\\text{CO}_2(aq) + 2\\text{NaOH}(aq) \\longrightarrow \\text{Na}_2 \\text{CO}_3(aq) + \\text{H}_2 \\text{O}(l)[\/latex]<\/div>\n<p id=\"fs-idp71749568\">The two dissolved ionic compounds, NaOH and Na<sub>2<\/sub>CO<sub>3<\/sub>, can be represented as dissociated ions to yield the complete ionic equation:<\/p>\n<div class=\"equation\" id=\"fs-idp41721264\" style=\"text-align: center\">[latex]\\text{CO}_2(aq) + 2\\text{Na}^{+}(aq) + 2\\text{OH}^{-}(aq) \\longrightarrow 2\\text{Na}^{+}(aq) + {\\text{CO}_3}^{2-}(aq) + \\text{H}_2 \\text{O}(l)[\/latex]<\/div>\n<p id=\"fs-idp111986480\">Finally, identify the spectator ion(s), in this case Na<sup>+<\/sup>(<em>aq<\/em>), and remove it from each side of the equation to generate the net ionic equation:<\/p>\n<div class=\"equation\" id=\"fs-idp56132768\">\n<p style=\"text-align: center\">[latex]\\text{CO}_2(aq) + \\rule[0.5ex]{4.25em}{0.1ex}\\hspace{-4.25em} 2\\text{Na}^{+}(aq) + 2\\text{OH}^{-}(aq) \\longrightarrow \\rule[0.5ex]{4.25em}{0.1ex}\\hspace{-4.25em} 2\\text{Na}^{+}(aq) + {\\text{CO}_3}^{2-}(aq) + \\text{H}_2 \\text{O}(l)[\/latex][latex]\\text{CO}_2(aq) + 2\\text{OH}^{-}(aq) \\longrightarrow {\\text{CO}_3}^{2-}(aq) + \\text{H}_2 \\text{O}(l)[\/latex]<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-idp71485312\"><em><strong>Test Yourself<\/strong><\/em><br \/>\nDiatomic chlorine and sodium hydroxide (lye) are commodity chemicals produced in large quantities, along with diatomic hydrogen, via the electrolysis of brine, according to the following unbalanced equation:<\/p>\n<div class=\"equation\" id=\"fs-idp47302160\" style=\"text-align: center\">[latex]\\text{NaCl}(aq) + \\text{H}_2 \\text{O}(l) \\;\\;\\xrightarrow{\\text{electricity}}\\;\\; \\text{NaOH}(aq) + \\text{H}_2(g) + \\text{Cl}_2(g)[\/latex]<\/div>\n<p id=\"fs-idp124894848\">Write balanced molecular, complete ionic, and net ionic equations for this process.<\/p>\n<p>&nbsp;<\/p>\n<p><em><strong>Answers<\/strong><\/em><\/p>\n<p>[latex]2\\text{NaCl}(aq) + 2\\text{H}_2 \\text{O} \\longrightarrow 2 \\text{NaOH}(aq) + \\text{H}_2(g) + \\text{Cl}_2(g) (\\text{molecular})[\/latex]<\/p>\n<p>[latex]2\\text{Na}^{+}(aq) + 2\\text{Cl}^{-}(aq) + 2\\text{H}_2 \\text{O} \\longrightarrow 2\\text{Na}^{+}(aq) + 2\\text{OH}^{-}(aq) + \\text{H}_2(g) + \\text{Cl}_2(g) (\\text{complete ionic})[\/latex]<\/p>\n<p>[latex]2\\text{Cl}^{-}(aq) + 2\\text{H}_2 \\text{O} \\longrightarrow 2\\text{OH}^{-}(aq) + 2\\text{H}_2(g) + \\text{Cl}_2(g) (\\text{net ionic})[\/latex]<\/p>\n<\/div>\n<\/section>\n<section id=\"fs-idp114420064\" class=\"summary\">\n<div class=\"callout block\" id=\"ball-ch04_s03_n05\">\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Chemistry Is Everywhere: Soluble and Insoluble Ionic Compounds<\/h3>\n<p id=\"ball-ch04_s03_p52\" class=\"para\">The concept of solubility versus insolubility in ionic compounds is a matter of degree. Some ionic compounds are very soluble, some are only moderately soluble, and some are soluble so little that they are considered insoluble. For most ionic compounds, there is also a limit to the amount of compound can be dissolved in a sample of water. For example, you can dissolve a maximum of 36.0 g of NaCl in 100 g of water at room temperature, but you can dissolve only 0.00019 g of AgCl in 100 g of water. We consider NaCl soluble but AgCl insoluble.<\/p>\n<figure id=\"attachment_3210\" aria-describedby=\"caption-attachment-3210\" style=\"width: 385px\" class=\"wp-caption aligncenter\"><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/800px-Grand_canyon_yavapal_point_2010.jpg\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/800px-Grand_canyon_yavapal_point_2010-1.jpg\" alt=\"The Grand Canyon was formed by water running through rock for billions of years, very slowly dissolving it. Note the Colorado River is still present in the lower part of the photo. \u201cGrand canyon yavapal point 2010\u2032\u2032 by chensiyuan is licensed under Creative Commons\" class=\"wp-image-3210\" height=\"251\" width=\"385\" \/><\/a><figcaption id=\"caption-attachment-3210\" class=\"wp-caption-text\"><strong>Figure 4.