{"id":1429,"date":"2018-04-11T22:51:50","date_gmt":"2018-04-12T02:51:50","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/chapter\/4-3-reaction-stoichiometry\/"},"modified":"2018-06-22T23:20:25","modified_gmt":"2018-06-23T03:20:25","slug":"4-3-reaction-stoichiometry","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/chapter\/4-3-reaction-stoichiometry\/","title":{"raw":"7.1 Reaction Stoichiometry","rendered":"7.1 Reaction Stoichiometry"},"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>Explain the concept of stoichiometry as it pertains to chemical reactions<\/li>\r\n \t<li>Use balanced chemical equations to derive stoichiometric factors relating amounts of reactants and products<\/li>\r\n \t<li>Perform stoichiometric calculations involving mass, moles, and solution molarity<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-idp65325120\">A balanced chemical equation provides a great deal of information in a very succinct format. Chemical formulas provide the identities of the reactants and products involved in the chemical change, allowing classification of the reaction. Coefficients provide the relative numbers of these chemical species, allowing a quantitative assessment of the relationships between the amounts of substances consumed and produced by the reaction. These quantitative relationships are known as the reaction\u2019s <strong>stoichiometry<\/strong>, a term derived from the Greek words <em>stoicheion<\/em> (meaning \u201celement\u201d) and <em>metron<\/em> (meaning \u201cmeasure\u201d). In this module, the use of balanced chemical equations for various stoichiometric applications is explored.<\/p>\r\n<p id=\"fs-idp92593072\">The general approach to using stoichiometric relationships is similar in concept to the way people go about many common activities. Food preparation, for example, offers an appropriate comparison. A recipe for making eight pancakes calls for 1 cup pancake mix, $latex \\frac{3}{4}$ cup milk, and one egg. The \u201cequation\u201d representing the preparation of pancakes per this recipe is<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp116126928\" style=\"text-align: center\">$latex 1 \\;\\text{cup mix} + \\frac{3}{4} \\;\\text{cup milk} + 1 \\;\\text{egg} \\longrightarrow 8 \\;\\text{pancakes}$<\/div>\r\n<p id=\"fs-idp78338704\">If two dozen pancakes are needed for a big family breakfast, the ingredient amounts must be increased proportionally according to the amounts given in the recipe. For example, the number of eggs required to make 24 pancakes is<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp58586688\" style=\"text-align: center\">$latex 24 \\;\\rule[0.5ex]{4em}{0.1ex}\\hspace{-4em}\\text{pancakes} \\times \\frac{1 \\;\\text{egg}}{8 \\;\\rule[0.25ex]{3em}{0.1ex}\\hspace{-3em}\\text{pancakes}} = 3 \\;\\text{eggs}$<\/div>\r\n<p id=\"fs-idp98270672\">Balanced chemical equations are used in much the same fashion to determine the amount of one reactant required to react with a given amount of another reactant, or to yield a given amount of product, and so forth. The coefficients in the balanced equation are used to derive <strong>stoichiometric factors<\/strong> that permit computation of the desired quantity. To illustrate this idea, consider the production of ammonia by reaction of hydrogen and nitrogen:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp124603552\" style=\"text-align: center\">$latex \\text{N}_2(g) + 3\\text{H}_2(g) \\longrightarrow 2\\text{NH}_3(g)$<\/div>\r\n<p id=\"fs-idp124955072\">This equation shows ammonia molecules are produced from hydrogen molecules in a 2:3 ratio, and stoichiometric factors may be derived using any amount (number) unit:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp64965808\" style=\"text-align: center\">$latex \\frac{2 \\;\\text{NH}_3 \\;\\text{molecules}}{3 \\;\\text{H}_2 \\;\\text{molecules}} \\;\\text{or} \\;\\frac{2 \\;\\text{doz NH}_3 \\;\\text{molecules}}{3 \\;\\text{doz H}_2 \\;\\text{molecules}} \\;\\text{or} \\;\\frac{2 \\;\\text{mol NH}_3 \\;\\text{molecules}}{3 \\;\\text{mol H}_2 \\;\\text{molecules}}$<\/div>\r\n<p id=\"fs-idp56404080\">These stoichiometric factors can be used to compute the number of ammonia molecules produced from a given number of hydrogen molecules, or the number of hydrogen molecules required to produce a given number of ammonia molecules. Similar factors may be derived for any pair of substances in any chemical equation.<\/p>\r\n\r\n<div class=\"textbox shaded\" id=\"fs-idp48900096\">\r\n<h3>Example 1<\/h3>\r\n<p id=\"fs-idp77269456\">How many moles of I<sub>2<\/sub> are required to react with 0.429 mol of Al according to the following equation (see <a href=\"#CNX_Chem_04_03_iodine\" class=\"autogenerated-content\">Figure 1<\/a>)?<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp147980704\" style=\"text-align: center\">$latex 2\\text{Al} + 3\\text{I}_2 \\longrightarrow 2\\text{AlI}_3$<\/div>\r\n<figure id=\"CNX_Chem_04_03_iodine\"><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_03_iodine.jpg\"><img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_03_iodine-2.jpg\" alt=\"This figure shows three photos with an arrow leading from one to the next. The first photo shows a small pile of iodine and aluminum on a white surface. The second photo shows a small amount of purple smoke coming from the pile. The third photo shows a large amount of purple and gray smoke coming from the pile.\" width=\"975\" height=\"182\" \/><\/a> <strong>Figure 1.<\/strong> Aluminum and iodine react to produce aluminum iodide. The heat of the reaction vaporizes some of the solid iodine as a purple vapor. (credit: modification of work by Mark Ott)[\/caption]\r\n\r\n<\/figcaption><\/figure>\r\n&nbsp;\r\n<p id=\"fs-idm34598272\"><strong>Solution<\/strong>\r\nReferring to the balanced chemical equation, the stoichiometric factor relating the two substances of interest is $latex \\frac{3 \\;\\text{mol I}_2}{2 \\;\\text{mol Al}}$. The molar amount of iodine is derived by multiplying the provided molar amount of aluminum by this factor:<\/p>\r\n<img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_03_moleratio1_img-2.jpg\" alt=\"This figure shows two pink rectangles. The first is labeled, \u201cMoles of A l.\u201d This rectangle is followed by an arrow pointing right to a second rectangle labeled, \u201cMoles of I subscript 2.\u201d\" width=\"490\" height=\"95\" class=\"aligncenter\" \/>\r\n<div class=\"equation\" id=\"fs-idp102857120\" style=\"text-align: center\">$latex \\begin{array}{r @{{}={}} l} \\text{mol I}_2 &amp; 0.429 \\;\\rule[0.5ex]{3.25em}{0.1ex}\\hspace{-3.25em}\\text{mol Al} \\times \\frac{3 \\;\\text{mol I}_2}{2 \\;\\rule[0.25ex]{2em}{0.1ex}\\hspace{-2em}\\text{mol Al}} \\\\[1em] &amp; 0.644 \\;\\text{mol I}_2 \\end{array}$<\/div>\r\n&nbsp;\r\n<p id=\"fs-idp30742160\"><em><strong>Test Yourself<\/strong><\/em>\r\nHow many moles of Ca(OH)<sub>2<\/sub> are required to react with 1.36 mol of H<sub>3<\/sub>PO<sub>4<\/sub> to produce Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub> according to the equation $latex 3\\text{Ca(OH)}_2 + 2\\text{H}_3 \\text{PO}_4 \\longrightarrow \\text{Ca}_3 \\text{(PO}_4)_2 + 6\\text{H}_2 \\text{O}$?<\/p>\r\n&nbsp;\r\n\r\n<em><strong>Answer<\/strong><\/em>\r\n\r\n2.04 mol\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\" id=\"fs-idp9124448\">\r\n<h3>Example 2<\/h3>\r\n<p id=\"fs-idp157170976\">How many carbon dioxide molecules are produced when 0.75 mol of propane is combusted according to this equation?<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp116053696\" style=\"text-align: center\">$latex \\text{C}_3 \\text{H}_8 + 5\\text{O}_2 \\longrightarrow 3\\text{CO}_2 + 4\\text{H}_2 \\text{O}$<\/div>\r\n&nbsp;\r\n<p id=\"fs-idp121952928\"><strong>Solution<\/strong>\r\nThe approach here is the same as for <a href=\"#fs-idp48900096\" class=\"autogenerated-content\">Example 1<\/a>, though the absolute number of molecules is requested, not the number of moles of molecules. This will simply require use of the moles-to-numbers conversion factor, Avogadro\u2019s number.<\/p>\r\n<p id=\"fs-idp229756528\">The balanced equation shows that carbon dioxide is produced from propane in a 3:1 ratio:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp114307520\" style=\"text-align: center\">$latex \\frac{3 \\;\\text{mol CO}_2}{1 \\;\\text{mol C}_3 \\text{H}_8}$<\/div>\r\n<p id=\"fs-idp83056800\">Using this stoichiometric factor, the provided molar amount of propane, and Avogadro\u2019s number,<\/p>\r\n<img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_03_moleratio2_img-2.jpg\" alt=\"This figure shows two pink rectangles. The first is labeled, \u201cMoles of C subscript 3 H subscript 8.\u201d This rectangle is followed by an arrow pointing right to a second rectangle labeled, \u201cMoles of C O subscript 2.\u201d\" width=\"679\" height=\"92\" class=\"aligncenter\" \/>\r\n<div class=\"equation\" id=\"fs-idp52373104\" style=\"text-align: center\">$latex 0.75 \\;\\rule[0.5ex]{4.5em}{0.1ex}\\hspace{-4.5em}\\text{mol C}_3 \\text{H}_8 \\times \\frac{3 \\;\\rule[0.25ex]{3em}{0.1ex}\\hspace{-3em} \\text{mol CO}_2}{1 \\;\\rule[0.5ex]{3em}{0.1ex}\\hspace{-3em}\\text{mol C}_3 \\text{H}_8} \\times \\frac{6.022 \\times 10^{23} \\;\\text{CO}_2 \\;\\text{molecules}}{\\rule[0.25ex]{3em}{0.1ex}\\hspace{-3em} \\text{mol CO}_2} = 1.4 \\times 10^{24} \\text{CO}_2 \\;\\text{molecules}$<\/div>\r\n&nbsp;\r\n<p id=\"fs-idm14771264\"><em><strong>Test Yourself<\/strong><\/em>\r\nHow many NH<sub>3<\/sub> molecules are produced by the reaction of 4.0 mol of Ca(OH)<sub>2<\/sub> according to the following equation:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp166213360\" style=\"text-align: center\">$latex (\\text{NH}_4)_2 \\text{SO}_4 + \\text{Ca(OH)}_2 \\longrightarrow 2\\text{NH}_3 + \\text{CaSO}_4 + 2\\text{H}_2 \\text{O}$<\/div>\r\n<div><\/div>\r\n<div><em><strong>Answer<\/strong><\/em><\/div>\r\n<div>4.8 \u00d7 10<sup>24<\/sup> NH<sub>3<\/sub> molecules<\/div>\r\n<\/div>\r\n<p id=\"fs-idp222914448\">These examples illustrate the ease with which the amounts of substances involved in a chemical reaction of known stoichiometry may be related. Directly measuring numbers of atoms and molecules is, however, not an easy task, and the practical application of stoichiometry requires that we use the more readily measured property of mass.<\/p>\r\n\r\n<div class=\"textbox shaded\" id=\"fs-idp113495744\">\r\n<h3>Example 3<\/h3>\r\n<p id=\"fs-idp56844080\">What mass of sodium hydroxide, NaOH, would be required to produce 16 g of the antacid milk of magnesia [magnesium hydroxide, Mg(OH)<sub>2<\/sub>] by the following reaction?<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp158099664\" style=\"text-align: center\">$latex \\text{MgCl}_2(aq) + 2\\text{NaOH}(aq) \\longrightarrow \\text{Mg(OH)}_2(s) + \\text{NaCl}(aq)$<\/div>\r\n&nbsp;\r\n<p id=\"fs-idm3520480\"><strong>Solution<\/strong>\r\nThe approach used previously in <a href=\"#fs-idp48900096\" class=\"autogenerated-content\">Example 1<\/a> and <a href=\"#fs-idp9124448\" class=\"autogenerated-content\">Example 2<\/a> is likewise used here; that is, we must derive an appropriate stoichiometric factor from the balanced chemical equation and use it to relate the amounts of the two substances of interest. In this case, however, masses (not molar amounts) are provided and requested, so additional steps of the sort learned in the previous chapter are required. The calculations required are outlined in this flowchart:<\/p>\r\n<img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_03_map2_img-2.jpg\" alt=\"This figure shows four rectangles. The first is shaded yellow and is labeled, \u201cMass of M g ( O H ) subscript 2.\u201d This rectangle is followed by an arrow pointing right to a second rectangle which is shaded pink and is labeled, \u201cMoles of M g ( O H ) subscript 2.\u201d This rectangle is followed by an arrow pointing right to a third rectangle which is shaded pink and is labeled, \u201cMoles of N a O H.\u201d This rectangle is followed by an arrow pointing right to a fourth rectangle which is shaded yellow and is labeled, \u201cMass of N a O H.\u201d\" width=\"655\" height=\"291\" class=\"aligncenter\" \/>\r\n<div class=\"equation\" id=\"fs-idp122145840\" style=\"text-align: center\">$latex 16 \\;\\rule[0.5ex]{5em}{0.1ex}\\hspace{-5em}\\text{g Mg(OH)}_2 \\times \\frac{1 \\;\\rule[0.25ex]{4.75em}{0.1ex}\\hspace{-4.75em}\\text{mol Mg(OH)}_2}{58.3197 \\;\\rule[0.25ex]{3.5em}{0.1ex}\\hspace{-3.5em}\\text{g Mg(OH)}_2} \\times \\frac{2 \\;\\rule[0.25ex]{3.5em}{0.1ex}\\hspace{-3.5em}\\text{mol NaOH}}{1 \\;\\rule[0.25ex]{4.5em}{0.1ex}\\hspace{-4.5em}\\text{mol Mg(OH)}_2} \\times \\frac{39.9971 \\;\\text{g NaOH}}{1 \\;\\rule[0.25ex]{3.25em}{0.1ex}\\hspace{-3.25em}\\text{mol NaOH}} = 22 \\;\\text{g NaOH}$<\/div>\r\n&nbsp;\r\n<p id=\"fs-idp52759328\"><em><strong>Test Yourself<\/strong><\/em>\r\nWhat mass of gallium oxide, Ga<sub>2<\/sub>O<sub>3<\/sub>, can be prepared from 29.0 g of gallium metal? The equation for the reaction is $latex 4 \\text{Ga} + 3\\text{O}_2 \\longrightarrow 2\\text{Ga}_2 \\text{O}_3.$<\/p>\r\n&nbsp;\r\n\r\n<em><strong>Answer<\/strong><\/em>\r\n\r\n39.0 g\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\" id=\"fs-idp274853984\">\r\n<h3>Example 4<\/h3>\r\n<p id=\"fs-idp64260528\">What mass of oxygen gas, O<sub>2<\/sub>, from the air is consumed in the combustion of 702 g of octane, C<sub>8<\/sub>H<sub>18<\/sub>, one of the principal components of gasoline?<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp87206624\" style=\"text-align: center\">$latex 2\\text{C}_8 \\text{H}_{18} + 25\\text{O}_2 \\longrightarrow 16\\text{CO}_2 + 18\\text{H}_2 \\text{O}$<\/div>\r\n&nbsp;\r\n<p id=\"fs-idp35508016\"><strong>Solution<\/strong>\r\nThe approach required here is the same as for the <a href=\"#fs-idp113495744\" class=\"autogenerated-content\">Example 3<\/a>, differing only in that the provided and requested masses are both for reactant species.<\/p>\r\n<img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_03_map3_img-2.jpg\" alt=\"This figure shows four rectangles. The first is shaded yellow and is labeled, \u201cMass of C subscript 8 H subscript 18.\u201d This rectangle is followed by an arrow pointing right to a second rectangle which is shaded pink and is labeled, \u201cMoles of C subscript 8 H subscript 18.\u201d This rectangle is followed by an arrow pointing right to a third rectangle which is shaded pink and is labeled, \u201cMoles of O subscript 2.\u201d This rectangle is followed by an arrow pointing right to a fourth rectangle which is shaded yellow and is labeled, \u201cMass of O subscript 2.\u201d\" width=\"650\" height=\"289\" class=\"aligncenter\" \/>\r\n<div class=\"equation\" id=\"fs-idm56417376\" style=\"text-align: center\">$latex 702 \\;\\rule[0.5ex]{3.5em}{0.1ex}\\hspace{-3.5em}\\text{g C}_8 \\text{H}_{18} \\times \\frac{1 \\;\\rule[0.25ex]{3.5em}{0.1ex}\\hspace{-3.5em}\\text{mol C}_8 \\text{H}_{18}}{114.231 \\;\\rule[0.25ex]{2.75em}{0.1ex}\\hspace{-2.75em}\\text{g C}_8 \\text{H}_{18}} \\times \\frac{25 \\;\\rule[0.25ex]{2.5em}{0.1ex}\\hspace{-2.5em}\\text{mol O}_2}{2 \\;\\rule[0.25ex]{3.5em}{0.1ex}\\hspace{-3.5em}\\text{mol C}_8 \\text{H}_{18}} \\times \\frac{31.9988 \\;\\text{g O}_2}{\\rule[0.25ex]{2.5em}{0.1ex}\\hspace{-2.5em}\\text{mol O}_2} = 2.46 \\times 10^3 \\;\\text{g O}_2$<\/div>\r\n&nbsp;\r\n<p id=\"fs-idp24638432\"><em><strong>Test Yourself<\/strong><\/em>\r\nWhat mass of CO is required to react with 25.13 g of Fe<sub>2<\/sub>O<sub>3<\/sub> according to the equation<\/p>\r\n<p style=\"text-align: center\">$latex \\text{Fe}_2 \\text{O}_3 + 3\\text{CO} \\longrightarrow 2\\text{Fe} + 3\\text{CO}_2$<\/p>\r\n&nbsp;\r\n\r\n<em><strong>Answer<\/strong><\/em>\r\n\r\n13.22 g\r\n\r\n<\/div>\r\n<p id=\"fs-idp90689472\">These examples illustrate just a few instances of reaction stoichiometry calculations. Numerous variations on the beginning and ending computational steps are possible depending upon what particular quantities are provided and sought (volumes, solution concentrations, and so forth). Regardless of the details, all these calculations share a common essential component: the use of stoichiometric factors derived from balanced chemical equations. <a href=\"#CNX_Chem_04_03_flowchart\" class=\"autogenerated-content\">Figure 2<\/a> provides a general outline of the various computational steps associated with many reaction stoichiometry calculations.