{"id":1439,"date":"2025-11-28T13:36:03","date_gmt":"2025-11-28T18:36:03","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/?post_type=chapter&#038;p=1439"},"modified":"2026-03-23T15:11:01","modified_gmt":"2026-03-23T19:11:01","slug":"2-mass-balance-stoichiommetry-calculations-2","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/chapter\/2-mass-balance-stoichiommetry-calculations-2\/","title":{"raw":"2. Mass Balance (Stoichiommetry) Calculations","rendered":"2. Mass Balance (Stoichiommetry) Calculations"},"content":{"raw":"Basic mass balance calculations go hand in hand with balancing chemical reactions. Just as the ability to balance chemical reactions is a crucial, basic skill for any process chemical industry, so is the ability to perform these calculations. The principle of conservation of mass is again the foundation.\r\n\r\nIf an ore consumes sulfuric acid, we can determine how much per tonne of ore based on the composition of the ore and the relevant chemical reactions. If we need oxygen to be fed into an [pb_glossary id=\"293\"]autoclave[\/pb_glossary] to treat a concentrate, we can calculate the flow rate required, based on the flow rate of concentrate, its composition and the pertinent chemical reactions. If we want to know how much copper metal we can produce from an ore, that depends on the amount of each copper mineral and the degree of reaction for each under specified conditions. These and myriad other mass balance questions can be addressed by straightforward calculations. The problem is that there is no single procedure available, because there is such a variety of types of calculations. Rather, we need to develop the basic principles and seek to apply them systematically. A degree of interpretation is needed to address each problem; to translate the information into a starting point for the calculations.\r\n\r\nFirst, amounts of matter can be measured in three basic ways: moles, mass and volume. In a chemical reaction like (in acid solution):\r\n\r\n\\[\\ce{Cu2S_{(s)} + \\frac{5}{2}O2_{(g)} + 2H+_{(aq)} = 2Cu^{2+}_{(aq)} + SO4^{2-}_{(aq)} + H2O_{(l)}} \\tag{115}\\]\r\n\r\n1 <em>mole<\/em> of Cu<sub>2<\/sub>S produces 2 <em>moles<\/em> of Cu<sup>+2<\/sup> and uses up <em>two<\/em> moles of H<sup>+<\/sup>. But, we do not buy or sell things in units of moles. Mostly we use mass and sometimes volume. How do we relate the mass of Cu<sup>+2<\/sup> formed to the mass of Cu<sub>2<\/sub>S available? This requires the atomic weights (denoted AW) of the elements in g\/mol, from which we can determine the formula weights of compounds or ions. Tables of atomic weights are available in materials engineering and chemistry texts.\r\n\r\n[table id=39 \/]\r\n\r\nFormula weight for Cu<sub>2<\/sub>S = 2 x 63.546 + 32.064 = 159.156 g\/mol. For every 1 kg of Cu<sub>2<\/sub>S we have:\r\n\r\n\\[ 1000\\ \\text{g Cu}_2\\text{S} \\times \\frac{1\\ \\text{mol Cu}_2\\text{S}}{159.156\\ \\text{g Cu}_2\\text{S}} = 6.28314\\ \\text{mol Cu}_2\\text{S} \\tag{116} \\]\r\n\r\nTo determine the amount of copper we can extract from 1000 g of Cu<sub>2<\/sub>S in kg we have to relate the moles of Cu<sup>+2<\/sup> to the moles of Cu<sub>2<\/sub>S, i.e. 1 mol Cu<sub>2<\/sub>S = 2 mol Cu<sup>+2<\/sup>. This yields:\r\n\r\n\\[\\ce{6.28314 mol Cu2S} \\times \\frac{\\ce{2 mol Cu^{+2}}}{\\ce{1 mol Cu2S}} \\times \\frac{\\ce{63.546 g Cu}}{\\ce{mol Cu^{+2}}} = \\ce{798.5 g Cu^{+2}} = \\ce{0.7985 kg} \\tag{117}\\]\r\n<p style=\"text-align: left\">This illustrates the point that at heart of all such calculations is a progression of conversions:<\/p>\r\n\\[\\text{mass} \\longrightarrow \\text{moles} \\longrightarrow \\dots \\longrightarrow \\text{moles} \\longrightarrow \\text{mass} \\tag{118}\\]\r\n\r\nHOWEVER, there are a number of things that must be borne in mind about this generalization if it is to be used correctly, i.e.\r\n<ul>\r\n \t<li>The calculation is begun on the left, with one of mass or moles units.<\/li>\r\n \t<li>There <em>might<\/em> be no moles conversions involved, i.e. mass to mass only.