<\/strong>\u00a0The Grand Canyon was formed by water running through rock for billions of years, very slowly dissolving it. Note the Colorado River is still present in the lower part of the photo. \u201cGrand canyon yavapal point 2010\u2032\u2032 by chensiyuan is licensed under Creative Commons<\/figcaption><\/figure>\n<p class=\"para\">One place where solubility is important is in the tank-type water heater found in many homes in the United States. Domestic water frequently contains small amounts of dissolved ionic compounds, including calcium carbonate (CaCO<sub class=\"subscript\">3<\/sub>). However, CaCO<sub class=\"subscript\">3<\/sub> has the relatively unusual property of being less soluble in hot water than in cold water. So as the water heater operates by heating water, CaCO<sub class=\"subscript\">3<\/sub> can precipitate if there is enough of it in the water. This precipitate, called <em class=\"emphasis\">limescale<\/em>, can also contain magnesium compounds, hydrogen carbonate compounds, and phosphate compounds. The problem is that too much limescale can impede the function of a water heater, requiring more energy to heat water to a specific temperature or even blocking water pipes into or out of the water heater, causing dysfunction.<\/p>\n<p id=\"ball-ch04_s03_p54\" class=\"para\">Another place where solubility versus insolubility is an issue is the Grand Canyon. We usually think of rock as insoluble. But it is actually ever so slightly soluble. This means that over a period of about two billion years, the Colorado River carved rock from the surface by slowly dissolving it, eventually generating a spectacular series of gorges and canyons. And all because of solubility!<\/p>\n<\/div>\n<h2><span style=\"font-family: Roboto, Helvetica, Arial, sans-serif\">Key Concepts and Summary<\/span><\/h2>\n<\/div>\n<p id=\"fs-idp43480096\">Chemical equations are symbolic representations of chemical and physical changes. Formulas for the substances undergoing the change (reactants) and substances generated by the change (products) are separated by an arrow and preceded by integer coefficients indicating their relative numbers. Balanced equations are those whose coefficients result in equal numbers of atoms for each element in the reactants and products. Chemical reactions in aqueous solution that involve ionic reactants or products may be represented more realistically by complete ionic equations and, more succinctly, by net ionic equations.<\/p>\n<\/section>\n<section id=\"fs-idm23851744\" class=\"exercises\">\n<div class=\"bcc-box bcc-info\">\n<h3>Exercises<\/h3>\n<p>1. What does it mean to say an equation is balanced? Why is it important for an equation to be balanced?<\/p>\n<p>2. Balance the following equations:<\/p>\n<p id=\"fs-idp156664864\">a) [latex]\\text{PCl}_5(s) + \\text{H}_2 \\text{O}(l) \\longrightarrow \\text{POCl}_3(l) + \\text{HCl}(aq)[\/latex]<\/p>\n<p id=\"fs-idm5703840\">b) [latex]\\text{Cu}(s) + \\text{HNO}_3(aq) \\longrightarrow \\text{Cu(NO}_3)_2(aq) + \\text{H}_2 \\text{O}(l) + \\text{NO}(g)[\/latex]<\/p>\n<p id=\"fs-idp26705376\">c) [latex]\\text{H}_2(g) + \\text{I}_2(s) \\longrightarrow \\text{HI}(s)[\/latex]<\/p>\n<p id=\"fs-idm2545856\">d) [latex]\\text{Fe}(s) + \\text{O}_2(g) \\longrightarrow \\text{Fe}_2 \\text{O}_3(s)[\/latex]<\/p>\n<p id=\"fs-idp38879216\">e) [latex]\\text{Na}(s) + \\text{H}_2 \\text{O}(l) \\longrightarrow \\text{NaOH}(aq) + \\text{H}_2(g)[\/latex]<\/p>\n<p id=\"fs-idp261699472\">f) [latex]\\text{(NH}_4)_2 \\text{Cr}_2\\text{O}_7(s) \\longrightarrow \\text{Cr}_2\\text{O}_3(s) + \\text{N}_2(g) + \\text{H}_2 \\text{O}(g)[\/latex]<\/p>\n<p id=\"fs-idp78017472\">g) [latex]\\text{P}_4(s) + \\text{Cl}_2(g) \\longrightarrow \\text{PCl}_3(l)[\/latex]<\/p>\n<p id=\"fs-idp44039696\">h) [latex]\\text{PtCl}_4(s) \\longrightarrow \\text{Pt}(s) + \\text{Cl}_2(g)[\/latex]<\/p>\n<p>3. Balance the following equations:<\/p>\n<p id=\"fs-idp14691792\">a) [latex]\\text{Ag}(s) + \\text{H}_2 \\text{S}(g) + \\text{O}_2(g) \\longrightarrow \\text{Ag}_2 \\text{S}(s) + \\text{H}_2 \\text{O}(l)[\/latex]<\/p>\n<p id=\"fs-idp41951632\">b) [latex]\\text{P}_4(s) + \\text{O}_2(g) \\longrightarrow \\text{P}_4 \\text{O}_{10}(s)[\/latex]<\/p>\n<p id=\"fs-idp215876720\">c) [latex]\\text{Pb}(s) + \\text{H}_2 \\text{O}(l) + \\text{O}_2(g) \\longrightarrow \\text{Pb(OH)}_2(s)[\/latex]<\/p>\n<p id=\"fs-idp7611408\">d) [latex]\\text{Fe}(s) + \\text{H}_2 \\text{O}(l) \\longrightarrow \\text{Fe}_3 \\text{O}_4(s) + \\text{H}_2(g)[\/latex]<\/p>\n<p id=\"fs-idp51158528\">e) [latex]\\text{Sc}_2 \\text{O}_3(s) + \\text{SO}_3(l) \\longrightarrow \\text{Sc}_2 \\text{(SO}_4)_3(s)[\/latex]<\/p>\n<p id=\"fs-idp77422352\">f) [latex]\\text{Ca}_3 \\text{(PO}_4)_2(aq) + \\text{H}_3 \\text{PO}_4(aq) \\longrightarrow \\text{Ca(H}_2 \\text{PO}_4)_2(aq)[\/latex]<\/p>\n<p id=\"fs-idp48923840\">g) [latex]\\text{Al}(s) + \\text{H}_2 \\text{SO}_4(aq) \\longrightarrow \\text{Al}_2 \\text{(SO}_4)_3(s) + \\text{H}_2(g)[\/latex]<\/p>\n<p id=\"fs-idp73552\">h) [latex]\\text{TiCl}_4(s) + \\text{H}_2 \\text{O}(g) \\longrightarrow \\text{TiO}_2(s) + \\text{HCl}(g)[\/latex]<\/p>\n<p>4. Write a balanced molecular equation describing each of the following chemical reactions.<\/p>\n<p id=\"fs-idp105395536\">a) Solid calcium carbonate is heated and decomposes to solid calcium oxide and carbon dioxide gas.<\/p>\n<p id=\"fs-idp105396032\">b) Gaseous butane, C<sub>4<\/sub>H<sub>10<\/sub>, reacts with diatomic oxygen gas to yield gaseous carbon dioxide and water vapor.<\/p>\n<p id=\"fs-idp45919792\">c) Aqueous solutions of magnesium chloride and sodium hydroxide react to produce solid magnesium hydroxide and aqueous sodium chloride.<\/p>\n<p id=\"fs-idp45920336\">d) Water vapor reacts with sodium metal to produce solid sodium hydroxide and hydrogen gas.<\/p>\n<p>5. Colorful fireworks often involve the decomposition of barium nitrate and potassium chlorate and the reaction of the metals magnesium, aluminum, and iron with oxygen.<\/p>\n<p id=\"fs-idm22159824\">a) Write the formulas of barium nitrate and potassium chlorate.<\/p>\n<p id=\"fs-idm22159440\">b) The decomposition of solid potassium chlorate leads to the formation of solid potassium chloride and diatomic oxygen gas. Write an equation for the reaction.<\/p>\n<p id=\"fs-idm22158880\">c) The decomposition of solid barium nitrate leads to the formation of solid barium oxide, diatomic nitrogen gas, and diatomic oxygen gas. Write an equation for the reaction.<\/p>\n<p id=\"fs-idp208977456\">d) Write separate equations for the reactions of the solid metals magnesium, aluminum, and iron with diatomic oxygen gas to yield the corresponding metal oxides. (Assume the iron oxide contains Fe<sup>3+<\/sup> ions.)