<\/p>\r\n\r\n<figure id=\"CNX_Chem_04_03_flowchart\"><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_03_flowchart.jpg\"><img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_03_flowchart-2.jpg\" alt=\"This flowchart shows 10 rectangles connected by double headed arrows. To the upper left, a rectangle is shaded lavender and is labeled, \u201cVolume of pure substance A.\u201d This rectangle is followed by a horizontal double headed arrow labeled, \u201cDensity.\u201d It connects to a second rectangle which is shaded yellow and is labeled, \u201cMass of A.\u201d This rectangle is followed by a double headed arrow which is labeled, \u201cMolar Mass,\u201d that connects to a third rectangle which is shaded pink and is labeled, \u201cMoles of A.\u201d To the left of this rectangle is a horizontal double headed arrow labeled, \u201cMolarity,\u201d which connects to a lavender rectangle which is labeled, \u201cVolume of solution A.\u201d The pink, \u201cMoles of A,\u201d rectangle is also connected with a double headed arrow below and to the left. This arrow is labeled \u201cAvogadro\u2019s number.\u201d It connects to a green shaded rectangle that is labeled, \u201cNumber of particles of A.\u201d To the right of the pink \u201cMoles of A,\u201d rectangle is a horizontal double headed arrow which is labeled, \u201cStoichiometric factor.\u201d It connects to a second pink rectangle which is labeled, \u201cMoles of B.\u201d A double headed arrow which is labeled, \u201cMolar mass,\u201d extends from the top of this rectangle above and to the right to a yellow shaded rectangle labeled, \u201cMass of B.\u201d A horizontal double headed arrow which is labeled, \u201cDensity\u201d links to a lavender rectangle labeled, \u201cVolume of substance B,\u201d to the right. A horizontal double headed arrow labeled, \u201cMolarity,\u201d extends right to the of the pink \u201cMoles of B\u201d rectangle. This arrow connects to a lavender rectangle that is labeled, \u201cVolume of substance B.\u201d Another double headed arrow extends below and to the right of the pink \u201cMoles of B\u201d rectangle. This arrow is labeled \u201cAvogadro\u2019s number,\u201d and it extends to a green rectangle which is labeled, \u201cNumber of particles of B.\u201d\" width=\"1300\" height=\"796\" \/><\/a> <strong>Figure 2.<\/strong> The flowchart depicts the various computational steps involved in most reaction stoichiometry calculations.[\/caption]\r\n\r\n<\/figcaption><\/figure>\r\n<div id=\"fs-idp84121360\" class=\"textbox shaded\">\r\n<h3 class=\"title\">Airbags<\/h3>\r\n<p id=\"fs-idp124943600\">Airbags (<a href=\"#CNX_Chem_04_03_airbag\" class=\"autogenerated-content\">Figure 3<\/a>) are a safety feature provided in most automobiles since the 1990s. The effective operation of an airbag requires that it be rapidly inflated with an appropriate amount (volume) of gas when the vehicle is involved in a collision. This requirement is satisfied in many automotive airbag systems through use of explosive chemical reactions, one common choice being the decomposition of sodium azide, NaN<sub>3<\/sub>. When sensors in the vehicle detect a collision, an electrical current is passed through a carefully measured amount of NaN<sub>3<\/sub> to initiate its decomposition:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp113504400\" style=\"text-align: center\">$latex 2 \\text{NaN}_3(s) \\longrightarrow 3\\text{N}_2(g) + 2\\text{Na}(s)$<\/div>\r\n<figure id=\"CNX_Chem_04_03_airbag\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"388\"]<a href=\"https:\/\/opentextbc.ca\/chemistry\/wp-content\/uploads\/sites\/150\/2016\/05\/CNX_Chem_04_03_airbag.jpg\"><img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_03_airbag-2.jpg\" alt=\"This photograph shows the inside of an automobile from the driver\u2019s side area. The image shows inflated airbags positioned just in front of the driver\u2019s and passenger\u2019s seats and along the length of the passenger side over the windows. A large, round airbag covers the steering wheel.\" width=\"388\" height=\"291\" class=\"\" \/><\/a> <strong>Figure 3.<\/strong> Airbags deploy upon impact to minimize serious injuries to passengers. (credit: Jon Seidman)[\/caption]\r\n\r\n&nbsp;\r\n<p id=\"fs-idp9645888\">This reaction is very rapid, generating gaseous nitrogen that can deploy and fully inflate a typical airbag in a fraction of a second (~0.03\u20130.1 s).<\/p>\r\nAmong many engineering considerations, the amount of sodium azide used must be appropriate for generating enough nitrogen gas to fully inflate the air bag and ensure its proper function.\r\n\r\nFor example, a small mass (~100 g) of NaN<sub>3<\/sub> will generate approximately 50 L of N<sub>2<\/sub>.<\/figure>\r\n<\/div>\r\n<h2><span style=\"font-family: Roboto, Helvetica, Arial, sans-serif\">More Worked Out Problems<\/span><\/h2>\r\n<section id=\"fs-idp102563712\" class=\"summary\">\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Example 5<\/h3>\r\n<p id=\"ball-ch05_s01_p19\" class=\"para\">How many molecules of SO<sub class=\"subscript\">3<\/sub> are needed to react with 144 molecules of Fe<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub> given this balanced chemical equation?<\/p>\r\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">Fe<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub>(s) +\u00a03SO<sub class=\"subscript\">3<\/sub>(g) <span style=\"font-size: 1em\">$latex \\longrightarrow$<\/span>\u00a0Fe<sub class=\"subscript\">2<\/sub>(SO<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">3<\/sub><\/span><\/span><\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\r\n<p id=\"ball-ch05_s01_p20\" class=\"para\">We use the balanced chemical equation to construct a conversion factor between Fe<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub> and SO<sub class=\"subscript\">3<\/sub>. The number of molecules of Fe<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub> goes on the bottom of our conversion factor so it cancels with our given amount, and the molecules of SO<sub class=\"subscript\">3<\/sub> go on the top. Thus, the appropriate conversion factor is<\/p>\r\n<a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/ss2.png\"><img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/ss2-1.png\" alt=\"ss2\" class=\"size-full wp-image-3631 aligncenter\" width=\"132\" height=\"59\" \/><\/a>\r\n<p id=\"ball-ch05_s01_p21\" class=\"para\">Starting with our given amount and applying the conversion factor, the result is<\/p>\r\n<p class=\"para\"><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/ss1.png\"><img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/ss1-1.png\" alt=\"ss1\" class=\"aligncenter wp-image-3630 size-full\" width=\"489\" height=\"79\" \/><\/a><\/p>\r\n<p id=\"ball-ch05_s01_p22\" class=\"para\">We need 432 molecules of SO<sub class=\"subscript\">3<\/sub> to react with 144 molecules of Fe<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub>.<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\r\n<p id=\"ball-ch05_s01_p23\" class=\"para\">How many molecules of H<sub class=\"subscript\">2<\/sub> are needed to react with 29 molecules of N<sub class=\"subscript\">2<\/sub> to make ammonia if the balanced chemical equation is N<sub class=\"subscript\">2<\/sub> +\u00a03H<sub class=\"subscript\">2<\/sub>\u00a0<span style=\"font-size: 1em\">$latex \\longrightarrow$<\/span> 2NH<sub class=\"subscript\">3<\/sub>?<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\r\n<p id=\"ball-ch05_s01_p24\" class=\"para\">87 molecules<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Example 6<\/h3>\r\n<p id=\"ball-ch05_s01_p27\" class=\"para\">How many molecules of NH<sub class=\"subscript\">3<\/sub> can you make if you have 228 atoms of H<sub class=\"subscript\">2<\/sub>?<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\r\n<p id=\"ball-ch05_s01_p28\" class=\"para\">From the formula, we know that one molecule of NH<sub class=\"subscript\">3<\/sub> has three H atoms. Use that fact as a conversion factor:<\/p>\r\n<a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/Screen-Shot-2014-07-17-at-10.49.40-AM1.png\"><img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/Screen-Shot-2014-07-17-at-10.49.40-AM1-1.png\" alt=\"Screen-Shot-2014-07-17-at-10.49.40-AM\" class=\"wp-image-3627 size-full aligncenter\" width=\"412\" height=\"79\" \/><\/a>\r\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\r\n<p id=\"ball-ch05_s01_p29\" class=\"para\">How many molecules of Fe<sub class=\"subscript\">2<\/sub>(SO<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">3<\/sub> can you make from 777 atoms of S?<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\r\n<p id=\"ball-ch05_s01_p30\" class=\"para\">259 molecules<\/p>\r\n\r\n<\/div>\r\n<div class=\"section\" id=\"ball-ch05_s04\" lang=\"en\">\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Example 7<\/h3>\r\n<p id=\"ball-ch05_s04_p09\" class=\"para\">How many moles of HCl will be produced when 249 g of AlCl<sub class=\"subscript\">3<\/sub> are reacted according to this chemical equation? \u00a0The molar mass of AlCl<sub class=\"subscript\">3<\/sub> is 133.34 g\/mol.<\/p>\r\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">2 AlCl<sub class=\"subscript\">3<\/sub> +\u00a03 H<sub class=\"subscript\">2<\/sub>O(\u2113) <span style=\"font-size: 1em\">$latex \\longrightarrow$<\/span>\u00a0Al<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub> +\u00a06 HCl(g)<\/span><\/span><\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\r\n<p id=\"ball-ch05_s04_p10\" class=\"para\">We will do this in two steps: convert the mass of AlCl<sub class=\"subscript\">3<\/sub> to moles and then use the balanced chemical equation to find the number of moles of HCl formed. The molar mass of AlCl<sub class=\"subscript\">3<\/sub> is 133.34 g\/mol, which we have to invert to get the appropriate conversion factor:<\/p>\r\n<p style=\"text-align: center\">$latex 249 \\;\\rule[0.5ex]{0.5em}{0.1ex}\\hspace{-0.5em}\\text{g AlCl}_3 \\times \\frac{1 \\;\\text{mol AlCl}_3}{133.34\\;\\rule[0.5ex]{0.5em}{0.1ex}\\hspace{-0.5em}\\text{g AlCl}_3} = \\underline{1.86}74 \\;\\text{mol of aluminium chloride with 3 sig figs} $<\/p>\r\n&nbsp;\r\n<p id=\"ball-ch05_s04_p11\" class=\"para\">Now we can use this quantity to determine the number of moles of HCl that will form. From the balanced chemical equation, we construct a conversion factor between the number of moles of AlCl<sub class=\"subscript\">3<\/sub> and the number of moles of HCl:<\/p>\r\n<p style=\"text-align: center\">$latex \\frac{6 \\; \\text{mol HCl}}{2 \\; \\text{mol AlCl}_3}$<\/p>\r\n<p id=\"ball-ch05_s04_p12\" class=\"para\">Applying this conversion factor to the quantity of AlCl<sub class=\"subscript\">3<\/sub>, we get<\/p>\r\n<p style=\"text-align: center\">$latex \\underline{1.86}74 \\;\\rule[0.5ex]{1.25em}{0.1ex}\\hspace{-1.25em}\\text{mol AlCl}_3 \\times \\frac{6 \\;\\text{mol HCl}}{2\\;\\rule[0.5ex]{1.25em}{0.1ex}\\hspace{-1.25em}\\text{mol AlCl}_3} =5.60 \\;\\text{mol HCl} $<\/p>\r\n<p id=\"ball-ch05_s04_p13\" class=\"para\">Alternatively, we could have done this in one line:<\/p>\r\n<p style=\"text-align: center\">$latex 249 \\;\\rule[0.5ex]{0.5em}{0.1ex}\\hspace{-0.5em}\\text{g AlCl}_3 \\times \\frac{1 \\;\\rule[0.5ex]{1.25em}{0.1ex}\\hspace{-1.25em}\\text{mol AlCl}_3}{133.34\\;\\rule[0.5ex]{0.5em}{0.1ex}\\hspace{-0.5em}\\text{g AlCl}_3} \\times \\frac{6 \\;\\text{mol HCl}}{2\\;\\rule[0.5ex]{1.25em}{0.1ex}\\hspace{-1.25em}\\text{mol AlCl}_3}=5.60 \\;\\text{mol HCl} $<\/p>\r\n<p id=\"ball-ch05_s04_p14\" class=\"para\">The last digit in our final answer is slightly different because of rounding differences, but the answer is essentially the same.<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\r\n<p id=\"ball-ch05_s04_p15\" class=\"para\">How many moles of Al<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub> will be produced when 23.9 g of H<sub class=\"subscript\">2<\/sub>O are reacted according to this chemical equation?<\/p>\r\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">2 AlCl<sub class=\"subscript\">3<\/sub> +\u00a03 H<sub class=\"subscript\">2<\/sub>O(\u2113) <span style=\"font-size: 1em\">$latex \\longrightarrow$<\/span>\u00a0Al<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub> +\u00a06 HCl(g)<\/span><\/span><\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\r\n<p id=\"ball-ch05_s04_p16\" class=\"para\">0.442 mol<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Example 8<\/h3>\r\n<p id=\"ball-ch05_s04_p18\" class=\"para\">How many grams of NH<sub class=\"subscript\">3<\/sub> will be produced when 33.9 mol of H<sub class=\"subscript\">2<\/sub> are reacted according to this chemical equation? \u00a0Use 17.03 g\/mol as the molar mass of NH<sub class=\"subscript\">3<\/sub>.<\/p>\r\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">N<sub class=\"subscript\">2<\/sub>(g) +\u00a03 H<sub class=\"subscript\">2<\/sub>(g) <span style=\"font-size: 1em\">$latex \\longrightarrow$<\/span>\u00a02 NH<sub class=\"subscript\">3<\/sub>(g)<\/span><\/span><\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\r\n<p id=\"ball-ch05_s04_p19\" class=\"para\">The conversions are the same, but they are applied in a different order. Start by using the balanced chemical equation to convert to moles of another substance and then use its molar mass to determine the mass of the final substance. In two steps, we have<\/p>\r\n<a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/339molh2.png\"><img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/339molh2-1.png\" alt=\"339molh2\" width=\"426\" height=\"93\" class=\"aligncenter wp-image-3738\" \/><\/a>Now, using the molar mass of NH<sub class=\"subscript\">3<\/sub>, which is 17.03 g\/mol, we get\r\n\r\n<a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/226molnh3.png\"><img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/226molnh3-1.png\" alt=\"226molnh3\" width=\"430\" height=\"99\" class=\"aligncenter wp-image-3739\" \/><\/a>\r\n\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\r\n<p id=\"ball-ch05_s04_p21\" class=\"para\">How many grams of N<sub class=\"subscript\">2<\/sub> are needed to produce 2.17 mol of NH<sub class=\"subscript\">3<\/sub> when reacted according to this chemical equation?<\/p>\r\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">N<sub class=\"subscript\">2<\/sub>(g) +\u00a03 H<sub class=\"subscript\">2<\/sub>(g) <span style=\"font-size: 1em\">$latex \\longrightarrow$<\/span>\u00a02 NH<sub class=\"subscript\">3<\/sub>(g)<\/span><\/span><\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\r\n<p id=\"ball-ch05_s04_p22\" class=\"para\">30.4 g (Note: here we go from a product to a reactant, showing that mole-mass problems can begin and end with any substance in the chemical equation.)<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Example 9<\/h3>\r\n<p id=\"ball-ch05_s04_p30\" class=\"para\">What mass of Mg will be produced when 86.4 g of K are reacted? \u00a0Use 39.09 g\/mol as the molar mass of potassium and 24.31 g\/mol as the molar mass of magnesium.