<\/li>\r\n \t<li>There <em>might<\/em> be one conversion to\/from moles required.<\/li>\r\n \t<li>There <em>might<\/em> be more than one moles to moles conversions required.<\/li>\r\n \t<li>The calculation sequence ends on the right, with one of moles or mass units.<\/li>\r\n \t<li>The following examples will serve to illustrate some of the points.<\/li>\r\n<\/ul>\r\nThe following examples will serve to illustrate some of the points.\r\n<h2>2.1 Examples<\/h2>\r\n<h1>Example 1<\/h1>\r\nAn ore contains 0.5% copper, in the form of Cu<sub>2<\/sub>S. Ore will be processed at a rate of 500,000 tonnes per year (t\/y). What is the maximum annual copper production rate in t\/y?\r\n<blockquote><strong>Note<\/strong>: 0.5% copper means 0.5% by weight, unless otherwise specified. Providing the copper content and the mineral form inevitably causes confusion. Students often assume that the question indicates that the ore contains 0.5% Cu<sub>2<\/sub>S. That is not what it says. It says, 0.5% Cu, i.e. 5 kg Cu\/tonne of ore, and that Cu is present as Cu<sub>2<\/sub>S. (In terms of Cu<sub>2<\/sub>S we have 0.626% Cu<sub>2<\/sub>S, which is equivalent to 0.5% Cu.)<\/blockquote>\r\nThe maximum copper production means that 100% of the Cu<sub>2<\/sub>S will be leached. Where do we start the calculations; mass or moles? Our data is in units of mass - 50 kg Cu\/t ore and 500,000 t ore\/y. On that basis we expect to start at the mass end of the sequence (118). Where do we end? We need Cu production in t\/y, i.e. mass units. Do we need to convert from mass to moles? We have mass Cu in ore and need to relate that to mass Cu produced. For both we have the same chemical species - Cu metal. Hence no moles conversions are needed. It is this sort of analysis that must always be done:\r\n<ul>\r\n \t<li>Does the data suggest mass or moles units for starting the calculations?<\/li>\r\n \t<li>Does the outcome require mass or moles units at the end of the calculations?<\/li>\r\n \t<li>Are conversions to\/from moles required?<\/li>\r\n<\/ul>\r\n\\[5\\ \\text{kg Cu} \\times \\frac{500{,}000\\ \\text{t ore}}{\\text{y}} \\times\\frac{10^{-3}\\ \\text{t Cu}}{\\text{kg Cu}} = 2500\\ \\text{t Cu\/y} \\tag{119}\\]\r\n<blockquote><strong>Note<\/strong>: Set up the calculation including the units, and include the chemical species in the units. Make sure the units cancel to get the correct units at the end. This is essential if mistakes are to be avoided.<\/blockquote>\r\n<h1>Example 2<\/h1>\r\nAssume conditions as in example 1 with the ore containing 0.5% copper, in the form of Cu<sub>2<\/sub>S. Ore will be processed at a rate of 500,000 tonnes per year (t\/y). Assume that 80% of the Cu<sub>2<\/sub>S is leached. How much sulfuric acid will be required in kg per tonne of ore? Refer to reaction (115).\r\n\r\nNow we introduce the idea of incomplete reaction: 80% of the Cu in Cu<sub>2<\/sub>S will dissolve. The remainder will not. Data is in mass units. Start at the mass end again. The answer is in mass units. Conclude at the mass end. We need to relate mass of Cu to mass of H<sub>2<\/sub>SO<sub>4<\/sub>. Conversions to\/from moles will be needed. Note that we need 2 mole H<sup>+<\/sup>\/mol Cu<sub>2<\/sub>S, and that 2 mol H<sup>+<\/sup> = 1 mol H<sub>2<\/sub>SO<sub>4<\/sub>. Also note that we have data in mass Cu\/t ore and the answer we want is mass H<sub>2<\/sub>SO<sub>4<\/sub>\/t ore. The \u201c\/t ore\u201d unit will follow through the calculations from start to finish. <em>Therefore we do not need the ore flow rate.<\/em>\r\n\r\n\\[\\begin{aligned}\r\n&amp;\\frac{\\ce{5 kg Cu}}{\\ce{t ore}} \\times \\frac{\\ce{1000 g Cu}}{\\ce{kg Cu}} \\times \\frac{\\ce{0.8 g Cu}}{\\ce{g Cu in ore}} \\times \\frac{\\ce{1 mol Cu}}{\\ce{63.546 g Cu}} \\times \\frac{\\ce{1 mol Cu2S}}{\\ce{2 mol Cu}} \\times \\frac{\\ce{1 mol H2SO4}}{\\ce{mol Cu2S}} \\\\\r\n&amp;\\times \\frac{\\ce{98.0775 g H2SO4}}{\\ce{mol H2SO4}} \\times \\frac{\\ce{10^{-3} kg H2SO4}}{\\ce{g H2SO4}} = \\ce{3.09 kg H2SO4 \/ t ore}\r\n\\end{aligned}\\tag{120}\\]\r\n\r\nThe 80% extent of Cu<sub>2<\/sub>S dissolution is expressed as [latex]\\frac{0.8\\,\\text{g Cu}_{\\text{aq}}}{\\text{g Cu in ore}}[\/latex], meaning 0.8 g Cu dissolved per 1 g Cu in the ore.