<\/p>\n<p>6. Fill in the blank with a single chemical formula for a covalent compound that will balance the equation:<img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_01_basehyd_img-2.jpg\" alt=\"This figure shows a chemical reaction. On the left side of the reaction arrow, a structural formula for a molecule is shown on the far left. It has a C atom on the left to which H atoms are bonded above, below, and to the left. To the right, another C atom is bonded which has H atoms bonded above and below. To the right, another C atom is bonded, which has a double bonded O atom above and another O atom singly bonded to the right. To the right of the singly bonded O atom, an H atom is bonded. This is followed by a plus sign and N a O H. A reaction arrow appears to the right, which is followed by another structural formula. It has a C atom on the left to which H atoms are bonded above, below, and to the left. To the right, another C atom is bonded which has H atoms bonded above and below. To the right, another C atom is bonded, which has a double bonded O atom above and another O atom singly bonded to the right. The singly bonded O atom is followed by a superscript negative sign. This is followed to the right by a plus sign, N a superscript positive sign, another plus sign, and a horizontal line segment, indicating a space for an answer to be written.\" \/><\/p>\n<p>7. Aqueous hydrogen fluoride (hydrofluoric acid) is used to etch glass and to analyze minerals for their silicon content. Hydrogen fluoride will also react with sand (silicon dioxide).<\/p>\n<p id=\"fs-idm98641232\">a) Write an equation for the reaction of solid silicon dioxide with hydrofluoric acid to yield gaseous silicon tetrafluoride and liquid water.<\/p>\n<p id=\"fs-idp126128736\">b) The mineral fluorite (calcium fluoride) occurs extensively in Illinois. Solid calcium fluoride can also be prepared by the reaction of aqueous solutions of calcium chloride and sodium fluoride, yielding aqueous sodium chloride as the other product. Write complete and net ionic equations for this reaction.<\/p>\n<p>8. From the balanced molecular equations, write the complete ionic and net ionic equations for the following:<\/p>\n<p id=\"fs-idp41636720\">a) [latex]\\text{K}_2 \\text{C}_2 \\text{O}_4(aq) + \\text{Ba(OH)}_2(aq) \\longrightarrow 2\\text{KOH}(aq) + \\text{BaC}_2 \\text{O}_2(s)[\/latex]<\/p>\n<p id=\"fs-idp46970864\">b) [latex]{\\text{Pb(NO}_3)}_2(aq) + \\text{H}_2 \\text{SO}_4(aq) \\longrightarrow \\text{PbSO}_4(s) + 2\\text{HNO}_3(aq)[\/latex]<\/p>\n<p id=\"fs-idm2481664\">c) [latex]\\text{CaCO}_3(s) + \\text{H}_2 \\text{SO}_4(aq) \\longrightarrow \\text{CaSO}_4(s) + \\text{CO}_2(g) + \\text{H}_2\\text{O}(l)[\/latex]<\/p>\n<p id=\"ball-ch04_s01_qs01_p1\" class=\"para\">9. From the statement \u201cnitrogen gas and hydrogen gas react to produce ammonia gas,\u201d identify the reactants and the products.<\/p>\n<p class=\"para\"><span style=\"font-size: 1em\">10. From the statement \u201ca solution of magnesium hydroxide reacts with a solution of nitric acid to produce a solution of magnesium nitrate and water,\u201d identify the reactants and the products.<\/span><\/p>\n<p class=\"para\"><span style=\"font-size: 1em\">11. Write and balance the chemical equation described by Exercise 1.<\/span><\/p>\n<p class=\"para\"><span style=\"font-size: 1em\">12. Write and balance the chemical equation described by Exercise 2.<\/span><\/p>\n<p class=\"para\"><span style=\"font-size: 1em\">13. Balance: ___NaClO<\/span><sub class=\"subscript\">3<\/sub><span style=\"font-size: 1em\"> [latex]\\longrightarrow[\/latex] ___NaCl +\u00a0___O<\/span><sub class=\"subscript\">2<\/sub><\/p>\n<p class=\"para\"><span style=\"font-size: 1em\">14. Balance: ___N<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\"> +\u00a0___H<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\"> [latex]\\longrightarrow[\/latex] ___N<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">H<\/span><sub class=\"subscript\">4<\/sub><\/p>\n<p class=\"para\"><span style=\"font-size: 1em\">15. Balance: ___Al +\u00a0___O<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\"> [latex]\\longrightarrow[\/latex] ___Al<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">O<\/span><sub class=\"subscript\">3<\/sub><\/p>\n<p class=\"para\"><span style=\"font-size: 1em\">16. Balance: ___C<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">H<\/span><sub class=\"subscript\">4<\/sub><span style=\"font-size: 1em\"> +\u00a0___O<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\"> [latex]\\longrightarrow[\/latex] ___CO<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\"> +\u00a0___H<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">O<\/span><\/p>\n<p class=\"para\"><span style=\"font-size: 1em\">17. Balance:\u00a0___N<sub class=\"subscript\">2<\/sub><sub>(g)<\/sub> +\u00a0___H<sub class=\"subscript\">2<\/sub><sub>(g)<\/sub> [latex]\\longrightarrow[\/latex] ___NH<sub class=\"subscript\">3<\/sub><sub>(g)<\/sub><\/span><\/p>\n<p id=\"ball-ch04_s03_qs01_p1\" class=\"para\">18. Write a chemical equation that represents NaBr(s) dissociating in water.<\/p>\n<p class=\"para\"><span style=\"font-size: 1em\">19. Write a chemical equation that represents (NH<\/span><sub class=\"subscript\">4<\/sub><span style=\"font-size: 1em\">)<\/span><sub class=\"subscript\">3<\/sub><span style=\"font-size: 1em\">PO<\/span><sub class=\"subscript\">4<\/sub><span style=\"font-size: 1em\">(s) dissociating in water.<\/span><\/p>\n<p class=\"para\"><span style=\"font-size: 1em\">20. Write the complete ionic equation for the reaction of FeCl<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">(aq) and AgNO<\/span><sub class=\"subscript\">3<\/sub><span style=\"font-size: 1em\">(aq). You may have to consult the solubility rules.<\/span><\/p>\n<p class=\"para\"><span style=\"font-size: 1em\">21. Write the complete ionic equation for the reaction of KCl(aq) and NaC<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">H<\/span><sub class=\"subscript\">3<\/sub><span style=\"font-size: 1em\">O<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">(aq). You may have to consult the solubility rules.<\/span><\/p>\n<p class=\"para\"><span style=\"font-size: 1em\">22. Write the net ionic equation for the reaction of FeCl<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">(aq) and AgNO<\/span><sub class=\"subscript\">3<\/sub><span style=\"font-size: 1em\">(aq). You may have to consult the solubility rules.<\/span><\/p>\n<p class=\"para\"><span style=\"font-size: 1em\">23. Write the net ionic equation for the reaction of KCl(aq) and NaC<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">H<\/span><sub class=\"subscript\">3<\/sub><span style=\"font-size: 1em\">O<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">(aq). You may have to consult the solubility rules.<\/span><\/p>\n<p class=\"para\"><span style=\"font-size: 1em\">24. Identify the spectator ions in Exercises 20 and 21.<\/span><\/p>\n<p>&nbsp;<\/p>\n<p><strong>Answers<\/strong><\/p>\n<p id=\"fs-idm10510096\">1. An equation is balanced when the same number of each element is represented on the reactant and product sides. Equations must be balanced to accurately reflect the law of conservation of matter.<\/p>\n<p id=\"fs-idm90785536\">2.\u00a0a) [latex]\\text{PCl}_5(s) + \\text{H}_2 \\text{O}(l) \\longrightarrow \\text{POCl}_3(l) + 2\\text{HCl}(aq)[\/latex]<\/p>\n<p>b) [latex]3\\text{Cu}(s) + 8\\text{HNO}_3(aq) \\longrightarrow 3\\text{Cu(NO}_3)_2(aq) + 4\\text{H}_2 \\text{O}(l) + 2\\text{NO}(g)[\/latex]<\/p>\n<p>c) [latex]\\text{H}_2(g) + \\text{I}_2(s) \\longrightarrow 2\\text{HI}(s)[\/latex]<\/p>\n<p>d) [latex]4\\text{Fe}(s) + 3\\text{O}_2(g) \\longrightarrow 2\\text{Fe}_2 \\text{O}_3(s)[\/latex]<\/p>\n<p>e) [latex]2\\text{Na}(s) + 2\\text{H}_2 \\text{O}(l) \\longrightarrow 2\\text{NaOH}(aq) + \\text{H}_2(g)[\/latex]<\/p>\n<p>f) [latex]\\text{(NH}_4)_2 \\text{Cr}_2\\text{2O}_7(s) \\longrightarrow \\text{Cr}_2\\text{O}_3(s) + \\text{N}_2(g) + 4\\text{H}_2 \\text{O}(l)[\/latex]<\/p>\n<p>g) [latex]\\text{P}_4(s) + 6\\text{Cl}_2(g) \\longrightarrow 4\\text{PCl}_3(l)[\/latex]<\/p>\n<p>h) [latex]\\text{PtCl}_4(s) \\longrightarrow \\text{Pt}(s) + 2\\text{Cl}_2(g)[\/latex]<\/p>\n<p>3.