<\/p>\r\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">MgCl<sub class=\"subscript\">2<\/sub>(s) +\u00a02 K(s) <span style=\"font-size: 1em\">$latex \\longrightarrow$<\/span>\u00a0Mg(s) +\u00a02 KCl(s)<\/span><\/span><\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\r\n<p id=\"ball-ch05_s04_p31\" class=\"para\">We will simply follow the steps<\/p>\r\n<span class=\"informalequation\"><span class=\"mathphrase\">mass K <span style=\"font-size: 1em\">$latex \\longrightarrow$<\/span>\u00a0mol K <span style=\"font-size: 1em\">$latex \\longrightarrow$<\/span>\u00a0mol Mg <span style=\"font-size: 1em\">$latex \\longrightarrow$<\/span>\u00a0mass Mg<\/span><\/span>\r\n<p id=\"ball-ch05_s04_p32\" class=\"para\">In addition to the balanced chemical equation, we need the molar masses of K (39.09 g\/mol) and Mg (24.31 g\/mol). In one line,<\/p>\r\n<a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/864gk.png\"><img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/864gk-1.png\" alt=\"864gk\" width=\"600\" height=\"89\" class=\"aligncenter wp-image-3740\" \/><\/a>\r\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\r\n<p id=\"ball-ch05_s04_p33\" class=\"para\">What mass of H<sub class=\"subscript\">2<\/sub> will be produced when 122 g of Zn are reacted?<\/p>\r\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">Zn(s) +\u00a02 HCl(aq) <span style=\"font-size: 1em\">$latex \\longrightarrow$<\/span>\u00a0ZnCl<sub class=\"subscript\">2<\/sub>(aq) +\u00a0H<sub class=\"subscript\">2<\/sub>(g)<\/span><\/span><\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\r\n<p id=\"ball-ch05_s04_p34\" class=\"para\">3.77 g<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Example 10<\/h3>\r\n<p id=\"ball-ch05_s03_p07\" class=\"para\">Interpret this balanced chemical equation in terms of moles.<\/p>\r\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">P<sub class=\"subscript\">4<\/sub> +\u00a05 O<sub class=\"subscript\">2<\/sub>\u00a0<span style=\"font-size: 1em\">$latex \\longrightarrow$<\/span> P<sub class=\"subscript\">4<\/sub>O<sub class=\"subscript\">10<\/sub><\/span><\/span><\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\r\n<p id=\"ball-ch05_s03_p08\" class=\"para\">The coefficients represent the number of moles that react, not just molecules. We would speak of this equation as \u201cone mole of molecular phosphorus reacts with five moles of elemental oxygen to make one mole of tetraphosphorus decoxide.\u201d<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\r\n<p id=\"ball-ch05_s03_p09\" class=\"para\">Interpret this balanced chemical equation in terms of moles.<\/p>\r\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">N<sub class=\"subscript\">2<\/sub> +\u00a03 H<sub class=\"subscript\">2<\/sub>\u00a0<span style=\"font-size: 1em\">$latex \\longrightarrow$<\/span> 2 NH<sub class=\"subscript\">3<\/sub><\/span><\/span><\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\r\n<p id=\"ball-ch05_s03_p10\" class=\"para\">One mole of elemental nitrogen reacts with three moles of elemental hydrogen to produce two moles of ammonia.<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Example 11<\/h3>\r\n<p id=\"ball-ch05_s03_p15\" class=\"para\">For the balanced chemical equation<\/p>\r\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">2 C<sub class=\"subscript\">4<\/sub>H<sub class=\"subscript\">10<\/sub>(g) +\u00a013 O<sub class=\"subscript\">2<\/sub>\u00a0<span style=\"font-size: 1em\">$latex \\longrightarrow$<\/span> 8 CO<sub class=\"subscript\">2<\/sub>(g) +\u00a010 H<sub class=\"subscript\">2<\/sub>O(\u2113)<\/span><\/span><\/p>\r\n<p id=\"ball-ch05_s03_p16\" class=\"para\">if 154 mol of O<sub class=\"subscript\">2<\/sub> are reacted, how many moles of CO<sub class=\"subscript\">2<\/sub> are produced?<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\r\n<p id=\"ball-ch05_s03_p17\" class=\"para\">We are relating an amount of oxygen to an amount of carbon dioxide, so we need the equivalence between these two substances. According to the balanced chemical equation, the equivalence is\u00a0<span class=\"informalequation\"><span class=\"mathphrase\">13 mol O<sub class=\"subscript\">2<\/sub>\u00a0to 8 mol CO<sub class=\"subscript\">2<\/sub><\/span><\/span><\/p>\r\n<p id=\"ball-ch05_s03_p18\" class=\"para\">We can use this equivalence to construct the proper conversion factor. We start with what we are given and apply the conversion factor:<\/p>\r\n<a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/ss8.png\"><img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/ss8-1.png\" alt=\"ss8\" class=\"wp-image-3642 size-full aligncenter\" width=\"358\" height=\"75\" \/><\/a>\r\n<p id=\"ball-ch05_s03_p19\" class=\"para\">The mol O<sub class=\"subscript\">2<\/sub> unit is in the denominator of the conversion factor so it cancels. Both the 8 and the 13 are exact numbers, so they don\u2019t contribute to the number of significant figures in the final answer.<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\r\n<p id=\"ball-ch05_s03_p20\" class=\"para\">Using the above equation, how many moles of H<sub class=\"subscript\">2<\/sub>O are produced when 154 mol of O<sub class=\"subscript\">2<\/sub> react?<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\r\n<p id=\"ball-ch05_s03_p21\" class=\"para\">118 mol<\/p>\r\n\r\n<\/div>\r\n<h2>Key Concepts and Summary<\/h2>\r\n<p id=\"fs-idm3728016\">A balanced chemical equation may be used to describe a reaction\u2019s stoichiometry (the relationships between amounts of reactants and products). Coefficients from the equation are used to derive stoichiometric factors that subsequently may be used for computations relating reactant and product masses, molar amounts, and other quantitative properties.<\/p>\r\n\r\n<\/section><section id=\"fs-idm1445952\" class=\"exercises\">\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Exercises<\/h3>\r\n1. Determine the number of moles and the mass requested for each of the following reactions: (Hint:\u00a0Write the balanced equation for each before attempting calculations.)\r\n<p id=\"fs-idp84195488\">a) The number of moles and the mass of chlorine, Cl<sub>2<\/sub>, required to react with 10.0 g of sodium metal, Na, to produce sodium chloride, NaCl.<\/p>\r\n<p id=\"fs-idp186895920\">b) The number of moles and the mass of oxygen formed by the decomposition of 1.252 g of mercury(II) oxide.<\/p>\r\n<p id=\"fs-idp92843168\">c) The number of moles and the mass of sodium nitrate, NaNO<sub>3<\/sub>, required to produce 128 g of oxygen. (NaNO<sub>2<\/sub> is the other product.)<\/p>\r\n<p id=\"fs-idp8293632\">d) The number of moles and the mass of carbon dioxide formed by the combustion of 20.0 kg of carbon in an excess of oxygen.<\/p>\r\n<p id=\"fs-idp56955104\">e) The number of moles and the mass of copper(II) carbonate needed to produce 1.500 kg of copper(II) oxide. (CO<sub>2<\/sub> is the other product.)<\/p>\r\n<p id=\"fs-idp241908400\">f)\r\n<img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_03_etheneBr_img-2.jpg\" alt=\"This figure includes two structural formulas. It reads, \u201cThe number of moles and the mass of,\u201d which is followed by a structure with two C atoms bonded with a single horizontal at the center. Both C atoms have H atoms bonded above and below. The C atom to the left has a B r atom bonded to its left. The C atom to the right has a B r atom bonded to its right. Following this structure, the figure reads, \u201cformed by the reaction of 12.85 g of,\u201d which is followed by a structure with two C atoms connected with a horizontal double bond. The C atom to the left has H atoms bonded above and to the left and below and to the left. The C atom to the right has H atoms bonded above and to the right and below and to the right. The figure ends with, \u201cwith an excess of B r subscript 2.\u201d\" \/><\/p>\r\n2.\u00a0Determine the number of moles and the mass requested for each of the following reactions: (Hint:\u00a0Write the balanced equation for each before attempting calculations.)\r\n<p id=\"fs-idp124413456\">a) The number of moles and the mass of Mg required to react with 5.00 g of HCl and produce MgCl<sub>2<\/sub> and H<sub>2<\/sub>.<\/p>\r\n<p id=\"fs-idp27637888\">b) The number of moles and the mass of oxygen formed by the decomposition of 1.252 g of silver(I) oxide.<\/p>\r\n<p id=\"fs-idp38709616\">c) The number of moles and the mass of magnesium carbonate, MgCO<sub>3<\/sub>, required to produce 283 g of carbon dioxide. (MgO is the other product.)<\/p>\r\n<p id=\"fs-idp56997136\">d) The number of moles and the mass of water formed by the combustion of 20.0 kg of acetylene, C<sub>2<\/sub>H<sub>2<\/sub>, in an excess of oxygen.<\/p>\r\n<p id=\"fs-idp149673136\">e) The number of moles and the mass of barium peroxide, BaO<sub>2<\/sub>, needed to produce 2.500 kg of barium oxide, BaO (O<sub>2<\/sub> is the other product.)<\/p>\r\n<p id=\"fs-idp83999440\">f)\r\n<img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_03_ethene_img-2.jpg\" alt=\"This figure includes two structural formulas. It reads, \u201cThe number of moles and the mass of,\u201d which is followed by a structure with two C atoms connected with a horizontal double bond at the center. The C atom to the left has H atoms bonded above and to the left and below and to the left. The C atom to the right has H atoms bonded above and to the right and below and to the right. Following this structure, the figure reads, \u201crequired to react with H subscript 2 O to produce 9.55 g of,\u201d which is followed by a structure with two C atoms connected with a horizontal single bond. The C atom to the left has H atoms bonded above, to the left, and below. The C atom to the right has H atoms bonded above and below. To the right, an O atom forms a single bond with the C atom. A single H atom is bonded to the right side of the O atom.\" \/><\/p>\r\n3. Gallium chloride is formed by the reaction of 2.6 L of a 1.44 M solution of HCl according to the following equation: $latex 2\\text{Ga} + 6\\text{HCl} \\longrightarrow 2\\text{GaCl}_3 + 3\\text{H}_2$.\r\n<p id=\"fs-idp7152848\">a) Outline the steps necessary to determine the number of moles and mass of gallium chloride.<\/p>\r\n<p id=\"fs-idp58298672\">b) Perform the calculations outlined.<\/p>\r\n4. Silver is often extracted from ores such as K[Ag(CN)<sub>2<\/sub>] and then recovered by the reaction\r\n$latex 2 \\text{K} [\\text{Ag(CN)}_2](aq) + \\text{Zn}(s) \\longrightarrow 2\\text{Ag}(s) + \\text{Zn(CN)}_2(aq) + 2\\text{KCN}(aq)$\r\n<p id=\"fs-idp124132784\">a) How many molecules of Zn(CN)<sub>2<\/sub> are produced by the reaction of 35.27 g of K[Ag(CN)<sub>2<\/sub>]?<\/p>\r\n<p id=\"fs-idp25006640\">b) What mass of Zn(CN)<sub>2<\/sub> is produced?<\/p>\r\n5. Carborundum is silicon carbide, SiC, a very hard material used as an abrasive on sandpaper and in other applications. It is prepared by the reaction of pure sand, SiO<sub>2<\/sub>, with carbon at high temperature. Carbon monoxide, CO, is the other product of this reaction. Write the balanced equation for the reaction, and calculate how much SiO<sub>2<\/sub> is required to produce 3.00 kg of SiC.\r\n\r\n6. Urea, CO(NH<sub>2<\/sub>)<sub>2<\/sub>, is manufactured on a large scale for use in producing urea-formaldehyde plastics and as a fertilizer. What is the maximum mass of urea that can be manufactured from the CO<sub>2<\/sub> produced by combustion of 1.00\u00d7103kg1.00\u00d7103kg of carbon followed by the reaction?\r\n$latex \\text{CO}_2(g) + 2\\text{NH}_3(g) \\longrightarrow {\\text{CO(NH}_2})_2(s) + \\text{H}_2 \\text{O}(l)$\r\n\r\n7. A compact car gets 37.5 miles per gallon on the highway. If gasoline contains 84.2% carbon by mass and has a density of 0.8205 g\/mL, determine the mass of carbon dioxide produced during a 500-mile trip (3.785 liters per gallon).\r\n\r\n8. What volume of a 0.2089 <em>M<\/em> KI solution contains enough KI to react exactly with the Cu(NO<sub>3<\/sub>)<sub>2<\/sub> in 43.88 mL of a 0.3842 <em>M<\/em> solution of Cu(NO<sub>3<\/sub>)<sub>2<\/sub>?$latex 2 \\text{Cu(NO}_3)_2 + 4\\text{KI} \\longrightarrow 2\\text{CuI} + \\text{I}_2 + 4{\\text{KNO}_3}$\r\n\r\n9. The toxic pigment called white lead, Pb<sub>3<\/sub>(OH)<sub>2<\/sub>(CO<sub>3<\/sub>)<sub>2<\/sub>, has been replaced in white paints by rutile, TiO<sub>2<\/sub>. How much rutile (g) can be prepared from 379 g of an ore that contains 88.3% ilmenite (FeTiO<sub>3<\/sub>) by mass?$latex 2\\text{FeTiO}_3 + 4\\text{HCl} + \\text{Cl}_2 \\longrightarrow 2\\text{FeCl}_3 + 2\\text{TiO}_2 + 2\\text{H}_2 \\text{O}$\r\n\r\n<span style=\"font-size: 1em\">10. Think back to the pound cake recipe. What possible conversion factors can you construct relating the components of the recipe?<\/span>\r\n\r\n<span style=\"font-size: 1em\">11. What are all the conversion factors that can be constructed from the balanced chemical reaction 2H<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">(g) +\u00a0O<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">(g) $latex \\longrightarrow$\u00a02H<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">O(\u2113)?<\/span>\r\n\r\n<span style=\"font-size: 1em\">12. Given the chemical equation<\/span>\r\n<div class=\"qandaset block\" id=\"ball-ch05_s01_qs01\">\r\n<div class=\"question\">\r\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">Na(s) +\u00a0H<sub class=\"subscript\">2<\/sub>O(\u2113) <span style=\"font-size: 1em\">$latex \\longrightarrow$<\/span>\u00a0NaOH(aq) +\u00a0H<sub class=\"subscript\">2<\/sub>(g)<\/span><\/span><\/p>\r\n\r\n<\/div>\r\na) \u00a0Balance the equation.\r\n\r\nb) \u00a0How many molecules of H<sub class=\"subscript\">2<\/sub> are produced when 332 atoms of Na react?\r\n<div class=\"question\"><span style=\"font-size: 1em\">13. For the balanced chemical equation<\/span><\/div>\r\n<div class=\"question\">\r\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">6 H<sup class=\"superscript\">+<\/sup>(aq) +\u00a02 MnO<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a05 H<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">2<\/sub>(\u2113) <span style=\"font-size: 1em\">$latex \\longrightarrow$<\/span>\u00a02 Mn<sup class=\"superscript\">2+<\/sup>(aq) +\u00a05 O<sub class=\"subscript\">2<\/sub>(g) +\u00a08 H<sub class=\"subscript\">2<\/sub>O(\u2113)<\/span><\/span><\/p>\r\n<p id=\"ball-ch05_s01_qs01_p12\" class=\"para\">how many molecules of H<sub class=\"subscript\">2<\/sub>O are produced when 75 molecules of H<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">2<\/sub> react?<\/p>\r\n\r\n<\/div>\r\n<div class=\"question\"><span style=\"font-size: 1em\">14. \u00a0Given the balanced chemical equation<\/span><\/div>\r\n<div class=\"question\">\r\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">Fe<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub>(s) +\u00a03SO<sub class=\"subscript\">3<\/sub>(g) <span style=\"font-size: 1em\">$latex \\longrightarrow$<\/span>\u00a0Fe<sub class=\"subscript\">2<\/sub>(SO<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">3<\/sub><\/span><\/span><\/p>\r\n<p id=\"ball-ch05_s01_qs01_p18\" class=\"para\">how many molecules of Fe<sub class=\"subscript\">2<\/sub>(SO<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">3<\/sub> are produced if 321 atoms of S are reacted?<\/p>\r\n\r\n<\/div>\r\n<div class=\"question\"><span style=\"font-size: 1em\">15. \u00a0For the balanced chemical equation<\/span><\/div>\r\n<div class=\"question\">\r\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">Fe<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub>(s) +\u00a03 SO<sub class=\"subscript\">3<\/sub>(g) <span style=\"font-size: 1em\">$latex \\longrightarrow$<\/span>\u00a0Fe<sub class=\"subscript\">2<\/sub>(SO<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">3<\/sub><\/span><\/span><\/p>\r\n<p id=\"ball-ch05_s01_qs01_p24\" class=\"para\">suppose we need to make 145,000 molecules of Fe<sub class=\"subscript\">2<\/sub>(SO<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">3<\/sub>. How many molecules of SO<sub class=\"subscript\">3<\/sub> do we need?