\r\n<blockquote><strong> Note<\/strong>: Stoichiommetric factors (such as 1 mol Cu<sub>2<\/sub>S\/2 mol Cu) are set up so as to make sure the units cancel. To get mol Cu in the numerator to cancel, we use 1 mol Cu<sub>2<\/sub>S\/mol Cu, rather than the other way round. This way there is no need to try to think through which way round the stoichiommetric factors have to go. This can prevent errors.\r\n\r\n&nbsp;<\/blockquote>\r\n<h1>Example 3<\/h1>\r\nAssume conditions as in example 1 with the ore containing 0.5% copper, in the form of Cu<sub>2<\/sub>S and with 80% of it leached. Ore will be processed at a rate of 500,000 tonnes per year (t\/y). How much oxygen will be needed in m<sup>3<\/sup>\/h at STP? Refer to reaction (115).\r\n\r\n\u201cOxygen\u201d refers to molecular oxygen, i.e. O<sub>2<\/sub>. (You must know this.) STP refers to standard temperature and pressure (0\u00b0C, or 273.15 K and 1 atm pressure). The data again is in units of mass, as per question 1. The required answer has units of m<sup>3<\/sup>\/h <em>for a gas<\/em>. We will assume the ideal gas law,\r\n\r\n\\[PV = nRT \\tag{121}\\]\r\n\r\nThis relates pressure, volume and temperature to amount of gas in <em>moles<\/em>. Hence the end of calculation sequence (118)\u00a0must be in moles. And since we have mass data and a moles output, mass to moles conversion is going to be required. Further, the answer we need is in volume per unit time. In order to obtain this result we need data that also involves time<sup>-1<\/sup>. The piece of data we have is the ore flow rate in t\/y. Then, once we have an answer in moles\/h of O<sub>2<\/sub>, we can use the ideal gas law to obtain volumetric flow rate. We start with mass of copper. In order to use reaction (115), we need to convert to moles of Cu, to moles of Cu<sub>2<\/sub>S, to moles of O<sub>2<\/sub>, as per the reaction.\r\n\r\n\\[\\begin{aligned}&amp;\\frac{\\ce{5 kg Cu}}{\\ce{t ore}} \\times \\frac{\\ce{1000 g Cu}}{\\ce{kg Cu}} \\times \\frac{\\ce{0.8 g Cu_{aq}}}{\\ce{g Cu in ore}} \\times \\frac{\\ce{500000 t ore}}{\\ce{y}} \\times \\frac{\\ce{1 y}}{\\ce{365 d}} \\times \\frac{\\ce{1 d}}{\\ce{24 hr}} \\times \\frac{\\ce{1 mol Cu}}{\\ce{63.546 g Cu}} \\\\ &amp;\\times \\frac{\\ce{5\/2 mol O2}}{\\ce{2 mol Cu}} = \\ce{4.49105 x 10^{4} mol O2 \/ h} \\end{aligned}\\tag{122}\\]\r\n\r\nThen convert mole O<sub>2<\/sub>\/h to volume\/h using the ideal gas law:\r\n\r\n\\[\r\n\\begin{aligned}\r\n&amp;\\frac{\\ce{4.49105 x 10^{4} mol O2}}{\\ce{h}}\r\n\\times \\frac{\\ce{0.08206 L atm}}{\\ce{mol K}}\r\n\\times \\frac{\\ce{273.15 K}}{\\ce{1 atm}}\r\n\\times \\frac{\\ce{1 x 10^{-3} m^{3}}}{\\ce{L}}\r\n= \\ce{100.7 m^{3} O2 \/ h}\r\n\\end{aligned}\r\n\\tag{123}\r\n\\]\r\n\r\n&nbsp;\r\n\r\nThe calculations can be schematically represented as:\r\n\r\n\\[\\text{mass Cu} \\rightarrow \\text{moles Cu} \\rightarrow \\text{moles O}_{2} \\rightarrow \\text{volume O}_{2} \\tag{124}\\]\r\n<h1>Quiz<\/h1>\r\nA gold ore contains 4 g Au\/t ore (as gold metal) and 0.012% copper as CuO. The ore will be leached with sodium cyanide and 95% of the CuO will be leached and will form [Cu(CN)<sub>3<\/sub>]<sup>2-<\/sup>. Gold forms [Au(CN)<sub>2<\/sub>]<sup>-<\/sup>, and 90% of it will be leached. The leach slurry contains 40% solids. The ore is processed at a rate of 2000 t\/h. If the initial sodium cyanide concentration is 0.5 g\/L, what will it be after the leaching process? The solution density is 1.01 g\/mL. The reactions involved are:\r\n\r\n\\[\\ce{2CuO_{(s)} + 7CN^{-}_{(aq)} + H2O_{(l)} = 2[Cu(CN)3]^{2-}_{(aq)} + CNO^{-}_{(aq)} + 2OH^{-}_{(aq)}} \\tag{125}\\]\r\n\r\n\\[\\ce{2Au_{(s)} + \\frac{1}{2}O2_{(g)} + H2O_{(l)} + 4CN^{-}_{(aq)} = 2[Au(CN)2]^{-}_{(aq)} + 2OH^{-}_{(aq)}} \\tag{126}\\]\r\n\r\nThis question has a lot of data. 0.012% Cu as CuO means 0.12 kg Cu\/t of ore. Ultimately we need to relate concentrations of Cu and Au (in mass units) in the ore to a concentration of NaCN in g\/L. In addition, the concentrations for Cu and Au are given for the ore (a solid), while the concentration for NaCN is needed in solution. So we also have to relate the solids and the solution. This is given by the % solids in the slurry: 40% solids means that for 1000 kg of slurry, 400 kg is solids and 600 kg is solution, i.e. 400 kg solids\/600 kg solution. Using this we can relate <em>mass of solution<\/em> to mass of ore. But we will need a <em>volume of solution<\/em> to get the required NaCN concentration in g\/L. Hence we will need the solution density. Finally, relating the masses of Cu and Au to mass NaCN will require conversion to\/from moles, as per the reactions above. Note that the ore mass flow rate has not entered into these considerations and will not be needed.\r\n\r\n&nbsp;\r\n\r\n[h5p id=\"3\"]\r\n\r\n[h5p id=\"4\"]\r\n\r\n[h5p id=\"5\"]\r\n\r\n[h5p id=\"6\"]\r\n\r\n[h5p id=\"7\"]\r\n\r\n<span style=\"background-color: #ffff00\">We can summarize the calculations as follows:<\/span>\r\n\r\n<img class=\"pb-hover-zoom aligncenter wp-image-1585 size-full\" src=\"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/wp-content\/uploads\/sites\/1991\/2025\/12\/Ch0-1_Mass_Moles_Volume_Conversion_Diagram-e1773257029375.png\" alt=\"Diagram showing the dimensional-analysis steps for converting a substance\u2019s mass to moles, then moles of oxygen, then volume of oxygen. The top line shows mass multiplied by conversion factors to obtain moles of material. The second line shows moles of material multiplied by the molar ratio to obtain moles of O\u2082. The final arrow points to \u201cvolume of O\u2082,\u201d indicating the final step in the calculation.\" width=\"605\" height=\"211\" \/>\r\n\r\n&nbsp;","rendered":"<p>Basic mass balance calculations go hand in hand with balancing chemical reactions. Just as the ability to balance chemical reactions is a crucial, basic skill for any process chemical industry, so is the ability to perform these calculations. The principle of conservation of mass is again the foundation.<\/p>\n<p>If an ore consumes sulfuric acid, we can determine how much per tonne of ore based on the composition of the ore and the relevant chemical reactions. If we need oxygen to be fed into an <a class=\"glossary-term\" aria-haspopup=\"dialog\" aria-describedby=\"definition\" href=\"#term_1439_293\">autoclave<\/a> to treat a concentrate, we can calculate the flow rate required, based on the flow rate of concentrate, its composition and the pertinent chemical reactions. If we want to know how much copper metal we can produce from an ore, that depends on the amount of each copper mineral and the degree of reaction for each under specified conditions. These and myriad other mass balance questions can be addressed by straightforward calculations. The problem is that there is no single procedure available, because there is such a variety of types of calculations. Rather, we need to develop the basic principles and seek to apply them systematically. A degree of interpretation is needed to address each problem; to translate the information into a starting point for the calculations.<\/p>\n<p>First, amounts of matter can be measured in three basic ways: moles, mass and volume. In a chemical reaction like (in acid solution):<\/p>\n<p>\\[\\ce{Cu2S_{(s)} + \\frac{5}{2}O2_{(g)} + 2H+_{(aq)} = 2Cu^{2+}_{(aq)} + SO4^{2-}_{(aq)} + H2O_{(l)}} \\tag{115}\\]<\/p>\n<p>1 <em>mole<\/em> of Cu<sub>2<\/sub>S produces 2 <em>moles<\/em> of Cu<sup>+2<\/sup> and uses up <em>two<\/em> moles of H<sup>+<\/sup>. But, we do not buy or sell things in units of moles. Mostly we use mass and sometimes volume. How do we relate the mass of Cu<sup>+2<\/sup> formed to the mass of Cu<sub>2<\/sub>S available? This requires the atomic weights (denoted AW) of the elements in g\/mol, from which we can determine the formula weights of compounds or ions. Tables of atomic weights are available in materials engineering and chemistry texts.