\u00a0a) [latex]4\\text{Ag}(s) + 2\\text{H}_2 \\text{S}(g) + \\text{O}_2(g) \\longrightarrow 2\\text{Ag}_2 \\text{S}(s) + 2\\text{H}_2 \\text{O}(l)[\/latex]<\/p>\n<p id=\"fs-idp41951632\">b) [latex]\\text{P}_4(s) + 5\\text{O}_2(g) \\longrightarrow \\text{P}_4 \\text{O}_{10}(s)[\/latex]<\/p>\n<p id=\"fs-idp215876720\">c) [latex]2\\text{Pb}(s) + 2\\text{H}_2 \\text{O}(l) + \\text{O}_2(g) \\longrightarrow 2\\text{Pb(OH)}_2(s)[\/latex]<\/p>\n<p id=\"fs-idp7611408\">d) [latex]3\\text{Fe}(s) + 4\\text{H}_2 \\text{O}(l) \\longrightarrow \\text{Fe}_3 \\text{O}_4(s) + 4\\text{H}_2(g)[\/latex]<\/p>\n<p id=\"fs-idp51158528\">e) [latex]\\text{Sc}_2 \\text{O}_3(s) + 3\\text{SO}_3(l) \\longrightarrow \\text{Sc}_2 \\text{(SO}_4)_3(s)[\/latex]<\/p>\n<p id=\"fs-idp77422352\">f) [latex]\\text{Ca}_3 \\text{(PO}_4)_2(aq) + 4\\text{H}_3 \\text{PO}_4(aq) \\longrightarrow 3\\text{Ca(H}_2 \\text{PO}_4)_2(aq)[\/latex]<\/p>\n<p id=\"fs-idp48923840\">g) [latex]2\\text{Al}(s) + 3\\text{H}_2 \\text{SO}_4(aq) \\longrightarrow \\text{Al}_2 \\text{(SO}_4)_3(s) + 3\\text{H}_2(g)[\/latex]<\/p>\n<p id=\"fs-idp73552\">h) [latex]\\text{TiCl}_4(s) + 2\\text{H}_2 \\text{O}(g) \\longrightarrow \\text{TiO}_2(s) + 4\\text{HCl}(g)[\/latex]<\/p>\n<p>4.a) [latex]\\text{CaCO}_3(s) \\longrightarrow \\text{CaO}(s) + \\text{CO}_2(g)[\/latex]<br \/>\nb) [latex]2\\text{C}_4 \\text{H}_{10}(g) + 13 \\text{O}_2(g) \\longrightarrow 8\\text{CO}_2(g) + 10\\text{H}_2\\text{O}(g)[\/latex]<br \/>\nc) [latex]\\text{MgCl}_{2}(aq) + 2 \\text{NaOH}(aq) \\longrightarrow \\text{Mg(OH)}_2(s) + 2 \\text{NaCl}(aq)[\/latex]<br \/>\nd) [latex]2\\text{H}_2 \\text{O}(g) + 2 \\text{Na}(s) \\longrightarrow 2\\text{NaOH}(s) + \\text{H}_2(g)[\/latex]<\/p>\n<p id=\"fs-idp208978816\">5.\u00a0a) [latex]\\text{Ba(NO}_3)_2[\/latex] , [latex]\\text{KClO}_3[\/latex]<br \/>\nb) [latex]2 \\text{KClO}_3(s) \\longrightarrow 2 \\text{KCl}(s) + 3\\text{O}_2(g)[\/latex]<br \/>\nc) [latex]2 \\text{Ba(NO}_3)_2(s) \\longrightarrow 2\\text{BaO}(s) + 2\\text{N}_2(g) + 5\\text{O}_2(g)[\/latex]<br \/>\nd) [latex]2 \\text{Mg}(s) + \\text{O}_2(g) \\longrightarrow 2 \\text{MgO}(s)[\/latex];\u00a0[latex]4\\text{Al}(s) + 3\\text{O}_2(g) \\longrightarrow 2\\text{Al}_2 \\text{O}_3(g)[\/latex];\u00a0<span style=\"line-height: 1.5\">[latex]4\\text{Fe}(s) + 3\\text{O}_2(g) \\longrightarrow 2\\text{Fe}_2 \\text{O}_3(s)[\/latex]<\/span><\/p>\n<p>6. H<sub>2<\/sub>O<\/p>\n<p>7.\u00a0a) [latex]4\\text{HF}(aq) + \\text{SiO}_2(s) \\longrightarrow \\text{SiF}_4(g) + 2\\text{H}_2 \\text{O}(l)[\/latex]<br \/>\nb) complete: [latex]2\\text{Na}^{+}(aq) + 2\\text{F}^{-}(aq) + \\text{Ca}^{2+}(aq) + 2\\text{Cl}^{-}(aq) \\longrightarrow \\text{CaF}_2(s) + 2\\text{Na}^{+}(aq) + 2\\text{Cl}^{-}(aq)[\/latex]<\/p>\n<p>net:\u00a0[latex]2\\text{F}^{-}(aq) + \\text{Ca}^{2+}(aq) \\longrightarrow \\text{CaF}_2(s)[\/latex]<\/p>\n<p id=\"fs-idp198600416\">8.