<\/p>\r\n\r\n<\/div>\r\n<div class=\"question\"><span style=\"font-size: 1em\">16. \u00a0Construct the three independent conversion factors possible for these two reactions:<\/span><\/div>\r\n<div class=\"question\">\r\n<p style=\"text-align: center\">2 H<sub class=\"subscript\">2<\/sub> +\u00a0O<sub class=\"subscript\">2<\/sub>\u00a0<span style=\"font-size: 1em\">$latex \\longrightarrow$<\/span> 2 H<sub class=\"subscript\">2<\/sub>O<\/p>\r\n<p style=\"text-align: center\">H<sub class=\"subscript\">2<\/sub> +\u00a0O<sub class=\"subscript\">2<\/sub>\u00a0<span style=\"font-size: 1em\">$latex \\longrightarrow$<\/span> H<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">2<\/sub><\/p>\r\n<p id=\"ball-ch05_s01_qs01_p30\" class=\"para\">Why are the ratios between H<sub class=\"subscript\">2<\/sub> and O<sub class=\"subscript\">2<\/sub> different?<\/p>\r\n<p id=\"ball-ch05_s01_qs01_p31\" class=\"para\">The conversion factors are different because the stoichiometries of the balanced chemical reactions are different.<\/p>\r\n<p class=\"para\"><span style=\"font-size: 1em\">17. What mass of CO<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\"> is produced by the combustion of 1.00 mol of CH<\/span><sub class=\"subscript\">4<\/sub><span style=\"font-size: 1em\">?<\/span><\/p>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<div class=\"qandaset block\" id=\"ball-ch05_s04_qs01\">\r\n<div class=\"question\">\r\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">CH<sub class=\"subscript\">4<\/sub>(g) +\u00a02 O<sub class=\"subscript\">2<\/sub>(g) <span style=\"font-size: 1em\">$latex \\longrightarrow$<\/span>\u00a0CO<sub class=\"subscript\">2<\/sub>(g) +\u00a02 H<sub class=\"subscript\">2<\/sub>O(\u2113)<\/span><\/span><\/p>\r\n<span style=\"font-size: 1em\">18. What mass of HgO is required to produce 0.692 mol of O<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">?<\/span>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">2 HgO(s) <span style=\"font-size: 1em\">$latex \\longrightarrow$<\/span>\u00a02 Hg(\u2113) +\u00a0O<sub class=\"subscript\">2<\/sub>(g)<\/span><\/span><\/p>\r\n<span style=\"font-size: 1em\">19. How many moles of Al can be produced from 10.87 g of Ag?<\/span>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">Al(NO<sub class=\"subscript\">3<\/sub>) <sub class=\"subscript\">3<\/sub>(s) +\u00a03 Ag <span style=\"font-size: 1em\">$latex \\longrightarrow$<\/span>\u00a0Al +\u00a03 AgNO<sub class=\"subscript\">3<\/sub><\/span><\/span><\/p>\r\n<span style=\"font-size: 1em\">20. How many moles of O<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\"> are needed to prepare 1.00 g of Ca(NO<\/span><sub class=\"subscript\">3<\/sub><span style=\"font-size: 1em\">)<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">?<\/span>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">Ca(s) +\u00a0N<sub class=\"subscript\">2<\/sub>(g) +\u00a03 O<sub class=\"subscript\">2<\/sub>(g) <span style=\"font-size: 1em\">$latex \\longrightarrow$<\/span>\u00a0Ca(NO<sub class=\"subscript\">3<\/sub>) <sub class=\"subscript\">2<\/sub>(s)<\/span><\/span><\/p>\r\n<span style=\"font-size: 1em\">21. What mass of O<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\"> can be generated by the decomposition of 100.0 g of NaClO<\/span><sub class=\"subscript\">3<\/sub><span style=\"font-size: 1em\">?<\/span>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">2 NaClO<sub class=\"subscript\">3<\/sub>\u00a0<span style=\"font-size: 1em\">$latex \\longrightarrow$<\/span> 2 NaCl(s) +\u00a03 O<sub class=\"subscript\">2<\/sub>(g)<\/span><\/span><\/p>\r\n<span style=\"font-size: 1em\">22. What mass of Fe<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">O<\/span><sub class=\"subscript\">3<\/sub><span style=\"font-size: 1em\"> must be reacted to generate 324 g of Al<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">O<\/span><sub class=\"subscript\">3<\/sub><span style=\"font-size: 1em\">?<\/span>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">Fe<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub>(s) +\u00a02 Al(s) <span style=\"font-size: 1em\">$latex \\longrightarrow$<\/span>\u00a02 Fe(s) +\u00a0Al<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub>(s)<\/span><\/span><\/p>\r\n<span style=\"font-size: 1em\">23. What mass of MnO<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\"> is produced when 445 g of H<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">O are reacted?<\/span>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">H<sub class=\"subscript\">2<\/sub>O(\u2113) +\u00a02 MnO<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a0Br<sup class=\"superscript\">\u2212<\/sup>(aq) <span style=\"font-size: 1em\">$latex \\longrightarrow$<\/span>\u00a0BrO<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a02 MnO<sub class=\"subscript\">2<\/sub>(s) +\u00a02 OH<sup class=\"superscript\">\u2212<\/sup>(aq)<\/span><\/span><\/p>\r\n<span style=\"font-size: 1em\">24. If 83.9 g of ZnO are formed, what mass of Mn<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">O<\/span><sub class=\"subscript\">3<\/sub><span style=\"font-size: 1em\"> is formed with it?<\/span>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">Zn(s) +\u00a02 MnO<sub class=\"subscript\">2<\/sub>(s) <span style=\"font-size: 1em\">$latex \\longrightarrow$<\/span>\u00a0ZnO(s) +\u00a0Mn<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub>(s)<\/span><\/span><\/p>\r\n<span style=\"font-size: 1em\">25. If 88.4 g of CH<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">S are reacted, what mass of HF is produced?<\/span>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">CH<sub class=\"subscript\">2<\/sub>S +\u00a06 F<sub class=\"subscript\">2<\/sub>\u00a0<span style=\"font-size: 1em\">$latex \\longrightarrow$<\/span> CF<sub class=\"subscript\">4<\/sub> +\u00a02 HF +\u00a0SF<sub class=\"subscript\">6<\/sub><\/span><\/span><\/p>\r\n<span style=\"font-size: 1em\">26. Express in mole terms what this chemical equation means.<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<div class=\"qandaset block\" id=\"ball-ch05_s03_qs01\">\r\n<div class=\"question\">\r\n\r\n<span class=\"informalequation\"><span class=\"mathphrase\">CH<sub class=\"subscript\">4<\/sub> +\u00a02O<sub class=\"subscript\">2<\/sub>\u00a0<span style=\"font-size: 1em\">$latex \\longrightarrow$<\/span> CO<sub class=\"subscript\">2<\/sub> +\u00a02H<sub class=\"subscript\">2<\/sub>O<\/span><\/span>\r\n\r\n<span style=\"font-size: 1em\">27. How many molecules of each substance are involved in the equation in Exercise 1 if it is interpreted in terms of moles?<\/span>\r\n\r\n<span style=\"font-size: 1em\">28. For the chemical equation<\/span>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">2 C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">6<\/sub> +\u00a07 O<sub class=\"subscript\">2<\/sub>\u00a0<span style=\"font-size: 1em\">$latex \\longrightarrow$<\/span> 4 CO<sub class=\"subscript\">2<\/sub> +\u00a06 H<sub class=\"subscript\">2<\/sub>O<\/span><\/span><\/p>\r\n<p id=\"ball-ch05_s03_qs01_p10\" class=\"para\">what equivalences can you write in terms of moles? Use the \u21d4 sign.<\/p>\r\n<p class=\"para\"><span style=\"font-size: 1em\">29. Write the balanced chemical reaction for the combustion of C<\/span><sub class=\"subscript\">5<\/sub><span style=\"font-size: 1em\">H<\/span><sub class=\"subscript\">12<\/sub><span style=\"font-size: 1em\"> (the products are CO<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\"> and H<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">O) and determine how many moles of H<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">O are formed when 5.8 mol of O<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\"> are reacted.<\/span><\/p>\r\n<p class=\"para\"><span style=\"font-size: 1em\">30. For the balanced chemical equation<\/span><\/p>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">3 Cu(s) +\u00a02 NO<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a08 H<sup class=\"superscript\">+<\/sup>(aq) <span style=\"font-size: 1em\">$latex \\longrightarrow$<\/span>\u00a03 Cu<sup class=\"superscript\">2+<\/sup>(aq) +\u00a04 H<sub class=\"subscript\">2<\/sub>O(\u2113) +\u00a02 NO(g)<\/span><\/span><\/p>\r\n<p id=\"ball-ch05_s03_qs01_p20\" class=\"para\">how many moles of Cu<sup class=\"superscript\">2+<\/sup> are formed when 55.7 mol of H<sup class=\"superscript\">+<\/sup> are reacted?<\/p>\r\n<p class=\"para\"><span style=\"font-size: 1em\">31. For the balanced chemical reaction<\/span><\/p>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">4 NH<sub class=\"subscript\">3<\/sub>(g) +\u00a05 O<sub class=\"subscript\">2<\/sub>(g) <span style=\"font-size: 1em\">$latex \\longrightarrow$<\/span>\u00a04 NO(g) +\u00a06 H<sub class=\"subscript\">2<\/sub>O(\u2113)<\/span><\/span><\/p>\r\n<p id=\"ball-ch05_s03_qs01_p26\" class=\"para\">how many moles of H<sub class=\"subscript\">2<\/sub>O are produced when 0.669 mol of NH<sub class=\"subscript\">3<\/sub> react?<\/p>\r\n<p class=\"para\"><span style=\"font-size: 1em\">32. For the balanced chemical reaction<\/span><\/p>\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">4 KO<sub class=\"subscript\">2<\/sub>(s) +\u00a02 CO<sub class=\"subscript\">2<\/sub>(g) <span style=\"font-size: 1em\">$latex \\longrightarrow$<\/span>\u00a02 K<sub class=\"subscript\">2<\/sub>CO<sub class=\"subscript\">3<\/sub>(s) +\u00a03 O<sub class=\"subscript\">2<\/sub>(g)<\/span><\/span><\/p>\r\n<p id=\"ball-ch05_s03_qs01_p32\" class=\"para\">determine the number of moles of both products formed when 6.88 mol of KO<sub class=\"subscript\">2<\/sub> react.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/div>\r\n<strong>Answers<\/strong>\r\n<p id=\"fs-idp75003744\">1. a) 0.435 mol Na, 0.217 mol Cl<sub>2<\/sub>, 15.4 g Cl<sub>2<\/sub><\/p>\r\nb) 0.005780 mol HgO, 2.890 \u00d7 10<sup>\u22123<\/sup> mol O<sub>2<\/sub>, 9.248 \u00d7 10<sup>\u22122<\/sup> g O<sub>2<\/sub>\r\n\r\nc) 8.00 mol NaNO<sub>3<\/sub>, 6.8 \u00d7 10<sup>2<\/sup> g NaNO<sub>3<\/sub>\r\n\r\nd) 1665 mol CO<sub>2<\/sub>, 73.3 kg CO<sub>2<\/sub>\r\n\r\ne) 18.86 mol CuO, 2.330 kg CuCO<sub>3<\/sub>\r\n\r\nf) 0.4580 mol C<sub>2<\/sub>H<sub>4<\/sub>Br<sub>2<\/sub>, 86.05 g C<sub>2<\/sub>H<sub>4<\/sub>Br<sub>2<\/sub>\r\n<p id=\"fs-idp103920384\">2. a) 0.0686 mol Mg, 1.67 g Mg<\/p>\r\nb) 2.701 \u00d7 10<sup>\u22123<\/sup> mol O<sub>2<\/sub>, 0.08644 g O<sub>2<\/sub>\r\n\r\nc) 6.43 mol MgCO<sub>3<\/sub>, 542 g MgCO<sub>3<\/sub>\r\n\r\nd) 713 mol H<sub>2<\/sub>O, 12.8 kg H<sub>2<\/sub>O\r\n\r\ne) 16.31 mol BaO<sub>2<\/sub>, 2762 g BaO<sub>2<\/sub>\r\n\r\nf) 0.207 mol C<sub>2<\/sub>H<sub>4<\/sub>, 5.81 g C<sub>2<\/sub>H<sub>4<\/sub>\r\n<p id=\"fs-idp32310528\">3. a) $latex \\text{volume HCl solution} \\longrightarrow \\text{mol HCl} \\longrightarrow \\text{mol GaCl}_3$; b) 1.25 mol GaCl<sub>3<\/sub>, 2.2 \u00d7 10<sup>2<\/sup> g GaCl<sub>3<\/sub><\/p>\r\n<p id=\"fs-idp18257312\">4. a) 5.337 \u00d7 10<sup>22<\/sup> molecules \u00a0 \u00a0 \u00a0b) 10.41 g Zn(CN)<sub>2<\/sub><\/p>\r\n<p id=\"fs-idp53851968\">5. $latex \\text{SiO}_2 + 3\\text{C} \\longrightarrow \\text{SiC} + 2\\text{CO}$, 4.50 kg SiO<sub>2<\/sub><\/p>\r\n<p id=\"fs-idp125512976\">6. 5.00 \u00d7 10<sup>3<\/sup> kg<\/p>\r\n<p id=\"fs-idp10693200\">7. 1.28 \u00d7 10<sup>5<\/sup> g CO<sub>2<\/sub><\/p>\r\n<p id=\"fs-idp74230176\">8. 161.40 mL KI solution<\/p>\r\n<p id=\"fs-idp100940480\">9. 176 g TiO<sub>2<\/sub><\/p>\r\n<span style=\"font-size: 1em\">10. <\/span><span style=\"font-size: 1em\"><\/span>\r\n\r\n<a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/Screen-Shot-2014-07-17-at-10.50.06-AM.png\" style=\"font-size: 1em\"><img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/Screen-Shot-2014-07-17-at-10.50.06-AM-1.png\" alt=\"Screen Shot 2014-07-17 at 10.50.06 AM\" class=\"alignnone wp-image-3511 size-full\" width=\"160\" height=\"40\" \/><\/a><span style=\"font-size: 1em\">\u00a0are two conversion factors that can be constructed from the pound cake recipe. Other conversion factors are also possible.<\/span>\r\n<div class=\"answer\">11.<\/div>\r\n<div class=\"answer\"><span style=\"font-size: 1em\"><\/span><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/Screen-Shot-2014-07-17-at-10.51.35-AM.png\" style=\"font-size: 1em\"><img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/Screen-Shot-2014-07-17-at-10.51.35-AM-1.png\" alt=\"Screen Shot 2014-07-17 at 10.51.35 AM\" class=\"alignnone size-full wp-image-3512\" width=\"247\" height=\"36\" \/><\/a><span style=\"font-size: 1em\">\u00a0and their reciprocals are the conversion factors that can be constructed.<\/span><\/div>\r\n<div class=\"answer\">\r\n<p class=\"para\">12. \u00a02Na(s) +\u00a02H<sub class=\"subscript\">2<\/sub>O(\u2113) <span style=\"font-size: 1em\">$latex \\longrightarrow$<\/span>\u00a02NaOH(aq) +\u00a0H<sub class=\"subscript\">2<\/sub>(g) and 166 molecules<\/p>\r\n\r\n<\/div>\r\n<div class=\"answer\">13.\u00a0120 molecules<\/div>\r\n<div class=\"answer\">14. 107 molecules<\/div>\r\n<div class=\"answer\">15. 435,000 molecules<\/div>\r\n<div class=\"answer\">16.<\/div>\r\n<div class=\"answer\"><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/Screen-Shot-2014-07-17-at-10.52.09-AM.png\"><img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/Screen-Shot-2014-07-17-at-10.52.09-AM-1.png\" alt=\"Screen Shot 2014-07-17 at 10.52.09 AM\" class=\"alignnone wp-image-3513 size-full\" width=\"303\" height=\"63\" \/><\/a><\/div>\r\n<div>\r\n\r\n17. 44.0 g\r\n\r\n18.\u00a03.00 \u00d7 10<sup class=\"superscript\">2<\/sup> g\r\n\r\n19.\u00a00.0336 mol\r\n\r\n20.\u00a00.0183 mol\r\n\r\n21.\u00a045.1 g\r\n\r\n22.\u00a0507 g\r\n\r\n23.\u00a04.30 \u00d7 10<sup class=\"superscript\">3<\/sup> g\r\n\r\n24.\u00a0163 g\r\n\r\n25.\u00a076.7 g\r\n\r\n26. One mole of CH<sub class=\"subscript\">4<\/sub> reacts with 2 mol of O<sub class=\"subscript\">2<\/sub> to make 1 mol of CO<sub class=\"subscript\">2<\/sub> and 2 mol of H<sub class=\"subscript\">2<\/sub>O.\r\n\r\n27.\u00a06.022 \u00d7 10<sup class=\"superscript\">23<\/sup> molecules of CH<sub class=\"subscript\">4<\/sub>, 1.2044 \u00d7 10<sup class=\"superscript\">24<\/sup> molecules of O<sub class=\"subscript\">2<\/sub>, 6.022 \u00d7 10<sup class=\"superscript\">23<\/sup> molecules of CO<sub class=\"subscript\">2<\/sub>, and 1.2044 \u00d7 10<sup class=\"superscript\">24<\/sup> molecules of H<sub class=\"subscript\">2<\/sub>O\r\n\r\n28.\u00a02 mol of C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">6<\/sub>\u00a0to 7 mol of O<sub class=\"subscript\">2<\/sub>\u00a0to 4 mol of CO<sub class=\"subscript\">2<\/sub>\u00a0to 6 mol of H<sub class=\"subscript\">2<\/sub>O\r\n\r\n29.\u00a0C<sub class=\"subscript\">5<\/sub>H<sub class=\"subscript\">12<\/sub> +\u00a08 O<sub class=\"subscript\">2\u00a0<\/sub>\u00a0<span style=\"font-size: 1em\">$latex \\longrightarrow$<\/span> 5CO<sub class=\"subscript\">2<\/sub> +\u00a06H<sub class=\"subscript\">2<\/sub>O; 4.4 mol\r\n\r\n30.\u00a020.9 mol\r\n\r\n31.\u00a01.00 mol\r\n\r\n32.\u00a03.44 mol of K<sub class=\"subscript\">2<\/sub>CO<sub class=\"subscript\">3<\/sub>; 5.16 mol of O<sub class=\"subscript\">2<\/sub>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/section>\r\n<div>\r\n<h2>Glossary<\/h2>\r\n<strong>stoichiometric factor:\u00a0<\/strong>ratio of coefficients in a balanced chemical equation, used in computations relating amounts of reactants and products\r\n\r\n<strong>stoichiometry:\u00a0<\/strong>relationships between the amounts of reactants and products of a chemical reaction\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>Explain the concept of stoichiometry as it pertains to chemical reactions<\/li>\n<li>Use balanced chemical equations to derive stoichiometric factors relating amounts of reactants and products<\/li>\n<li>Perform stoichiometric calculations involving mass, moles, and solution molarity<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-idp65325120\">A balanced chemical equation provides a great deal of information in a very succinct format. Chemical formulas provide the identities of the reactants and products involved in the chemical change, allowing classification of the reaction. Coefficients provide the relative numbers of these chemical species, allowing a quantitative assessment of the relationships between the amounts of substances consumed and produced by the reaction. These quantitative relationships are known as the reaction\u2019s <strong>stoichiometry<\/strong>, a term derived from the Greek words <em>stoicheion<\/em> (meaning \u201celement\u201d) and <em>metron<\/em> (meaning \u201cmeasure\u201d). In this module, the use of balanced chemical equations for various stoichiometric applications is explored.