<\/p>\n<table id=\"tablepress-39\" class=\"tablepress tablepress-id-39\">\n<thead>\n<tr class=\"row-1\">\n<th class=\"column-1\">Element<\/th>\n<th class=\"column-2\">Atomic weight (g\/mol)<\/th>\n<\/tr>\n<\/thead>\n<tbody class=\"row-striping row-hover\">\n<tr class=\"row-2\">\n<td class=\"column-1\">Cu<\/td>\n<td class=\"column-2\">63.546<\/td>\n<\/tr>\n<tr class=\"row-3\">\n<td class=\"column-1\">S<\/td>\n<td class=\"column-2\">32.064<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><!-- #tablepress-39 from cache --><\/p>\n<p>Formula weight for Cu<sub>2<\/sub>S = 2 x 63.546 + 32.064 = 159.156 g\/mol. For every 1 kg of Cu<sub>2<\/sub>S we have:<\/p>\n<p>\\[ 1000\\ \\text{g Cu}_2\\text{S} \\times \\frac{1\\ \\text{mol Cu}_2\\text{S}}{159.156\\ \\text{g Cu}_2\\text{S}} = 6.28314\\ \\text{mol Cu}_2\\text{S} \\tag{116} \\]<\/p>\n<p>To determine the amount of copper we can extract from 1000 g of Cu<sub>2<\/sub>S in kg we have to relate the moles of Cu<sup>+2<\/sup> to the moles of Cu<sub>2<\/sub>S, i.e. 1 mol Cu<sub>2<\/sub>S = 2 mol Cu<sup>+2<\/sup>. This yields:<\/p>\n<p>\\[\\ce{6.28314 mol Cu2S} \\times \\frac{\\ce{2 mol Cu^{+2}}}{\\ce{1 mol Cu2S}} \\times \\frac{\\ce{63.546 g Cu}}{\\ce{mol Cu^{+2}}} = \\ce{798.5 g Cu^{+2}} = \\ce{0.7985 kg} \\tag{117}\\]<\/p>\n<p style=\"text-align: left\">This illustrates the point that at heart of all such calculations is a progression of conversions:<\/p>\n<p>\\[\\text{mass} \\longrightarrow \\text{moles} \\longrightarrow \\dots \\longrightarrow \\text{moles} \\longrightarrow \\text{mass} \\tag{118}\\]<\/p>\n<p>HOWEVER, there are a number of things that must be borne in mind about this generalization if it is to be used correctly, i.e.<\/p>\n<ul>\n<li>The calculation is begun on the left, with one of mass or moles units.<\/li>\n<li>There <em>might<\/em> be no moles conversions involved, i.e. mass to mass only.<\/li>\n<li>There <em>might<\/em> be one conversion to\/from moles required.<\/li>\n<li>There <em>might<\/em> be more than one moles to moles conversions required.<\/li>\n<li>The calculation sequence ends on the right, with one of moles or mass units.<\/li>\n<li>The following examples will serve to illustrate some of the points.<\/li>\n<\/ul>\n<p>The following examples will serve to illustrate some of the points.<\/p>\n<h2>2.1 Examples<\/h2>\n<h1>Example 1<\/h1>\n<p>An ore contains 0.5% copper, in the form of Cu<sub>2<\/sub>S. Ore will be processed at a rate of 500,000 tonnes per year (t\/y). What is the maximum annual copper production rate in t\/y?<\/p>\n<blockquote><p><strong>Note<\/strong>: 0.5% copper means 0.5% by weight, unless otherwise specified. Providing the copper content and the mineral form inevitably causes confusion. Students often assume that the question indicates that the ore contains 0.5% Cu<sub>2<\/sub>S. That is not what it says. It says, 0.5% Cu, i.e. 5 kg Cu\/tonne of ore, and that Cu is present as Cu<sub>2<\/sub>S. (In terms of Cu<sub>2<\/sub>S we have 0.626% Cu<sub>2<\/sub>S, which is equivalent to 0.5% Cu.)<\/p><\/blockquote>\n<p>The maximum copper production means that 100% of the Cu<sub>2<\/sub>S will be leached. Where do we start the calculations; mass or moles? Our data is in units of mass &#8211; 50 kg Cu\/t ore and 500,000 t ore\/y. On that basis we expect to start at the mass end of the sequence (118). Where do we end? We need Cu production in t\/y, i.e. mass units. Do we need to convert from mass to moles? We have mass Cu in ore and need to relate that to mass Cu produced. For both we have the same chemical species &#8211; Cu metal. Hence no moles conversions are needed. It is this sort of analysis that must always be done:<\/p>\n<ul>\n<li>Does the data suggest mass or moles units for starting the calculations?<\/li>\n<li>Does the outcome require mass or moles units at the end of the calculations?<\/li>\n<li>Are conversions to\/from moles required?