\u00a0a) complete:\u00a0[latex]2\\text{K}^{+}(aq) + {\\text{C}_2 \\text{O}_4}^{2-}(aq) + \\text{Ba}^{2+}(aq) + 2\\text{OH}^{-}(aq) \\longrightarrow 2\\text{K}^{+}(aq) + 2\\text{OH}^{-}(aq) + \\text{BaC}_2 \\text{O}_4(s)[\/latex]<br \/>\nnet: [latex]\\text{Ba}^{2+}(aq) + {\\text{C}_2 {\\text{O}_4}}^{2-}(aq) \\longrightarrow \\text{BaC}_2 \\text{O}_4(s)[\/latex]<\/p>\n<p>b) complete: [latex]\\text{Pb}^{2+}(aq) + 2{\\text{NO}_3}^{-}(aq) + 2\\text{H}^{+}(aq) + {\\text{SO}_4}^{2-}(aq) \\longrightarrow \\text{PbSO}_4(s) + 2\\text{H}^{+}(aq) + 2{\\text{NO}_3}^{-}(aq)[\/latex]<\/p>\n<p>net: [latex]\\text{Pb}^{2+}(aq) + {\\text{SO}_4}^{2-}(aq) \\longrightarrow \\text{PbSO}_4(s)[\/latex]<\/p>\n<p>c) complete:\u00a0[latex]\\text{CaCO}_3(s) + 2\\text{H}^{+}(aq) + {\\text{SO}_4}^{2-}(aq) \\longrightarrow \\text{CaSO}_4(s) + \\text{CO}_2(g) + \\text{H}_2 \\text{O}(l)[\/latex]<\/p>\n<p>net: [latex]\\text{CaCO}_3(s) + 2\\text{H}^{+}(aq) + {\\text{SO}_4}^{2-}(aq) \\longrightarrow \\text{CaSO}_4(s) + \\text{CO}_2(g) + \\text{H}_2 \\text{O}(l)[\/latex]<\/p>\n<div class=\"qandaset block\" id=\"ball-ch04_s01_qs01_ans\">\n<p>9.\u00a0<span style=\"font-size: 1em\">reactants: nitrogen and hydrogen; product: ammonia<\/span><\/p>\n<\/div>\n<div class=\"qandaset block\">\n<p>10.\u00a0reactants: magnesium hydroxide and nitric acid; products: magnesium nitrate and water<\/p>\n<p>11.\u00a0N<sub class=\"subscript\">2<\/sub><sub>(g)<\/sub> +\u00a03 H<sub class=\"subscript\">2<\/sub><sub>(g)<\/sub> [latex]\\longrightarrow[\/latex] 2 NH<sub class=\"subscript\">3<\/sub><sub>(g)<\/sub><\/p>\n<p>12.\u00a0Mg(OH)<sub class=\"subscript\">2<\/sub><sub>(aq)<\/sub> +\u00a02 HNO<sub class=\"subscript\">3<\/sub><sub>(aq)<\/sub> [latex]\\longrightarrow[\/latex] Mg(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub><sub>(aq)<\/sub> +\u00a02 H<sub class=\"subscript\">2<\/sub>O<span class=\"informalequation block\"><sub>(\u2113)<\/sub><\/span><\/p>\n<p>13.\u00a02 NaClO<sub class=\"subscript\">3<\/sub> [latex]\\longrightarrow[\/latex] 2 NaCl +\u00a03 O<sub class=\"subscript\">2<\/sub><\/p>\n<p>14.\u00a0<span style=\"font-size: 1em\">N<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\"> +\u00a02 H<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\"> [latex]\\longrightarrow[\/latex] N<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">H<\/span><sub class=\"subscript\">4<\/sub><\/p>\n<p>15.\u00a04Al +\u00a03O<sub class=\"subscript\">2<\/sub> [latex]\\longrightarrow[\/latex] 2Al<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub><\/p>\n<p>16.\u00a0<span style=\"font-size: 1em\">C<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">H<\/span><sub class=\"subscript\">4<\/sub><span style=\"font-size: 1em\"> +\u00a03 O<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\"> [latex]\\longrightarrow[\/latex] 2 CO<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\"> +\u00a02 H<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">O<\/span><\/p>\n<p>17.\u00a0N<sub class=\"subscript\">2<\/sub><sub>(g)<\/sub> +\u00a03 H<sub class=\"subscript\">2<\/sub><sub>(g)<\/sub> [latex]\\longrightarrow[\/latex] 2 NH<sub class=\"subscript\">3<\/sub><sub>(g)<\/sub><\/p>\n<p>18.\u00a0NaBr(s) [latex]\\longrightarrow[\/latex]\u00a0Na<sup class=\"superscript\">+<\/sup>(aq) +\u00a0Br<sup class=\"superscript\">\u2212<\/sup>(aq)<\/p>\n<p>19.\u00a0(NH<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">3<\/sub>PO<sub class=\"subscript\">4<\/sub>(s) [latex]\\longrightarrow[\/latex]\u00a03 NH<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">+<\/sup>(aq) +\u00a0PO<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">3\u2212<\/sup>(aq)<\/p>\n<p>20.\u00a0Fe<sup class=\"superscript\">2+<\/sup>(aq) +\u00a02 Cl<sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a02 Ag<sup class=\"superscript\">+<\/sup>(aq) +\u00a02 NO<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq) [latex]\\longrightarrow[\/latex] Fe<sup class=\"superscript\">2+<\/sup>(aq) +\u00a02 NO<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a02 AgCl(s)<\/p>\n<p>21.