<\/p>\n<p id=\"fs-idp92593072\">The general approach to using stoichiometric relationships is similar in concept to the way people go about many common activities. Food preparation, for example, offers an appropriate comparison. A recipe for making eight pancakes calls for 1 cup pancake mix, [latex]\\frac{3}{4}[\/latex] cup milk, and one egg. The \u201cequation\u201d representing the preparation of pancakes per this recipe is<\/p>\n<div class=\"equation\" id=\"fs-idp116126928\" style=\"text-align: center\">[latex]1 \\;\\text{cup mix} + \\frac{3}{4} \\;\\text{cup milk} + 1 \\;\\text{egg} \\longrightarrow 8 \\;\\text{pancakes}[\/latex]<\/div>\n<p id=\"fs-idp78338704\">If two dozen pancakes are needed for a big family breakfast, the ingredient amounts must be increased proportionally according to the amounts given in the recipe. For example, the number of eggs required to make 24 pancakes is<\/p>\n<div class=\"equation\" id=\"fs-idp58586688\" style=\"text-align: center\">[latex]24 \\;\\rule[0.5ex]{4em}{0.1ex}\\hspace{-4em}\\text{pancakes} \\times \\frac{1 \\;\\text{egg}}{8 \\;\\rule[0.25ex]{3em}{0.1ex}\\hspace{-3em}\\text{pancakes}} = 3 \\;\\text{eggs}[\/latex]<\/div>\n<p id=\"fs-idp98270672\">Balanced chemical equations are used in much the same fashion to determine the amount of one reactant required to react with a given amount of another reactant, or to yield a given amount of product, and so forth. The coefficients in the balanced equation are used to derive <strong>stoichiometric factors<\/strong> that permit computation of the desired quantity. To illustrate this idea, consider the production of ammonia by reaction of hydrogen and nitrogen:<\/p>\n<div class=\"equation\" id=\"fs-idp124603552\" style=\"text-align: center\">[latex]\\text{N}_2(g) + 3\\text{H}_2(g) \\longrightarrow 2\\text{NH}_3(g)[\/latex]<\/div>\n<p id=\"fs-idp124955072\">This equation shows ammonia molecules are produced from hydrogen molecules in a 2:3 ratio, and stoichiometric factors may be derived using any amount (number) unit:<\/p>\n<div class=\"equation\" id=\"fs-idp64965808\" style=\"text-align: center\">[latex]\\frac{2 \\;\\text{NH}_3 \\;\\text{molecules}}{3 \\;\\text{H}_2 \\;\\text{molecules}} \\;\\text{or} \\;\\frac{2 \\;\\text{doz NH}_3 \\;\\text{molecules}}{3 \\;\\text{doz H}_2 \\;\\text{molecules}} \\;\\text{or} \\;\\frac{2 \\;\\text{mol NH}_3 \\;\\text{molecules}}{3 \\;\\text{mol H}_2 \\;\\text{molecules}}[\/latex]<\/div>\n<p id=\"fs-idp56404080\">These stoichiometric factors can be used to compute the number of ammonia molecules produced from a given number of hydrogen molecules, or the number of hydrogen molecules required to produce a given number of ammonia molecules. Similar factors may be derived for any pair of substances in any chemical equation.<\/p>\n<div class=\"textbox shaded\" id=\"fs-idp48900096\">\n<h3>Example 1<\/h3>\n<p id=\"fs-idp77269456\">How many moles of I<sub>2<\/sub> are required to react with 0.429 mol of Al according to the following equation (see <a href=\"#CNX_Chem_04_03_iodine\" class=\"autogenerated-content\">Figure 1<\/a>)?<\/p>\n<div class=\"equation\" id=\"fs-idp147980704\" style=\"text-align: center\">[latex]2\\text{Al} + 3\\text{I}_2 \\longrightarrow 2\\text{AlI}_3[\/latex]<\/div>\n<figure id=\"CNX_Chem_04_03_iodine\"><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_03_iodine.jpg\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_03_iodine-2.jpg\" alt=\"This figure shows three photos with an arrow leading from one to the next. The first photo shows a small pile of iodine and aluminum on a white surface. The second photo shows a small amount of purple smoke coming from the pile. The third photo shows a large amount of purple and gray smoke coming from the pile.\" width=\"975\" height=\"182\" \/><\/a><figcaption class=\"wp-caption-text\"><strong>Figure 1.<\/strong> Aluminum and iodine react to produce aluminum iodide. The heat of the reaction vaporizes some of the solid iodine as a purple vapor. (credit: modification of work by Mark Ott)<\/figcaption><\/figure>\n<\/figcaption><\/figure>\n<p>&nbsp;<\/p>\n<p id=\"fs-idm34598272\"><strong>Solution<\/strong><br \/>\nReferring to the balanced chemical equation, the stoichiometric factor relating the two substances of interest is [latex]\\frac{3 \\;\\text{mol I}_2}{2 \\;\\text{mol Al}}[\/latex]. The molar amount of iodine is derived by multiplying the provided molar amount of aluminum by this factor:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_03_moleratio1_img-2.jpg\" alt=\"This figure shows two pink rectangles. The first is labeled, \u201cMoles of A l.\u201d This rectangle is followed by an arrow pointing right to a second rectangle labeled, \u201cMoles of I subscript 2.\u201d\" width=\"490\" height=\"95\" class=\"aligncenter\" \/><\/p>\n<div class=\"equation\" id=\"fs-idp102857120\" style=\"text-align: center\">[latex]\\begin{array}{r @{{}={}} l} \\text{mol I}_2 & 0.429 \\;\\rule[0.5ex]{3.25em}{0.1ex}\\hspace{-3.25em}\\text{mol Al} \\times \\frac{3 \\;\\text{mol I}_2}{2 \\;\\rule[0.25ex]{2em}{0.1ex}\\hspace{-2em}\\text{mol Al}} \\\\[1em] & 0.644 \\;\\text{mol I}_2 \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-idp30742160\"><em><strong>Test Yourself<\/strong><\/em><br \/>\nHow many moles of Ca(OH)<sub>2<\/sub> are required to react with 1.36 mol of H<sub>3<\/sub>PO<sub>4<\/sub> to produce Ca<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub> according to the equation [latex]3\\text{Ca(OH)}_2 + 2\\text{H}_3 \\text{PO}_4 \\longrightarrow \\text{Ca}_3 \\text{(PO}_4)_2 + 6\\text{H}_2 \\text{O}[\/latex]?<\/p>\n<p>&nbsp;<\/p>\n<p><em><strong>Answer<\/strong><\/em><\/p>\n<p>2.04 mol<\/p>\n<\/div>\n<div class=\"textbox shaded\" id=\"fs-idp9124448\">\n<h3>Example 2<\/h3>\n<p id=\"fs-idp157170976\">How many carbon dioxide molecules are produced when 0.75 mol of propane is combusted according to this equation?<\/p>\n<div class=\"equation\" id=\"fs-idp116053696\" style=\"text-align: center\">[latex]\\text{C}_3 \\text{H}_8 + 5\\text{O}_2 \\longrightarrow 3\\text{CO}_2 + 4\\text{H}_2 \\text{O}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-idp121952928\"><strong>Solution<\/strong><br \/>\nThe approach here is the same as for <a href=\"#fs-idp48900096\" class=\"autogenerated-content\">Example 1<\/a>, though the absolute number of molecules is requested, not the number of moles of molecules. This will simply require use of the moles-to-numbers conversion factor, Avogadro\u2019s number.<\/p>\n<p id=\"fs-idp229756528\">The balanced equation shows that carbon dioxide is produced from propane in a 3:1 ratio:<\/p>\n<div class=\"equation\" id=\"fs-idp114307520\" style=\"text-align: center\">[latex]\\frac{3 \\;\\text{mol CO}_2}{1 \\;\\text{mol C}_3 \\text{H}_8}[\/latex]<\/div>\n<p id=\"fs-idp83056800\">Using this stoichiometric factor, the provided molar amount of propane, and Avogadro\u2019s number,<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_03_moleratio2_img-2.jpg\" alt=\"This figure shows two pink rectangles. The first is labeled, \u201cMoles of C subscript 3 H subscript 8.\u201d This rectangle is followed by an arrow pointing right to a second rectangle labeled, \u201cMoles of C O subscript 2.\u201d\" width=\"679\" height=\"92\" class=\"aligncenter\" \/><\/p>\n<div class=\"equation\" id=\"fs-idp52373104\" style=\"text-align: center\">[latex]0.75 \\;\\rule[0.5ex]{4.5em}{0.1ex}\\hspace{-4.5em}\\text{mol C}_3 \\text{H}_8 \\times \\frac{3 \\;\\rule[0.25ex]{3em}{0.1ex}\\hspace{-3em} \\text{mol CO}_2}{1 \\;\\rule[0.5ex]{3em}{0.1ex}\\hspace{-3em}\\text{mol C}_3 \\text{H}_8} \\times \\frac{6.022 \\times 10^{23} \\;\\text{CO}_2 \\;\\text{molecules}}{\\rule[0.25ex]{3em}{0.1ex}\\hspace{-3em} \\text{mol CO}_2} = 1.4 \\times 10^{24} \\text{CO}_2 \\;\\text{molecules}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-idm14771264\"><em><strong>Test Yourself<\/strong><\/em><br \/>\nHow many NH<sub>3<\/sub> molecules are produced by the reaction of 4.0 mol of Ca(OH)<sub>2<\/sub> according to the following equation:<\/p>\n<div class=\"equation\" id=\"fs-idp166213360\" style=\"text-align: center\">[latex](\\text{NH}_4)_2 \\text{SO}_4 + \\text{Ca(OH)}_2 \\longrightarrow 2\\text{NH}_3 + \\text{CaSO}_4 + 2\\text{H}_2 \\text{O}[\/latex]<\/div>\n<div><\/div>\n<div><em><strong>Answer<\/strong><\/em><\/div>\n<div>4.8 \u00d7 10<sup>24<\/sup> NH<sub>3<\/sub> molecules<\/div>\n<\/div>\n<p id=\"fs-idp222914448\">These examples illustrate the ease with which the amounts of substances involved in a chemical reaction of known stoichiometry may be related. Directly measuring numbers of atoms and molecules is, however, not an easy task, and the practical application of stoichiometry requires that we use the more readily measured property of mass.<\/p>\n<div class=\"textbox shaded\" id=\"fs-idp113495744\">\n<h3>Example 3<\/h3>\n<p id=\"fs-idp56844080\">What mass of sodium hydroxide, NaOH, would be required to produce 16 g of the antacid milk of magnesia [magnesium hydroxide, Mg(OH)<sub>2<\/sub>] by the following reaction?<\/p>\n<div class=\"equation\" id=\"fs-idp158099664\" style=\"text-align: center\">[latex]\\text{MgCl}_2(aq) + 2\\text{NaOH}(aq) \\longrightarrow \\text{Mg(OH)}_2(s) + \\text{NaCl}(aq)[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-idm3520480\"><strong>Solution<\/strong><br \/>\nThe approach used previously in <a href=\"#fs-idp48900096\" class=\"autogenerated-content\">Example 1<\/a> and <a href=\"#fs-idp9124448\" class=\"autogenerated-content\">Example 2<\/a> is likewise used here; that is, we must derive an appropriate stoichiometric factor from the balanced chemical equation and use it to relate the amounts of the two substances of interest. In this case, however, masses (not molar amounts) are provided and requested, so additional steps of the sort learned in the previous chapter are required. The calculations required are outlined in this flowchart:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_03_map2_img-2.jpg\" alt=\"This figure shows four rectangles. The first is shaded yellow and is labeled, \u201cMass of M g ( O H ) subscript 2.\u201d This rectangle is followed by an arrow pointing right to a second rectangle which is shaded pink and is labeled, \u201cMoles of M g ( O H ) subscript 2.\u201d This rectangle is followed by an arrow pointing right to a third rectangle which is shaded pink and is labeled, \u201cMoles of N a O H.\u201d This rectangle is followed by an arrow pointing right to a fourth rectangle which is shaded yellow and is labeled, \u201cMass of N a O H.\u201d\" width=\"655\" height=\"291\" class=\"aligncenter\" \/><\/p>\n<div class=\"equation\" id=\"fs-idp122145840\" style=\"text-align: center\">[latex]16 \\;\\rule[0.5ex]{5em}{0.1ex}\\hspace{-5em}\\text{g Mg(OH)}_2 \\times \\frac{1 \\;\\rule[0.25ex]{4.75em}{0.1ex}\\hspace{-4.75em}\\text{mol Mg(OH)}_2}{58.3197 \\;\\rule[0.25ex]{3.5em}{0.1ex}\\hspace{-3.5em}\\text{g Mg(OH)}_2} \\times \\frac{2 \\;\\rule[0.25ex]{3.5em}{0.1ex}\\hspace{-3.5em}\\text{mol NaOH}}{1 \\;\\rule[0.25ex]{4.5em}{0.1ex}\\hspace{-4.5em}\\text{mol Mg(OH)}_2} \\times \\frac{39.9971 \\;\\text{g NaOH}}{1 \\;\\rule[0.25ex]{3.25em}{0.1ex}\\hspace{-3.25em}\\text{mol NaOH}} = 22 \\;\\text{g NaOH}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-idp52759328\"><em><strong>Test Yourself<\/strong><\/em><br \/>\nWhat mass of gallium oxide, Ga<sub>2<\/sub>O<sub>3<\/sub>, can be prepared from 29.0 g of gallium metal? The equation for the reaction is [latex]4 \\text{Ga} + 3\\text{O}_2 \\longrightarrow 2\\text{Ga}_2 \\text{O}_3.[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p><em><strong>Answer<\/strong><\/em><\/p>\n<p>39.0 g<\/p>\n<\/div>\n<div class=\"textbox shaded\" id=\"fs-idp274853984\">\n<h3>Example 4<\/h3>\n<p id=\"fs-idp64260528\">What mass of oxygen gas, O<sub>2<\/sub>, from the air is consumed in the combustion of 702 g of octane, C<sub>8<\/sub>H<sub>18<\/sub>, one of the principal components of gasoline?<\/p>\n<div class=\"equation\" id=\"fs-idp87206624\" style=\"text-align: center\">[latex]2\\text{C}_8 \\text{H}_{18} + 25\\text{O}_2 \\longrightarrow 16\\text{CO}_2 + 18\\text{H}_2 \\text{O}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-idp35508016\"><strong>Solution<\/strong><br \/>\nThe approach required here is the same as for the <a href=\"#fs-idp113495744\" class=\"autogenerated-content\">Example 3<\/a>, differing only in that the provided and requested masses are both for reactant species.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_03_map3_img-2.jpg\" alt=\"This figure shows four rectangles. The first is shaded yellow and is labeled, \u201cMass of C subscript 8 H subscript 18.\u201d This rectangle is followed by an arrow pointing right to a second rectangle which is shaded pink and is labeled, \u201cMoles of C subscript 8 H subscript 18.\u201d This rectangle is followed by an arrow pointing right to a third rectangle which is shaded pink and is labeled, \u201cMoles of O subscript 2.\u201d This rectangle is followed by an arrow pointing right to a fourth rectangle which is shaded yellow and is labeled, \u201cMass of O subscript 2.\u201d\" width=\"650\" height=\"289\" class=\"aligncenter\" \/><\/p>\n<div class=\"equation\" id=\"fs-idm56417376\" style=\"text-align: center\">[latex]702 \\;\\rule[0.5ex]{3.5em}{0.1ex}\\hspace{-3.5em}\\text{g C}_8 \\text{H}_{18} \\times \\frac{1 \\;\\rule[0.25ex]{3.5em}{0.1ex}\\hspace{-3.5em}\\text{mol C}_8 \\text{H}_{18}}{114.231 \\;\\rule[0.25ex]{2.75em}{0.1ex}\\hspace{-2.75em}\\text{g C}_8 \\text{H}_{18}} \\times \\frac{25 \\;\\rule[0.25ex]{2.5em}{0.1ex}\\hspace{-2.5em}\\text{mol O}_2}{2 \\;\\rule[0.25ex]{3.5em}{0.1ex}\\hspace{-3.5em}\\text{mol C}_8 \\text{H}_{18}} \\times \\frac{31.9988 \\;\\text{g O}_2}{\\rule[0.25ex]{2.5em}{0.1ex}\\hspace{-2.5em}\\text{mol O}_2} = 2.46 \\times 10^3 \\;\\text{g O}_2[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-idp24638432\"><em><strong>Test Yourself<\/strong><\/em><br \/>\nWhat mass of CO is required to react with 25.13 g of Fe<sub>2<\/sub>O<sub>3<\/sub> according to the equation<\/p>\n<p style=\"text-align: center\">[latex]\\text{Fe}_2 \\text{O}_3 + 3\\text{CO} \\longrightarrow 2\\text{Fe} + 3\\text{CO}_2[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p><em><strong>Answer<\/strong><\/em><\/p>\n<p>13.22 g<\/p>\n<\/div>\n<p id=\"fs-idp90689472\">These examples illustrate just a few instances of reaction stoichiometry calculations. Numerous variations on the beginning and ending computational steps are possible depending upon what particular quantities are provided and sought (volumes, solution concentrations, and so forth). Regardless of the details, all these calculations share a common essential component: the use of stoichiometric factors derived from balanced chemical equations. <a href=\"#CNX_Chem_04_03_flowchart\" class=\"autogenerated-content\">Figure 2<\/a> provides a general outline of the various computational steps associated with many reaction stoichiometry calculations.<\/p>\n<figure id=\"CNX_Chem_04_03_flowchart\"><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_03_flowchart.jpg\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_03_flowchart-2.jpg\" alt=\"This flowchart shows 10 rectangles connected by double headed arrows. To the upper left, a rectangle is shaded lavender and is labeled, \u201cVolume of pure substance A.\u201d This rectangle is followed by a horizontal double headed arrow labeled, \u201cDensity.\u201d It connects to a second rectangle which is shaded yellow and is labeled, \u201cMass of A.