<\/li>\n<\/ul>\n<p>\\[5\\ \\text{kg Cu} \\times \\frac{500{,}000\\ \\text{t ore}}{\\text{y}} \\times\\frac{10^{-3}\\ \\text{t Cu}}{\\text{kg Cu}} = 2500\\ \\text{t Cu\/y} \\tag{119}\\]<\/p>\n<blockquote><p><strong>Note<\/strong>: Set up the calculation including the units, and include the chemical species in the units. Make sure the units cancel to get the correct units at the end. This is essential if mistakes are to be avoided.<\/p><\/blockquote>\n<h1>Example 2<\/h1>\n<p>Assume conditions as in example 1 with the ore containing 0.5% copper, in the form of Cu<sub>2<\/sub>S. Ore will be processed at a rate of 500,000 tonnes per year (t\/y). Assume that 80% of the Cu<sub>2<\/sub>S is leached. How much sulfuric acid will be required in kg per tonne of ore? Refer to reaction (115).<\/p>\n<p>Now we introduce the idea of incomplete reaction: 80% of the Cu in Cu<sub>2<\/sub>S will dissolve. The remainder will not. Data is in mass units. Start at the mass end again. The answer is in mass units. Conclude at the mass end. We need to relate mass of Cu to mass of H<sub>2<\/sub>SO<sub>4<\/sub>. Conversions to\/from moles will be needed. Note that we need 2 mole H<sup>+<\/sup>\/mol Cu<sub>2<\/sub>S, and that 2 mol H<sup>+<\/sup> = 1 mol H<sub>2<\/sub>SO<sub>4<\/sub>. Also note that we have data in mass Cu\/t ore and the answer we want is mass H<sub>2<\/sub>SO<sub>4<\/sub>\/t ore. The \u201c\/t ore\u201d unit will follow through the calculations from start to finish. <em>Therefore we do not need the ore flow rate.<\/em><\/p>\n<p>\\[\\begin{aligned}<br \/>\n&amp;\\frac{\\ce{5 kg Cu}}{\\ce{t ore}} \\times \\frac{\\ce{1000 g Cu}}{\\ce{kg Cu}} \\times \\frac{\\ce{0.8 g Cu}}{\\ce{g Cu in ore}} \\times \\frac{\\ce{1 mol Cu}}{\\ce{63.546 g Cu}} \\times \\frac{\\ce{1 mol Cu2S}}{\\ce{2 mol Cu}} \\times \\frac{\\ce{1 mol H2SO4}}{\\ce{mol Cu2S}} \\\\<br \/>\n&amp;\\times \\frac{\\ce{98.0775 g H2SO4}}{\\ce{mol H2SO4}} \\times \\frac{\\ce{10^{-3} kg H2SO4}}{\\ce{g H2SO4}} = \\ce{3.09 kg H2SO4 \/ t ore}<br \/>\n\\end{aligned}\\tag{120}\\]<\/p>\n<p>The 80% extent of Cu<sub>2<\/sub>S dissolution is expressed as [latex]\\frac{0.8\\,\\text{g Cu}_{\\text{aq}}}{\\text{g Cu in ore}}[\/latex], meaning 0.8 g Cu dissolved per 1 g Cu in the ore.<\/p>\n<blockquote><p><strong> Note<\/strong>: Stoichiommetric factors (such as 1 mol Cu<sub>2<\/sub>S\/2 mol Cu) are set up so as to make sure the units cancel. To get mol Cu in the numerator to cancel, we use 1 mol Cu<sub>2<\/sub>S\/mol Cu, rather than the other way round. This way there is no need to try to think through which way round the stoichiommetric factors have to go. This can prevent errors.<\/p>\n<p>&nbsp;<\/p><\/blockquote>\n<h1>Example 3<\/h1>\n<p>Assume conditions as in example 1 with the ore containing 0.5% copper, in the form of Cu<sub>2<\/sub>S and with 80% of it leached. Ore will be processed at a rate of 500,000 tonnes per year (t\/y). How much oxygen will be needed in m<sup>3<\/sup>\/h at STP? Refer to reaction (115).<\/p>\n<p>\u201cOxygen\u201d refers to molecular oxygen, i.e. O<sub>2<\/sub>. (You must know this.) STP refers to standard temperature and pressure (0\u00b0C, or 273.15 K and 1 atm pressure). The data again is in units of mass, as per question 1. The required answer has units of m<sup>3<\/sup>\/h <em>for a gas<\/em>. We will assume the ideal gas law,<\/p>\n<p>\\[PV = nRT \\tag{121}\\]<\/p>\n<p>This relates pressure, volume and temperature to amount of gas in <em>moles<\/em>. Hence the end of calculation sequence (118)\u00a0must be in moles. And since we have mass data and a moles output, mass to moles conversion is going to be required. Further, the answer we need is in volume per unit time. In order to obtain this result we need data that also involves time<sup>-1<\/sup>. The piece of data we have is the ore flow rate in t\/y. Then, once we have an answer in moles\/h of O<sub>2<\/sub>, we can use the ideal gas law to obtain volumetric flow rate. We start with mass of copper. In order to use reaction (115), we need to convert to moles of Cu, to moles of Cu<sub>2<\/sub>S, to moles of O<sub>2<\/sub>, as per the reaction.