\u00a0K<sup class=\"superscript\">+<\/sup>(aq) +\u00a0Cl<sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a0Na<sup class=\"superscript\">+<\/sup>(aq) +\u00a0C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq) [latex]\\longrightarrow[\/latex] Na<sup class=\"superscript\">+<\/sup>(aq) +\u00a0Cl<sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a0K<sup class=\"superscript\">+<\/sup>(aq) +\u00a0C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq)<\/p>\n<p>22.\u00a02 Cl<sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a02 Ag<sup class=\"superscript\">+<\/sup>(aq) [latex]\\longrightarrow[\/latex] 2 AgCl(s)<\/p>\n<p>23.\u00a0There is no overall reaction.<\/p>\n<p>24.\u00a0In Exercise 20, Fe<sup class=\"superscript\">2+<\/sup>(aq) and NO<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq) are spectator ions; in Exercise 21, Na<sup class=\"superscript\">+<\/sup>(aq) and Cl<sup class=\"superscript\">\u2212<\/sup>(aq) are spectator ions.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<div>\n<dl id=\"fs-idp25650896\" class=\"definition\"><\/dl>\n<h2>Glossary<\/h2>\n<p><strong>balanced equation:<\/strong>\u00a0chemical equation with equal numbers of atoms for each element in the reactant and product<\/p>\n<p><strong>chemical equation:<\/strong>\u00a0symbolic representation of a chemical reaction<\/p>\n<p><strong>coefficient:<\/strong>\u00a0number placed in front of symbols or formulas in a chemical equation to indicate their relative amount<\/p>\n<p><strong>complete ionic equation:<\/strong>\u00a0chemical equation in which all dissolved ionic reactants and products, including spectator ions, are explicitly represented by formulas for their dissociated ions<\/p>\n<p><strong>molecular equation:<\/strong>\u00a0chemical equation in which all reactants and products are represented as neutral substances<\/p>\n<p><strong>net ionic equation:<\/strong>\u00a0chemical equation in which only those dissolved ionic reactants and products that undergo a chemical or physical change are represented (excludes spectator ions)<\/p>\n<p><strong>product:<\/strong>\u00a0substance formed by a chemical or physical change; shown on the right side of the arrow in a chemical equation<\/p>\n<p><strong>reactant:<\/strong>\u00a0substance undergoing a chemical or physical change; shown on the left side of the arrow in a chemical equation<\/p>\n<p><strong>spectator ion:<\/strong>\u00a0ion that does not undergo a chemical or physical change during a reaction, but its presence is required to maintain charge neutrality<\/p>\n<\/div>\n","protected":false},"author":330,"menu_order":2,"template":"","meta":{"pb_show_title":"on","pb_short_title":"6.1 Writing and Balancing Chemical Equations","pb_subtitle":"","pb_authors":[],"pb_section_license":"cc-by-nc-sa"},"chapter-type":[],"contributor":[],"license":[54],"class_list":["post-1412","chapter","type-chapter","status-publish","hentry","license-cc-by-nc-sa"],"part":2162,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/chapters\/1412","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/wp\/v2\/users\/330"}],"version-history":[{"count":25,"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/chapters\/1412\/revisions"}],"predecessor-version":[{"id":4172,"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/chapters\/1412\/revisions\/4172"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/parts\/2162"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/chapters\/1412\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/wp\/v2\/media?parent=1412"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/chapter-type?post=1412"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/wp\/v2\/contributor?post=1412"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/wp\/v2\/license?post=1412"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}