\u201d This rectangle is followed by a double headed arrow which is labeled, \u201cMolar Mass,\u201d that connects to a third rectangle which is shaded pink and is labeled, \u201cMoles of A.\u201d To the left of this rectangle is a horizontal double headed arrow labeled, \u201cMolarity,\u201d which connects to a lavender rectangle which is labeled, \u201cVolume of solution A.\u201d The pink, \u201cMoles of A,\u201d rectangle is also connected with a double headed arrow below and to the left. This arrow is labeled \u201cAvogadro\u2019s number.\u201d It connects to a green shaded rectangle that is labeled, \u201cNumber of particles of A.\u201d To the right of the pink \u201cMoles of A,\u201d rectangle is a horizontal double headed arrow which is labeled, \u201cStoichiometric factor.\u201d It connects to a second pink rectangle which is labeled, \u201cMoles of B.\u201d A double headed arrow which is labeled, \u201cMolar mass,\u201d extends from the top of this rectangle above and to the right to a yellow shaded rectangle labeled, \u201cMass of B.\u201d A horizontal double headed arrow which is labeled, \u201cDensity\u201d links to a lavender rectangle labeled, \u201cVolume of substance B,\u201d to the right. A horizontal double headed arrow labeled, \u201cMolarity,\u201d extends right to the of the pink \u201cMoles of B\u201d rectangle. This arrow connects to a lavender rectangle that is labeled, \u201cVolume of substance B.\u201d Another double headed arrow extends below and to the right of the pink \u201cMoles of B\u201d rectangle. This arrow is labeled \u201cAvogadro\u2019s number,\u201d and it extends to a green rectangle which is labeled, \u201cNumber of particles of B.\u201d\" width=\"1300\" height=\"796\" \/><\/a><figcaption class=\"wp-caption-text\"><strong>Figure 2.<\/strong> The flowchart depicts the various computational steps involved in most reaction stoichiometry calculations.<\/figcaption><\/figure>\n<\/figcaption><\/figure>\n<div id=\"fs-idp84121360\" class=\"textbox shaded\">\n<h3 class=\"title\">Airbags<\/h3>\n<p id=\"fs-idp124943600\">Airbags (<a href=\"#CNX_Chem_04_03_airbag\" class=\"autogenerated-content\">Figure 3<\/a>) are a safety feature provided in most automobiles since the 1990s. The effective operation of an airbag requires that it be rapidly inflated with an appropriate amount (volume) of gas when the vehicle is involved in a collision. This requirement is satisfied in many automotive airbag systems through use of explosive chemical reactions, one common choice being the decomposition of sodium azide, NaN<sub>3<\/sub>. When sensors in the vehicle detect a collision, an electrical current is passed through a carefully measured amount of NaN<sub>3<\/sub> to initiate its decomposition:<\/p>\n<div class=\"equation\" id=\"fs-idp113504400\" style=\"text-align: center\">[latex]2 \\text{NaN}_3(s) \\longrightarrow 3\\text{N}_2(g) + 2\\text{Na}(s)[\/latex]<\/div>\n<figure id=\"CNX_Chem_04_03_airbag\">\n<figure style=\"width: 388px\" class=\"wp-caption aligncenter\"><a href=\"https:\/\/opentextbc.ca\/chemistry\/wp-content\/uploads\/sites\/150\/2016\/05\/CNX_Chem_04_03_airbag.jpg\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_03_airbag-2.jpg\" alt=\"This photograph shows the inside of an automobile from the driver\u2019s side area. The image shows inflated airbags positioned just in front of the driver\u2019s and passenger\u2019s seats and along the length of the passenger side over the windows. A large, round airbag covers the steering wheel.\" width=\"388\" height=\"291\" class=\"\" \/><\/a><figcaption class=\"wp-caption-text\"><strong>Figure 3.<\/strong> Airbags deploy upon impact to minimize serious injuries to passengers. (credit: Jon Seidman)<\/figcaption><\/figure>\n<p>&nbsp;<\/p>\n<p id=\"fs-idp9645888\">This reaction is very rapid, generating gaseous nitrogen that can deploy and fully inflate a typical airbag in a fraction of a second (~0.03\u20130.1 s).<\/p>\n<p>Among many engineering considerations, the amount of sodium azide used must be appropriate for generating enough nitrogen gas to fully inflate the air bag and ensure its proper function.<\/p>\n<p>For example, a small mass (~100 g) of NaN<sub>3<\/sub> will generate approximately 50 L of N<sub>2<\/sub>.<\/figure>\n<\/div>\n<h2><span style=\"font-family: Roboto, Helvetica, Arial, sans-serif\">More Worked Out Problems<\/span><\/h2>\n<section id=\"fs-idp102563712\" class=\"summary\">\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 5<\/h3>\n<p id=\"ball-ch05_s01_p19\" class=\"para\">How many molecules of SO<sub class=\"subscript\">3<\/sub> are needed to react with 144 molecules of Fe<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub> given this balanced chemical equation?<\/p>\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">Fe<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub>(s) +\u00a03SO<sub class=\"subscript\">3<\/sub>(g) <span style=\"font-size: 1em\">[latex]\\longrightarrow[\/latex]<\/span>\u00a0Fe<sub class=\"subscript\">2<\/sub>(SO<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">3<\/sub><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\n<p id=\"ball-ch05_s01_p20\" class=\"para\">We use the balanced chemical equation to construct a conversion factor between Fe<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub> and SO<sub class=\"subscript\">3<\/sub>. The number of molecules of Fe<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub> goes on the bottom of our conversion factor so it cancels with our given amount, and the molecules of SO<sub class=\"subscript\">3<\/sub> go on the top. Thus, the appropriate conversion factor is<\/p>\n<p><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/ss2.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/ss2-1.png\" alt=\"ss2\" class=\"size-full wp-image-3631 aligncenter\" width=\"132\" height=\"59\" \/><\/a><\/p>\n<p id=\"ball-ch05_s01_p21\" class=\"para\">Starting with our given amount and applying the conversion factor, the result is<\/p>\n<p class=\"para\"><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/ss1.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/ss1-1.png\" alt=\"ss1\" class=\"aligncenter wp-image-3630 size-full\" width=\"489\" height=\"79\" \/><\/a><\/p>\n<p id=\"ball-ch05_s01_p22\" class=\"para\">We need 432 molecules of SO<sub class=\"subscript\">3<\/sub> to react with 144 molecules of Fe<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub>.<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\n<p id=\"ball-ch05_s01_p23\" class=\"para\">How many molecules of H<sub class=\"subscript\">2<\/sub> are needed to react with 29 molecules of N<sub class=\"subscript\">2<\/sub> to make ammonia if the balanced chemical equation is N<sub class=\"subscript\">2<\/sub> +\u00a03H<sub class=\"subscript\">2<\/sub>\u00a0<span style=\"font-size: 1em\">[latex]\\longrightarrow[\/latex]<\/span> 2NH<sub class=\"subscript\">3<\/sub>?<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\n<p id=\"ball-ch05_s01_p24\" class=\"para\">87 molecules<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 6<\/h3>\n<p id=\"ball-ch05_s01_p27\" class=\"para\">How many molecules of NH<sub class=\"subscript\">3<\/sub> can you make if you have 228 atoms of H<sub class=\"subscript\">2<\/sub>?<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\n<p id=\"ball-ch05_s01_p28\" class=\"para\">From the formula, we know that one molecule of NH<sub class=\"subscript\">3<\/sub> has three H atoms. Use that fact as a conversion factor:<\/p>\n<p><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/Screen-Shot-2014-07-17-at-10.49.40-AM1.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/Screen-Shot-2014-07-17-at-10.49.40-AM1-1.png\" alt=\"Screen-Shot-2014-07-17-at-10.49.40-AM\" class=\"wp-image-3627 size-full aligncenter\" width=\"412\" height=\"79\" \/><\/a><\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\n<p id=\"ball-ch05_s01_p29\" class=\"para\">How many molecules of Fe<sub class=\"subscript\">2<\/sub>(SO<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">3<\/sub> can you make from 777 atoms of S?<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\n<p id=\"ball-ch05_s01_p30\" class=\"para\">259 molecules<\/p>\n<\/div>\n<div class=\"section\" id=\"ball-ch05_s04\" lang=\"en\">\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 7<\/h3>\n<p id=\"ball-ch05_s04_p09\" class=\"para\">How many moles of HCl will be produced when 249 g of AlCl<sub class=\"subscript\">3<\/sub> are reacted according to this chemical equation? \u00a0The molar mass of AlCl<sub class=\"subscript\">3<\/sub> is 133.34 g\/mol.<\/p>\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">2 AlCl<sub class=\"subscript\">3<\/sub> +\u00a03 H<sub class=\"subscript\">2<\/sub>O(\u2113) <span style=\"font-size: 1em\">[latex]\\longrightarrow[\/latex]<\/span>\u00a0Al<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub> +\u00a06 HCl(g)<\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\n<p id=\"ball-ch05_s04_p10\" class=\"para\">We will do this in two steps: convert the mass of AlCl<sub class=\"subscript\">3<\/sub> to moles and then use the balanced chemical equation to find the number of moles of HCl formed. The molar mass of AlCl<sub class=\"subscript\">3<\/sub> is 133.34 g\/mol, which we have to invert to get the appropriate conversion factor:<\/p>\n<p style=\"text-align: center\">[latex]249 \\;\\rule[0.5ex]{0.5em}{0.1ex}\\hspace{-0.5em}\\text{g AlCl}_3 \\times \\frac{1 \\;\\text{mol AlCl}_3}{133.34\\;\\rule[0.5ex]{0.5em}{0.1ex}\\hspace{-0.5em}\\text{g AlCl}_3} = \\underline{1.86}74 \\;\\text{mol of aluminium chloride with 3 sig figs}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p id=\"ball-ch05_s04_p11\" class=\"para\">Now we can use this quantity to determine the number of moles of HCl that will form. From the balanced chemical equation, we construct a conversion factor between the number of moles of AlCl<sub class=\"subscript\">3<\/sub> and the number of moles of HCl:<\/p>\n<p style=\"text-align: center\">[latex]\\frac{6 \\; \\text{mol HCl}}{2 \\; \\text{mol AlCl}_3}[\/latex]<\/p>\n<p id=\"ball-ch05_s04_p12\" class=\"para\">Applying this conversion factor to the quantity of AlCl<sub class=\"subscript\">3<\/sub>, we get<\/p>\n<p style=\"text-align: center\">[latex]\\underline{1.86}74 \\;\\rule[0.5ex]{1.25em}{0.1ex}\\hspace{-1.25em}\\text{mol AlCl}_3 \\times \\frac{6 \\;\\text{mol HCl}}{2\\;\\rule[0.5ex]{1.25em}{0.1ex}\\hspace{-1.25em}\\text{mol AlCl}_3} =5.60 \\;\\text{mol HCl}[\/latex]<\/p>\n<p id=\"ball-ch05_s04_p13\" class=\"para\">Alternatively, we could have done this in one line:<\/p>\n<p style=\"text-align: center\">[latex]249 \\;\\rule[0.5ex]{0.5em}{0.1ex}\\hspace{-0.5em}\\text{g AlCl}_3 \\times \\frac{1 \\;\\rule[0.5ex]{1.25em}{0.1ex}\\hspace{-1.25em}\\text{mol AlCl}_3}{133.34\\;\\rule[0.5ex]{0.5em}{0.1ex}\\hspace{-0.5em}\\text{g AlCl}_3} \\times \\frac{6 \\;\\text{mol HCl}}{2\\;\\rule[0.5ex]{1.25em}{0.1ex}\\hspace{-1.25em}\\text{mol AlCl}_3}=5.60 \\;\\text{mol HCl}[\/latex]<\/p>\n<p id=\"ball-ch05_s04_p14\" class=\"para\">The last digit in our final answer is slightly different because of rounding differences, but the answer is essentially the same.<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\n<p id=\"ball-ch05_s04_p15\" class=\"para\">How many moles of Al<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub> will be produced when 23.9 g of H<sub class=\"subscript\">2<\/sub>O are reacted according to this chemical equation?<\/p>\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">2 AlCl<sub class=\"subscript\">3<\/sub> +\u00a03 H<sub class=\"subscript\">2<\/sub>O(\u2113) <span style=\"font-size: 1em\">[latex]\\longrightarrow[\/latex]<\/span>\u00a0Al<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub> +\u00a06 HCl(g)<\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\n<p id=\"ball-ch05_s04_p16\" class=\"para\">0.442 mol<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 8<\/h3>\n<p id=\"ball-ch05_s04_p18\" class=\"para\">How many grams of NH<sub class=\"subscript\">3<\/sub> will be produced when 33.9 mol of H<sub class=\"subscript\">2<\/sub> are reacted according to this chemical equation? \u00a0Use 17.03 g\/mol as the molar mass of NH<sub class=\"subscript\">3<\/sub>.<\/p>\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">N<sub class=\"subscript\">2<\/sub>(g) +\u00a03 H<sub class=\"subscript\">2<\/sub>(g) <span style=\"font-size: 1em\">[latex]\\longrightarrow[\/latex]<\/span>\u00a02 NH<sub class=\"subscript\">3<\/sub>(g)<\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\n<p id=\"ball-ch05_s04_p19\" class=\"para\">The conversions are the same, but they are applied in a different order. Start by using the balanced chemical equation to convert to moles of another substance and then use its molar mass to determine the mass of the final substance. In two steps, we have<\/p>\n<p><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/339molh2.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/339molh2-1.png\" alt=\"339molh2\" width=\"426\" height=\"93\" class=\"aligncenter wp-image-3738\" \/><\/a>Now, using the molar mass of NH<sub class=\"subscript\">3<\/sub>, which is 17.03 g\/mol, we get<\/p>\n<p><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/226molnh3.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/226molnh3-1.png\" alt=\"226molnh3\" width=\"430\" height=\"99\" class=\"aligncenter wp-image-3739\" \/><\/a><\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\n<p id=\"ball-ch05_s04_p21\" class=\"para\">How many grams of N<sub class=\"subscript\">2<\/sub> are needed to produce 2.17 mol of NH<sub class=\"subscript\">3<\/sub> when reacted according to this chemical equation?<\/p>\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">N<sub class=\"subscript\">2<\/sub>(g) +\u00a03 H<sub class=\"subscript\">2<\/sub>(g) <span style=\"font-size: 1em\">[latex]\\longrightarrow[\/latex]<\/span>\u00a02 NH<sub class=\"subscript\">3<\/sub>(g)<\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\n<p id=\"ball-ch05_s04_p22\" class=\"para\">30.4 g (Note: here we go from a product to a reactant, showing that mole-mass problems can begin and end with any substance in the chemical equation.)<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 9<\/h3>\n<p id=\"ball-ch05_s04_p30\" class=\"para\">What mass of Mg will be produced when 86.4 g of K are reacted? \u00a0Use 39.09 g\/mol as the molar mass of potassium and 24.31 g\/mol as the molar mass of magnesium.<\/p>\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">MgCl<sub class=\"subscript\">2<\/sub>(s) +\u00a02 K(s) <span style=\"font-size: 1em\">[latex]\\longrightarrow[\/latex]<\/span>\u00a0Mg(s) +\u00a02 KCl(s)<\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\n<p id=\"ball-ch05_s04_p31\" class=\"para\">We will simply follow the steps<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">mass K <span style=\"font-size: 1em\">[latex]\\longrightarrow[\/latex]<\/span>\u00a0mol K <span style=\"font-size: 1em\">[latex]\\longrightarrow[\/latex]<\/span>\u00a0mol Mg <span style=\"font-size: 1em\">[latex]\\longrightarrow[\/latex]<\/span>\u00a0mass Mg<\/span><\/span><\/p>\n<p id=\"ball-ch05_s04_p32\" class=\"para\">In addition to the balanced chemical equation, we need the molar masses of K (39.09 g\/mol) and Mg (24.31 g\/mol). In one line,<\/p>\n<p><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/864gk.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/864gk-1.png\" alt=\"864gk\" width=\"600\" height=\"89\" class=\"aligncenter wp-image-3740\" \/><\/a><\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\n<p id=\"ball-ch05_s04_p33\" class=\"para\">What mass of H<sub class=\"subscript\">2<\/sub> will be produced when 122 g of Zn are reacted?