<\/p>\n<p>\\[\\begin{aligned}&amp;\\frac{\\ce{5 kg Cu}}{\\ce{t ore}} \\times \\frac{\\ce{1000 g Cu}}{\\ce{kg Cu}} \\times \\frac{\\ce{0.8 g Cu_{aq}}}{\\ce{g Cu in ore}} \\times \\frac{\\ce{500000 t ore}}{\\ce{y}} \\times \\frac{\\ce{1 y}}{\\ce{365 d}} \\times \\frac{\\ce{1 d}}{\\ce{24 hr}} \\times \\frac{\\ce{1 mol Cu}}{\\ce{63.546 g Cu}} \\\\ &amp;\\times \\frac{\\ce{5\/2 mol O2}}{\\ce{2 mol Cu}} = \\ce{4.49105 x 10^{4} mol O2 \/ h} \\end{aligned}\\tag{122}\\]<\/p>\n<p>Then convert mole O<sub>2<\/sub>\/h to volume\/h using the ideal gas law:<\/p>\n<p>\\[<br \/>\n\\begin{aligned}<br \/>\n&amp;\\frac{\\ce{4.49105 x 10^{4} mol O2}}{\\ce{h}}<br \/>\n\\times \\frac{\\ce{0.08206 L atm}}{\\ce{mol K}}<br \/>\n\\times \\frac{\\ce{273.15 K}}{\\ce{1 atm}}<br \/>\n\\times \\frac{\\ce{1 x 10^{-3} m^{3}}}{\\ce{L}}<br \/>\n= \\ce{100.7 m^{3} O2 \/ h}<br \/>\n\\end{aligned}<br \/>\n\\tag{123}<br \/>\n\\]<\/p>\n<p>&nbsp;<\/p>\n<p>The calculations can be schematically represented as:<\/p>\n<p>\\[\\text{mass Cu} \\rightarrow \\text{moles Cu} \\rightarrow \\text{moles O}_{2} \\rightarrow \\text{volume O}_{2} \\tag{124}\\]<\/p>\n<h1>Quiz<\/h1>\n<p>A gold ore contains 4 g Au\/t ore (as gold metal) and 0.012% copper as CuO. The ore will be leached with sodium cyanide and 95% of the CuO will be leached and will form [Cu(CN)<sub>3<\/sub>]<sup>2-<\/sup>. Gold forms [Au(CN)<sub>2<\/sub>]<sup>&#8211;<\/sup>, and 90% of it will be leached. The leach slurry contains 40% solids. The ore is processed at a rate of 2000 t\/h. If the initial sodium cyanide concentration is 0.5 g\/L, what will it be after the leaching process? The solution density is 1.01 g\/mL. The reactions involved are:<\/p>\n<p>\\[\\ce{2CuO_{(s)} + 7CN^{-}_{(aq)} + H2O_{(l)} = 2[Cu(CN)3]^{2-}_{(aq)} + CNO^{-}_{(aq)} + 2OH^{-}_{(aq)}} \\tag{125}\\]<\/p>\n<p>\\[\\ce{2Au_{(s)} + \\frac{1}{2}O2_{(g)} + H2O_{(l)} + 4CN^{-}_{(aq)} = 2[Au(CN)2]^{-}_{(aq)} + 2OH^{-}_{(aq)}} \\tag{126}\\]<\/p>\n<p>This question has a lot of data. 0.012% Cu as CuO means 0.12 kg Cu\/t of ore. Ultimately we need to relate concentrations of Cu and Au (in mass units) in the ore to a concentration of NaCN in g\/L. In addition, the concentrations for Cu and Au are given for the ore (a solid), while the concentration for NaCN is needed in solution. So we also have to relate the solids and the solution. This is given by the % solids in the slurry: 40% solids means that for 1000 kg of slurry, 400 kg is solids and 600 kg is solution, i.e. 400 kg solids\/600 kg solution. Using this we can relate <em>mass of solution<\/em> to mass of ore. But we will need a <em>volume of solution<\/em> to get the required NaCN concentration in g\/L. Hence we will need the solution density. Finally, relating the masses of Cu and Au to mass NaCN will require conversion to\/from moles, as per the reactions above. Note that the ore mass flow rate has not entered into these considerations and will not be needed.<\/p>\n<p>&nbsp;<\/p>\n<div id=\"h5p-3\">\n<div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-3\" class=\"h5p-iframe\" data-content-id=\"3\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Ch0-1 2.1.1 Gold and Cyanide Example\"><\/iframe><\/div>\n<\/div>\n<div id=\"h5p-4\">\n<div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-4\" class=\"h5p-iframe\" data-content-id=\"4\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Ch0-1 2.1.2 Gold and Cyanide Example\"><\/iframe><\/div>\n<\/div>\n<div id=\"h5p-5\">\n<div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-5\" class=\"h5p-iframe\" data-content-id=\"5\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Ch0-1 2.1.3 Gold and Cyanide Example\"><\/iframe><\/div>\n<\/div>\n<div id=\"h5p-6\">\n<div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-6\" class=\"h5p-iframe\" data-content-id=\"6\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Ch0-1 2.1.4 Gold and Cyanide Example\"><\/iframe><\/div>\n<\/div>\n<div id=\"h5p-7\">\n<div class=\"h5p-iframe-wrapper\"><iframe id=\"h5p-iframe-7\" class=\"h5p-iframe\" data-content-id=\"7\" style=\"height:1px\" src=\"about:blank\" frameBorder=\"0\" scrolling=\"no\" title=\"Ch0-1 2.1.5 Gold and Cyanide