<\/p>\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">Zn(s) +\u00a02 HCl(aq) <span style=\"font-size: 1em\">[latex]\\longrightarrow[\/latex]<\/span>\u00a0ZnCl<sub class=\"subscript\">2<\/sub>(aq) +\u00a0H<sub class=\"subscript\">2<\/sub>(g)<\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\n<p id=\"ball-ch05_s04_p34\" class=\"para\">3.77 g<\/p>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 10<\/h3>\n<p id=\"ball-ch05_s03_p07\" class=\"para\">Interpret this balanced chemical equation in terms of moles.<\/p>\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">P<sub class=\"subscript\">4<\/sub> +\u00a05 O<sub class=\"subscript\">2<\/sub>\u00a0<span style=\"font-size: 1em\">[latex]\\longrightarrow[\/latex]<\/span> P<sub class=\"subscript\">4<\/sub>O<sub class=\"subscript\">10<\/sub><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\n<p id=\"ball-ch05_s03_p08\" class=\"para\">The coefficients represent the number of moles that react, not just molecules. We would speak of this equation as \u201cone mole of molecular phosphorus reacts with five moles of elemental oxygen to make one mole of tetraphosphorus decoxide.\u201d<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\n<p id=\"ball-ch05_s03_p09\" class=\"para\">Interpret this balanced chemical equation in terms of moles.<\/p>\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">N<sub class=\"subscript\">2<\/sub> +\u00a03 H<sub class=\"subscript\">2<\/sub>\u00a0<span style=\"font-size: 1em\">[latex]\\longrightarrow[\/latex]<\/span> 2 NH<sub class=\"subscript\">3<\/sub><\/span><\/span><\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\n<p id=\"ball-ch05_s03_p10\" class=\"para\">One mole of elemental nitrogen reacts with three moles of elemental hydrogen to produce two moles of ammonia.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 11<\/h3>\n<p id=\"ball-ch05_s03_p15\" class=\"para\">For the balanced chemical equation<\/p>\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">2 C<sub class=\"subscript\">4<\/sub>H<sub class=\"subscript\">10<\/sub>(g) +\u00a013 O<sub class=\"subscript\">2<\/sub>\u00a0<span style=\"font-size: 1em\">[latex]\\longrightarrow[\/latex]<\/span> 8 CO<sub class=\"subscript\">2<\/sub>(g) +\u00a010 H<sub class=\"subscript\">2<\/sub>O(\u2113)<\/span><\/span><\/p>\n<p id=\"ball-ch05_s03_p16\" class=\"para\">if 154 mol of O<sub class=\"subscript\">2<\/sub> are reacted, how many moles of CO<sub class=\"subscript\">2<\/sub> are produced?<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\n<p id=\"ball-ch05_s03_p17\" class=\"para\">We are relating an amount of oxygen to an amount of carbon dioxide, so we need the equivalence between these two substances. According to the balanced chemical equation, the equivalence is\u00a0<span class=\"informalequation\"><span class=\"mathphrase\">13 mol O<sub class=\"subscript\">2<\/sub>\u00a0to 8 mol CO<sub class=\"subscript\">2<\/sub><\/span><\/span><\/p>\n<p id=\"ball-ch05_s03_p18\" class=\"para\">We can use this equivalence to construct the proper conversion factor. We start with what we are given and apply the conversion factor:<\/p>\n<p><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/ss8.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/ss8-1.png\" alt=\"ss8\" class=\"wp-image-3642 size-full aligncenter\" width=\"358\" height=\"75\" \/><\/a><\/p>\n<p id=\"ball-ch05_s03_p19\" class=\"para\">The mol O<sub class=\"subscript\">2<\/sub> unit is in the denominator of the conversion factor so it cancels. Both the 8 and the 13 are exact numbers, so they don\u2019t contribute to the number of significant figures in the final answer.<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\n<p id=\"ball-ch05_s03_p20\" class=\"para\">Using the above equation, how many moles of H<sub class=\"subscript\">2<\/sub>O are produced when 154 mol of O<sub class=\"subscript\">2<\/sub> react?<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\n<p id=\"ball-ch05_s03_p21\" class=\"para\">118 mol<\/p>\n<\/div>\n<h2>Key Concepts and Summary<\/h2>\n<p id=\"fs-idm3728016\">A balanced chemical equation may be used to describe a reaction\u2019s stoichiometry (the relationships between amounts of reactants and products). Coefficients from the equation are used to derive stoichiometric factors that subsequently may be used for computations relating reactant and product masses, molar amounts, and other quantitative properties.<\/p>\n<\/section>\n<section id=\"fs-idm1445952\" class=\"exercises\">\n<div class=\"bcc-box bcc-info\">\n<h3>Exercises<\/h3>\n<p>1. Determine the number of moles and the mass requested for each of the following reactions: (Hint:\u00a0Write the balanced equation for each before attempting calculations.)<\/p>\n<p id=\"fs-idp84195488\">a) The number of moles and the mass of chlorine, Cl<sub>2<\/sub>, required to react with 10.0 g of sodium metal, Na, to produce sodium chloride, NaCl.<\/p>\n<p id=\"fs-idp186895920\">b) The number of moles and the mass of oxygen formed by the decomposition of 1.252 g of mercury(II) oxide.<\/p>\n<p id=\"fs-idp92843168\">c) The number of moles and the mass of sodium nitrate, NaNO<sub>3<\/sub>, required to produce 128 g of oxygen. (NaNO<sub>2<\/sub> is the other product.)<\/p>\n<p id=\"fs-idp8293632\">d) The number of moles and the mass of carbon dioxide formed by the combustion of 20.0 kg of carbon in an excess of oxygen.<\/p>\n<p id=\"fs-idp56955104\">e) The number of moles and the mass of copper(II) carbonate needed to produce 1.500 kg of copper(II) oxide. (CO<sub>2<\/sub> is the other product.)<\/p>\n<p id=\"fs-idp241908400\">f)<br \/>\n<img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_03_etheneBr_img-2.jpg\" alt=\"This figure includes two structural formulas. It reads, \u201cThe number of moles and the mass of,\u201d which is followed by a structure with two C atoms bonded with a single horizontal at the center. Both C atoms have H atoms bonded above and below. The C atom to the left has a B r atom bonded to its left. The C atom to the right has a B r atom bonded to its right. Following this structure, the figure reads, \u201cformed by the reaction of 12.85 g of,\u201d which is followed by a structure with two C atoms connected with a horizontal double bond. The C atom to the left has H atoms bonded above and to the left and below and to the left. The C atom to the right has H atoms bonded above and to the right and below and to the right. The figure ends with, \u201cwith an excess of B r subscript 2.\u201d\" \/><\/p>\n<p>2.\u00a0Determine the number of moles and the mass requested for each of the following reactions: (Hint:\u00a0Write the balanced equation for each before attempting calculations.)<\/p>\n<p id=\"fs-idp124413456\">a) The number of moles and the mass of Mg required to react with 5.00 g of HCl and produce MgCl<sub>2<\/sub> and H<sub>2<\/sub>.<\/p>\n<p id=\"fs-idp27637888\">b) The number of moles and the mass of oxygen formed by the decomposition of 1.252 g of silver(I) oxide.<\/p>\n<p id=\"fs-idp38709616\">c) The number of moles and the mass of magnesium carbonate, MgCO<sub>3<\/sub>, required to produce 283 g of carbon dioxide. (MgO is the other product.)<\/p>\n<p id=\"fs-idp56997136\">d) The number of moles and the mass of water formed by the combustion of 20.0 kg of acetylene, C<sub>2<\/sub>H<sub>2<\/sub>, in an excess of oxygen.<\/p>\n<p id=\"fs-idp149673136\">e) The number of moles and the mass of barium peroxide, BaO<sub>2<\/sub>, needed to produce 2.500 kg of barium oxide, BaO (O<sub>2<\/sub> is the other product.)<\/p>\n<p id=\"fs-idp83999440\">f)<br \/>\n<img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_03_ethene_img-2.jpg\" alt=\"This figure includes two structural formulas. It reads, \u201cThe number of moles and the mass of,\u201d which is followed by a structure with two C atoms connected with a horizontal double bond at the center. The C atom to the left has H atoms bonded above and to the left and below and to the left. The C atom to the right has H atoms bonded above and to the right and below and to the right. Following this structure, the figure reads, \u201crequired to react with H subscript 2 O to produce 9.55 g of,\u201d which is followed by a structure with two C atoms connected with a horizontal single bond. The C atom to the left has H atoms bonded above, to the left, and below. The C atom to the right has H atoms bonded above and below. To the right, an O atom forms a single bond with the C atom. A single H atom is bonded to the right side of the O atom.\" \/><\/p>\n<p>3. Gallium chloride is formed by the reaction of 2.6 L of a 1.44 M solution of HCl according to the following equation: [latex]2\\text{Ga} + 6\\text{HCl} \\longrightarrow 2\\text{GaCl}_3 + 3\\text{H}_2[\/latex].<\/p>\n<p id=\"fs-idp7152848\">a) Outline the steps necessary to determine the number of moles and mass of gallium chloride.<\/p>\n<p id=\"fs-idp58298672\">b) Perform the calculations outlined.<\/p>\n<p>4. Silver is often extracted from ores such as K[Ag(CN)<sub>2<\/sub>] and then recovered by the reaction<br \/>\n[latex]2 \\text{K} [\\text{Ag(CN)}_2](aq) + \\text{Zn}(s) \\longrightarrow 2\\text{Ag}(s) + \\text{Zn(CN)}_2(aq) + 2\\text{KCN}(aq)[\/latex]<\/p>\n<p id=\"fs-idp124132784\">a) How many molecules of Zn(CN)<sub>2<\/sub> are produced by the reaction of 35.27 g of K[Ag(CN)<sub>2<\/sub>]?<\/p>\n<p id=\"fs-idp25006640\">b) What mass of Zn(CN)<sub>2<\/sub> is produced?<\/p>\n<p>5. Carborundum is silicon carbide, SiC, a very hard material used as an abrasive on sandpaper and in other applications. It is prepared by the reaction of pure sand, SiO<sub>2<\/sub>, with carbon at high temperature. Carbon monoxide, CO, is the other product of this reaction. Write the balanced equation for the reaction, and calculate how much SiO<sub>2<\/sub> is required to produce 3.00 kg of SiC.<\/p>\n<p>6. Urea, CO(NH<sub>2<\/sub>)<sub>2<\/sub>, is manufactured on a large scale for use in producing urea-formaldehyde plastics and as a fertilizer. What is the maximum mass of urea that can be manufactured from the CO<sub>2<\/sub> produced by combustion of 1.00\u00d7103kg1.00\u00d7103kg of carbon followed by the reaction?<br \/>\n[latex]\\text{CO}_2(g) + 2\\text{NH}_3(g) \\longrightarrow {\\text{CO(NH}_2})_2(s) + \\text{H}_2 \\text{O}(l)[\/latex]<\/p>\n<p>7. A compact car gets 37.5 miles per gallon on the highway. If gasoline contains 84.2% carbon by mass and has a density of 0.8205 g\/mL, determine the mass of carbon dioxide produced during a 500-mile trip (3.785 liters per gallon).<\/p>\n<p>8. What volume of a 0.2089 <em>M<\/em> KI solution contains enough KI to react exactly with the Cu(NO<sub>3<\/sub>)<sub>2<\/sub> in 43.88 mL of a 0.3842 <em>M<\/em> solution of Cu(NO<sub>3<\/sub>)<sub>2<\/sub>?[latex]2 \\text{Cu(NO}_3)_2 + 4\\text{KI} \\longrightarrow 2\\text{CuI} + \\text{I}_2 + 4{\\text{KNO}_3}[\/latex]<\/p>\n<p>9. The toxic pigment called white lead, Pb<sub>3<\/sub>(OH)<sub>2<\/sub>(CO<sub>3<\/sub>)<sub>2<\/sub>, has been replaced in white paints by rutile, TiO<sub>2<\/sub>. How much rutile (g) can be prepared from 379 g of an ore that contains 88.3% ilmenite (FeTiO<sub>3<\/sub>) by mass?[latex]2\\text{FeTiO}_3 + 4\\text{HCl} + \\text{Cl}_2 \\longrightarrow 2\\text{FeCl}_3 + 2\\text{TiO}_2 + 2\\text{H}_2 \\text{O}[\/latex]<\/p>\n<p><span style=\"font-size: 1em\">10. Think back to the pound cake recipe. What possible conversion factors can you construct relating the components of the recipe?<\/span><\/p>\n<p><span style=\"font-size: 1em\">11. What are all the conversion factors that can be constructed from the balanced chemical reaction 2H<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">(g) +\u00a0O<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">(g) [latex]\\longrightarrow[\/latex]\u00a02H<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">O(\u2113)?<\/span><\/p>\n<p><span style=\"font-size: 1em\">12. Given the chemical equation<\/span><\/p>\n<div class=\"qandaset block\" id=\"ball-ch05_s01_qs01\">\n<div class=\"question\">\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">Na(s) +\u00a0H<sub class=\"subscript\">2<\/sub>O(\u2113) <span style=\"font-size: 1em\">[latex]\\longrightarrow[\/latex]<\/span>\u00a0NaOH(aq) +\u00a0H<sub class=\"subscript\">2<\/sub>(g)<\/span><\/span><\/p>\n<\/div>\n<p>a) \u00a0Balance the equation.<\/p>\n<p>b) \u00a0How many molecules of H<sub class=\"subscript\">2<\/sub> are produced when 332 atoms of Na react?<\/p>\n<div class=\"question\"><span style=\"font-size: 1em\">13. For the balanced chemical equation<\/span><\/div>\n<div class=\"question\">\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">6 H<sup class=\"superscript\">+<\/sup>(aq) +\u00a02 MnO<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a05 H<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">2<\/sub>(\u2113) <span style=\"font-size: 1em\">[latex]\\longrightarrow[\/latex]<\/span>\u00a02 Mn<sup class=\"superscript\">2+<\/sup>(aq) +\u00a05 O<sub class=\"subscript\">2<\/sub>(g) +\u00a08 H<sub class=\"subscript\">2<\/sub>O(\u2113)<\/span><\/span><\/p>\n<p id=\"ball-ch05_s01_qs01_p12\" class=\"para\">how many molecules of H<sub class=\"subscript\">2<\/sub>O are produced when 75 molecules of H<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">2<\/sub> react?<\/p>\n<\/div>\n<div class=\"question\"><span style=\"font-size: 1em\">14. \u00a0Given the balanced chemical equation<\/span><\/div>\n<div class=\"question\">\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">Fe<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub>(s) +\u00a03SO<sub class=\"subscript\">3<\/sub>(g) <span style=\"font-size: 1em\">[latex]\\longrightarrow[\/latex]<\/span>\u00a0Fe<sub class=\"subscript\">2<\/sub>(SO<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">3<\/sub><\/span><\/span><\/p>\n<p id=\"ball-ch05_s01_qs01_p18\" class=\"para\">how many molecules of Fe<sub class=\"subscript\">2<\/sub>(SO<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">3<\/sub> are produced if 321 atoms of S are reacted?<\/p>\n<\/div>\n<div class=\"question\"><span style=\"font-size: 1em\">15. \u00a0For the balanced chemical equation<\/span><\/div>\n<div class=\"question\">\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">Fe<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub>(s) +\u00a03 SO<sub class=\"subscript\">3<\/sub>(g) <span style=\"font-size: 1em\">[latex]\\longrightarrow[\/latex]<\/span>\u00a0Fe<sub class=\"subscript\">2<\/sub>(SO<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">3<\/sub><\/span><\/span><\/p>\n<p id=\"ball-ch05_s01_qs01_p24\" class=\"para\">suppose we need to make 145,000 molecules of Fe<sub class=\"subscript\">2<\/sub>(SO<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">3<\/sub>. How many molecules of SO<sub class=\"subscript\">3<\/sub> do we need?<\/p>\n<\/div>\n<div class=\"question\"><span style=\"font-size: 1em\">16. \u00a0Construct the three independent conversion factors possible for these two reactions:<\/span><\/div>\n<div class=\"question\">\n<p style=\"text-align: center\">2 H<sub class=\"subscript\">2<\/sub> +\u00a0O<sub class=\"subscript\">2<\/sub>\u00a0<span style=\"font-size: 1em\">[latex]\\longrightarrow[\/latex]<\/span> 2 H<sub class=\"subscript\">2<\/sub>O<\/p>\n<p style=\"text-align: center\">H<sub class=\"subscript\">2<\/sub> +\u00a0O<sub class=\"subscript\">2<\/sub>\u00a0<span style=\"font-size: 1em\">[latex]\\longrightarrow[\/latex]<\/span> H<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">2<\/sub><\/p>\n<p id=\"ball-ch05_s01_qs01_p30\" class=\"para\">Why are the ratios between H<sub class=\"subscript\">2<\/sub> and O<sub class=\"subscript\">2<\/sub> different?<\/p>\n<p id=\"ball-ch05_s01_qs01_p31\" class=\"para\">The conversion factors are different because the stoichiometries of the balanced chemical reactions are different.<\/p>\n<p class=\"para\"><span style=\"font-size: 1em\">17. What mass of CO<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\"> is produced by the combustion of 1.00 mol of CH<\/span><sub class=\"subscript\">4<\/sub><span style=\"font-size: 1em\">?