Example\"><\/iframe><\/div>\n<\/div>\n<p><span style=\"background-color: #ffff00\">We can summarize the calculations as follows:<\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"pb-hover-zoom aligncenter wp-image-1585 size-full\" src=\"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/wp-content\/uploads\/sites\/1991\/2025\/12\/Ch0-1_Mass_Moles_Volume_Conversion_Diagram-e1773257029375.png\" alt=\"Diagram showing the dimensional-analysis steps for converting a substance\u2019s mass to moles, then moles of oxygen, then volume of oxygen. The top line shows mass multiplied by conversion factors to obtain moles of material. The second line shows moles of material multiplied by the molar ratio to obtain moles of O\u2082. The final arrow points to \u201cvolume of O\u2082,\u201d indicating the final step in the calculation.\" width=\"605\" height=\"211\" srcset=\"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/wp-content\/uploads\/sites\/1991\/2025\/12\/Ch0-1_Mass_Moles_Volume_Conversion_Diagram-e1773257029375.png 605w, https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/wp-content\/uploads\/sites\/1991\/2025\/12\/Ch0-1_Mass_Moles_Volume_Conversion_Diagram-e1773257029375-300x105.png 300w, https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/wp-content\/uploads\/sites\/1991\/2025\/12\/Ch0-1_Mass_Moles_Volume_Conversion_Diagram-e1773257029375-65x23.png 65w, https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/wp-content\/uploads\/sites\/1991\/2025\/12\/Ch0-1_Mass_Moles_Volume_Conversion_Diagram-e1773257029375-225x78.png 225w, https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/wp-content\/uploads\/sites\/1991\/2025\/12\/Ch0-1_Mass_Moles_Volume_Conversion_Diagram-e1773257029375-350x122.png 350w\" sizes=\"auto, (max-width: 605px) 100vw, 605px\" \/><\/p>\n<p>&nbsp;<\/p>\n<div class=\"media-attributions clear\" prefix:cc=\"http:\/\/creativecommons.org\/ns#\" prefix:dc=\"http:\/\/purl.org\/dc\/terms\/\"><h2>Media Attributions<\/h2><ul><li >Ch0-1_Mass_Moles_Volume_Conversion_Diagram  &copy;  B\u00e9 Wassink and Amir M. Dehkoda    is licensed under a  <a rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc\/4.0\/\">CC BY-NC (Attribution NonCommercial)<\/a> license<\/li><\/ul><\/div><div class=\"glossary\"><span class=\"screen-reader-text\" id=\"definition\">definition<\/span><template id=\"term_1439_293\"><div class=\"glossary__definition\" role=\"dialog\" data-id=\"term_1439_293\"><div tabindex=\"-1\"><p>An autoclave is a machine that uses steam under pressure to kill harmful bacteria, viruses, fungi, and spores on items that are placed inside a pressure vessel.<\/p>\n<\/div><button><span aria-hidden=\"true\">&times;<\/span><span class=\"screen-reader-text\">Close definition<\/span><\/button><\/div><\/template><\/div>","protected":false},"author":1076,"menu_order":2,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1439","chapter","type-chapter","status-publish","hentry"],"part":1126,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/wp-json\/pressbooks\/v2\/chapters\/1439","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/wp-json\/wp\/v2\/users\/1076"}],"version-history":[{"count":26,"href":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/wp-json\/pressbooks\/v2\/chapters\/1439\/revisions"}],"predecessor-version":[{"id":3856,"href":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/wp-json\/pressbooks\/v2\/chapters\/1439\/revisions\/3856"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/wp-json\/pressbooks\/v2\/parts\/1126"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/wp-json\/pressbooks\/v2\/chapters\/1439\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/wp-json\/wp\/v2\/media?parent=1439"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/wp-json\/pressbooks\/v2\/chapter-type?post=1439"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/wp-json\/wp\/v2\/contributor?post=1439"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/wp-json\/wp\/v2\/license?post=1439"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}