<\/span><\/p>\n<\/div>\n<div class=\"question\">\n<div class=\"qandaset block\" id=\"ball-ch05_s04_qs01\">\n<div class=\"question\">\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">CH<sub class=\"subscript\">4<\/sub>(g) +\u00a02 O<sub class=\"subscript\">2<\/sub>(g) <span style=\"font-size: 1em\">[latex]\\longrightarrow[\/latex]<\/span>\u00a0CO<sub class=\"subscript\">2<\/sub>(g) +\u00a02 H<sub class=\"subscript\">2<\/sub>O(\u2113)<\/span><\/span><\/p>\n<p><span style=\"font-size: 1em\">18. What mass of HgO is required to produce 0.692 mol of O<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">?<\/span><\/p>\n<\/div>\n<div class=\"question\">\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">2 HgO(s) <span style=\"font-size: 1em\">[latex]\\longrightarrow[\/latex]<\/span>\u00a02 Hg(\u2113) +\u00a0O<sub class=\"subscript\">2<\/sub>(g)<\/span><\/span><\/p>\n<p><span style=\"font-size: 1em\">19. How many moles of Al can be produced from 10.87 g of Ag?<\/span><\/p>\n<\/div>\n<div class=\"question\">\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">Al(NO<sub class=\"subscript\">3<\/sub>) <sub class=\"subscript\">3<\/sub>(s) +\u00a03 Ag <span style=\"font-size: 1em\">[latex]\\longrightarrow[\/latex]<\/span>\u00a0Al +\u00a03 AgNO<sub class=\"subscript\">3<\/sub><\/span><\/span><\/p>\n<p><span style=\"font-size: 1em\">20. How many moles of O<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\"> are needed to prepare 1.00 g of Ca(NO<\/span><sub class=\"subscript\">3<\/sub><span style=\"font-size: 1em\">)<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">?<\/span><\/p>\n<\/div>\n<div class=\"question\">\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">Ca(s) +\u00a0N<sub class=\"subscript\">2<\/sub>(g) +\u00a03 O<sub class=\"subscript\">2<\/sub>(g) <span style=\"font-size: 1em\">[latex]\\longrightarrow[\/latex]<\/span>\u00a0Ca(NO<sub class=\"subscript\">3<\/sub>) <sub class=\"subscript\">2<\/sub>(s)<\/span><\/span><\/p>\n<p><span style=\"font-size: 1em\">21. What mass of O<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\"> can be generated by the decomposition of 100.0 g of NaClO<\/span><sub class=\"subscript\">3<\/sub><span style=\"font-size: 1em\">?<\/span><\/p>\n<\/div>\n<div class=\"question\">\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">2 NaClO<sub class=\"subscript\">3<\/sub>\u00a0<span style=\"font-size: 1em\">[latex]\\longrightarrow[\/latex]<\/span> 2 NaCl(s) +\u00a03 O<sub class=\"subscript\">2<\/sub>(g)<\/span><\/span><\/p>\n<p><span style=\"font-size: 1em\">22. What mass of Fe<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">O<\/span><sub class=\"subscript\">3<\/sub><span style=\"font-size: 1em\"> must be reacted to generate 324 g of Al<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">O<\/span><sub class=\"subscript\">3<\/sub><span style=\"font-size: 1em\">?<\/span><\/p>\n<\/div>\n<div class=\"question\">\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">Fe<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub>(s) +\u00a02 Al(s) <span style=\"font-size: 1em\">[latex]\\longrightarrow[\/latex]<\/span>\u00a02 Fe(s) +\u00a0Al<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub>(s)<\/span><\/span><\/p>\n<p><span style=\"font-size: 1em\">23. What mass of MnO<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\"> is produced when 445 g of H<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">O are reacted?<\/span><\/p>\n<\/div>\n<div class=\"question\">\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">H<sub class=\"subscript\">2<\/sub>O(\u2113) +\u00a02 MnO<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a0Br<sup class=\"superscript\">\u2212<\/sup>(aq) <span style=\"font-size: 1em\">[latex]\\longrightarrow[\/latex]<\/span>\u00a0BrO<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a02 MnO<sub class=\"subscript\">2<\/sub>(s) +\u00a02 OH<sup class=\"superscript\">\u2212<\/sup>(aq)<\/span><\/span><\/p>\n<p><span style=\"font-size: 1em\">24. If 83.9 g of ZnO are formed, what mass of Mn<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">O<\/span><sub class=\"subscript\">3<\/sub><span style=\"font-size: 1em\"> is formed with it?<\/span><\/p>\n<\/div>\n<div class=\"question\">\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">Zn(s) +\u00a02 MnO<sub class=\"subscript\">2<\/sub>(s) <span style=\"font-size: 1em\">[latex]\\longrightarrow[\/latex]<\/span>\u00a0ZnO(s) +\u00a0Mn<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">3<\/sub>(s)<\/span><\/span><\/p>\n<p><span style=\"font-size: 1em\">25. If 88.4 g of CH<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">S are reacted, what mass of HF is produced?<\/span><\/p>\n<\/div>\n<div class=\"question\">\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">CH<sub class=\"subscript\">2<\/sub>S +\u00a06 F<sub class=\"subscript\">2<\/sub>\u00a0<span style=\"font-size: 1em\">[latex]\\longrightarrow[\/latex]<\/span> CF<sub class=\"subscript\">4<\/sub> +\u00a02 HF +\u00a0SF<sub class=\"subscript\">6<\/sub><\/span><\/span><\/p>\n<p><span style=\"font-size: 1em\">26. Express in mole terms what this chemical equation means.<\/span><\/p>\n<\/div>\n<\/div>\n<div class=\"qandaset block\" id=\"ball-ch05_s03_qs01\">\n<div class=\"question\">\n<p><span class=\"informalequation\"><span class=\"mathphrase\">CH<sub class=\"subscript\">4<\/sub> +\u00a02O<sub class=\"subscript\">2<\/sub>\u00a0<span style=\"font-size: 1em\">[latex]\\longrightarrow[\/latex]<\/span> CO<sub class=\"subscript\">2<\/sub> +\u00a02H<sub class=\"subscript\">2<\/sub>O<\/span><\/span><\/p>\n<p><span style=\"font-size: 1em\">27. How many molecules of each substance are involved in the equation in Exercise 1 if it is interpreted in terms of moles?<\/span><\/p>\n<p><span style=\"font-size: 1em\">28. For the chemical equation<\/span><\/p>\n<\/div>\n<div class=\"question\">\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">2 C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">6<\/sub> +\u00a07 O<sub class=\"subscript\">2<\/sub>\u00a0<span style=\"font-size: 1em\">[latex]\\longrightarrow[\/latex]<\/span> 4 CO<sub class=\"subscript\">2<\/sub> +\u00a06 H<sub class=\"subscript\">2<\/sub>O<\/span><\/span><\/p>\n<p id=\"ball-ch05_s03_qs01_p10\" class=\"para\">what equivalences can you write in terms of moles? Use the \u21d4 sign.<\/p>\n<p class=\"para\"><span style=\"font-size: 1em\">29. Write the balanced chemical reaction for the combustion of C<\/span><sub class=\"subscript\">5<\/sub><span style=\"font-size: 1em\">H<\/span><sub class=\"subscript\">12<\/sub><span style=\"font-size: 1em\"> (the products are CO<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\"> and H<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">O) and determine how many moles of H<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">O are formed when 5.8 mol of O<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\"> are reacted.<\/span><\/p>\n<p class=\"para\"><span style=\"font-size: 1em\">30. For the balanced chemical equation<\/span><\/p>\n<\/div>\n<div class=\"question\">\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">3 Cu(s) +\u00a02 NO<sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a08 H<sup class=\"superscript\">+<\/sup>(aq) <span style=\"font-size: 1em\">[latex]\\longrightarrow[\/latex]<\/span>\u00a03 Cu<sup class=\"superscript\">2+<\/sup>(aq) +\u00a04 H<sub class=\"subscript\">2<\/sub>O(\u2113) +\u00a02 NO(g)<\/span><\/span><\/p>\n<p id=\"ball-ch05_s03_qs01_p20\" class=\"para\">how many moles of Cu<sup class=\"superscript\">2+<\/sup> are formed when 55.7 mol of H<sup class=\"superscript\">+<\/sup> are reacted?<\/p>\n<p class=\"para\"><span style=\"font-size: 1em\">31. For the balanced chemical reaction<\/span><\/p>\n<\/div>\n<div class=\"question\">\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">4 NH<sub class=\"subscript\">3<\/sub>(g) +\u00a05 O<sub class=\"subscript\">2<\/sub>(g) <span style=\"font-size: 1em\">[latex]\\longrightarrow[\/latex]<\/span>\u00a04 NO(g) +\u00a06 H<sub class=\"subscript\">2<\/sub>O(\u2113)<\/span><\/span><\/p>\n<p id=\"ball-ch05_s03_qs01_p26\" class=\"para\">how many moles of H<sub class=\"subscript\">2<\/sub>O are produced when 0.669 mol of NH<sub class=\"subscript\">3<\/sub> react?<\/p>\n<p class=\"para\"><span style=\"font-size: 1em\">32. For the balanced chemical reaction<\/span><\/p>\n<\/div>\n<div class=\"question\">\n<p style=\"text-align: center\"><span class=\"informalequation\"><span class=\"mathphrase\">4 KO<sub class=\"subscript\">2<\/sub>(s) +\u00a02 CO<sub class=\"subscript\">2<\/sub>(g) <span style=\"font-size: 1em\">[latex]\\longrightarrow[\/latex]<\/span>\u00a02 K<sub class=\"subscript\">2<\/sub>CO<sub class=\"subscript\">3<\/sub>(s) +\u00a03 O<sub class=\"subscript\">2<\/sub>(g)<\/span><\/span><\/p>\n<p id=\"ball-ch05_s03_qs01_p32\" class=\"para\">determine the number of moles of both products formed when 6.88 mol of KO<sub class=\"subscript\">2<\/sub> react.<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<p><strong>Answers<\/strong><\/p>\n<p id=\"fs-idp75003744\">1. a) 0.435 mol Na, 0.217 mol Cl<sub>2<\/sub>, 15.4 g Cl<sub>2<\/sub><\/p>\n<p>b) 0.005780 mol HgO, 2.890 \u00d7 10<sup>\u22123<\/sup> mol O<sub>2<\/sub>, 9.248 \u00d7 10<sup>\u22122<\/sup> g O<sub>2<\/sub><\/p>\n<p>c) 8.00 mol NaNO<sub>3<\/sub>, 6.8 \u00d7 10<sup>2<\/sup> g NaNO<sub>3<\/sub><\/p>\n<p>d) 1665 mol CO<sub>2<\/sub>, 73.3 kg CO<sub>2<\/sub><\/p>\n<p>e) 18.86 mol CuO, 2.330 kg CuCO<sub>3<\/sub><\/p>\n<p>f) 0.4580 mol C<sub>2<\/sub>H<sub>4<\/sub>Br<sub>2<\/sub>, 86.05 g C<sub>2<\/sub>H<sub>4<\/sub>Br<sub>2<\/sub><\/p>\n<p id=\"fs-idp103920384\">2. a) 0.0686 mol Mg, 1.67 g Mg<\/p>\n<p>b) 2.701 \u00d7 10<sup>\u22123<\/sup> mol O<sub>2<\/sub>, 0.08644 g O<sub>2<\/sub><\/p>\n<p>c) 6.43 mol MgCO<sub>3<\/sub>, 542 g MgCO<sub>3<\/sub><\/p>\n<p>d) 713 mol H<sub>2<\/sub>O, 12.8 kg H<sub>2<\/sub>O<\/p>\n<p>e) 16.31 mol BaO<sub>2<\/sub>, 2762 g BaO<sub>2<\/sub><\/p>\n<p>f) 0.207 mol C<sub>2<\/sub>H<sub>4<\/sub>, 5.81 g C<sub>2<\/sub>H<sub>4<\/sub><\/p>\n<p id=\"fs-idp32310528\">3. a) [latex]\\text{volume HCl solution} \\longrightarrow \\text{mol HCl} \\longrightarrow \\text{mol GaCl}_3[\/latex]; b) 1.25 mol GaCl<sub>3<\/sub>, 2.2 \u00d7 10<sup>2<\/sup> g GaCl<sub>3<\/sub><\/p>\n<p id=\"fs-idp18257312\">4. a) 5.337 \u00d7 10<sup>22<\/sup> molecules \u00a0 \u00a0 \u00a0b) 10.41 g Zn(CN)<sub>2<\/sub><\/p>\n<p id=\"fs-idp53851968\">5. [latex]\\text{SiO}_2 + 3\\text{C} \\longrightarrow \\text{SiC} + 2\\text{CO}[\/latex], 4.50 kg SiO<sub>2<\/sub><\/p>\n<p id=\"fs-idp125512976\">6. 5.00 \u00d7 10<sup>3<\/sup> kg<\/p>\n<p id=\"fs-idp10693200\">7. 1.28 \u00d7 10<sup>5<\/sup> g CO<sub>2<\/sub><\/p>\n<p id=\"fs-idp74230176\">8. 161.40 mL KI solution<\/p>\n<p id=\"fs-idp100940480\">9. 176 g TiO<sub>2<\/sub><\/p>\n<p><span style=\"font-size: 1em\">10. <\/span><span style=\"font-size: 1em\"><\/span><\/p>\n<p><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/Screen-Shot-2014-07-17-at-10.50.06-AM.png\" style=\"font-size: 1em\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/Screen-Shot-2014-07-17-at-10.50.06-AM-1.png\" alt=\"Screen Shot 2014-07-17 at 10.50.06 AM\" class=\"alignnone wp-image-3511 size-full\" width=\"160\" height=\"40\" \/><\/a><span style=\"font-size: 1em\">\u00a0are two conversion factors that can be constructed from the pound cake recipe. Other conversion factors are also possible.<\/span><\/p>\n<div class=\"answer\">11.<\/div>\n<div class=\"answer\"><span style=\"font-size: 1em\"><\/span><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/Screen-Shot-2014-07-17-at-10.51.35-AM.png\" style=\"font-size: 1em\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/Screen-Shot-2014-07-17-at-10.51.35-AM-1.png\" alt=\"Screen Shot 2014-07-17 at 10.51.35 AM\" class=\"alignnone size-full wp-image-3512\" width=\"247\" height=\"36\" \/><\/a><span style=\"font-size: 1em\">\u00a0and their reciprocals are the conversion factors that can be constructed.<\/span><\/div>\n<div class=\"answer\">\n<p class=\"para\">12. \u00a02Na(s) +\u00a02H<sub class=\"subscript\">2<\/sub>O(\u2113) <span style=\"font-size: 1em\">[latex]\\longrightarrow[\/latex]<\/span>\u00a02NaOH(aq) +\u00a0H<sub class=\"subscript\">2<\/sub>(g) and 166 molecules<\/p>\n<\/div>\n<div class=\"answer\">13.\u00a0120 molecules<\/div>\n<div class=\"answer\">14. 107 molecules<\/div>\n<div class=\"answer\">15. 435,000 molecules<\/div>\n<div class=\"answer\">16.<\/div>\n<div class=\"answer\"><a href=\"http:\/\/opentextbc.ca\/introductorychemistry\/wp-content\/uploads\/sites\/17\/2014\/07\/Screen-Shot-2014-07-17-at-10.52.09-AM.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/Screen-Shot-2014-07-17-at-10.52.09-AM-1.png\" alt=\"Screen Shot 2014-07-17 at 10.52.09 AM\" class=\"alignnone wp-image-3513 size-full\" width=\"303\" height=\"63\" \/><\/a><\/div>\n<div>\n<p>17. 44.0 g<\/p>\n<p>18.\u00a03.00 \u00d7 10<sup class=\"superscript\">2<\/sup> g<\/p>\n<p>19.\u00a00.0336 mol<\/p>\n<p>20.\u00a00.0183 mol<\/p>\n<p>21.\u00a045.1 g<\/p>\n<p>22.\u00a0507 g<\/p>\n<p>23.\u00a04.30 \u00d7 10<sup class=\"superscript\">3<\/sup> g<\/p>\n<p>24.\u00a0163 g<\/p>\n<p>25.\u00a076.7 g<\/p>\n<p>26. One mole of CH<sub class=\"subscript\">4<\/sub> reacts with 2 mol of O<sub class=\"subscript\">2<\/sub> to make 1 mol of CO<sub class=\"subscript\">2<\/sub> and 2 mol of H<sub class=\"subscript\">2<\/sub>O.<\/p>\n<p>27.\u00a06.022 \u00d7 10<sup class=\"superscript\">23<\/sup> molecules of CH<sub class=\"subscript\">4<\/sub>, 1.2044 \u00d7 10<sup class=\"superscript\">24<\/sup> molecules of O<sub class=\"subscript\">2<\/sub>, 6.022 \u00d7 10<sup class=\"superscript\">23<\/sup> molecules of CO<sub class=\"subscript\">2<\/sub>, and 1.2044 \u00d7 10<sup class=\"superscript\">24<\/sup> molecules of H<sub class=\"subscript\">2<\/sub>O<\/p>\n<p>28.\u00a02 mol of C<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">6<\/sub>\u00a0to 7 mol of O<sub class=\"subscript\">2<\/sub>\u00a0to 4 mol of CO<sub class=\"subscript\">2<\/sub>\u00a0to 6 mol of H<sub class=\"subscript\">2<\/sub>O<\/p>\n<p>29.\u00a0C<sub class=\"subscript\">5<\/sub>H<sub class=\"subscript\">12<\/sub> +\u00a08 O<sub class=\"subscript\">2\u00a0<\/sub>\u00a0<span style=\"font-size: 1em\">[latex]\\longrightarrow[\/latex]<\/span> 5CO<sub class=\"subscript\">2<\/sub> +\u00a06H<sub class=\"subscript\">2<\/sub>O; 4.4 mol<\/p>\n<p>30.\u00a020.9 mol<\/p>\n<p>31.\u00a01.00 mol<\/p>\n<p>32.\u00a03.44 mol of K<sub class=\"subscript\">2<\/sub>CO<sub class=\"subscript\">3<\/sub>; 5.16 mol of O<sub class=\"subscript\">2<\/sub><\/p>\n<\/div>\n<\/div>\n<\/section>\n<div>\n<h2>Glossary<\/h2>\n<p><strong>stoichiometric factor:\u00a0<\/strong>ratio of coefficients in a balanced chemical equation, used in computations relating amounts of reactants and products<\/p>\n<p><strong>stoichiometry:\u00a0<\/strong>relationships between the amounts of reactants and products of a chemical reaction<\/p>\n<\/div>\n","protected":false},"author":330,"menu_order":2,"template":"","meta":{"pb_show_title":"on","pb_short_title":"7.1 Reaction Stoichiometry","pb_subtitle":"","pb_authors":[],"pb_section_license":"cc-by-nc-sa"},"chapter-type":[],"contributor":[],"license":[54],"class_list":["post-1429","chapter","type-chapter","status-publish","hentry","license-cc-by-nc-sa"],"part":1405,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/chapters\/1429","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\/1429\/revisions"}],"predecessor-version":[{"id":4758,"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/chapters\/1429\/revisions\/4758"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/parts\/1405"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/chapters\/1429\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/wp\/v2\/media?parent=1429"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/pressbooks\/v2\/chapter-type?post=1429"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/wp\/v2\/contributor?post=1429"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-json\/wp\/v2\/license?post=1429"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}