{"id":1685,"date":"2015-04-22T18:49:46","date_gmt":"2015-04-22T18:49:46","guid":{"rendered":"https:\/\/courses.candelalearning.com\/oschemtemp\/?post_type=chapter&#038;p=1685"},"modified":"2015-08-28T18:00:14","modified_gmt":"2015-08-28T18:00:14","slug":"reaction-stoichiometry-several-formula-issues","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-buffstate-chemistryformajorsxmaster\/chapter\/reaction-stoichiometry-several-formula-issues\/","title":{"raw":"Reaction Stoichiometry","rendered":"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\nA 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><span data-type=\"term\">stoichiometry<\/span><\/strong>, a term derived from the Greek words <em data-effect=\"italics\">stoicheion<\/em> (meaning \u201celement\u201d) and <em data-effect=\"italics\">metron<\/em> (meaning \u201cmeasure\u201d). In this module, the use of balanced chemical equations for various stoichiometric applications is explored.\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}[\/latex] cup milk, and one egg. The \u201cequation\u201d representing the preparation of pancakes per this recipe is<\/p>\r\n\r\n<div id=\"fs-idp116126928\" data-type=\"equation\">[latex]1\\text{cup mix}+\\frac{3}{4}\\text{cup milk}+1\\text{egg}\\rightarrow 8\\text{pancakes}[\/latex]<\/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 id=\"fs-idp58586688\" data-type=\"equation\">[latex]24\\cancel{\\text{pancakes}}\\times \\frac{1\\text{egg}}{8\\cancel{\\text{pancakes}}}=3\\text{eggs}[\/latex]<\/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><span data-type=\"term\">stoichiometric factors<\/span><\/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 id=\"fs-idp124603552\" data-type=\"equation\">[latex]{\\text{N}}_{2}\\text{(}g\\text{)}+3{\\text{H}}_{2}\\text{(}g\\text{)}\\rightarrow 2{\\text{NH}}_{3}\\text{(}g\\text{)}[\/latex]<\/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 id=\"fs-idp64965808\" data-type=\"equation\">[latex]\\frac{2{\\text{NH}}_{3}\\text{molecules}}{3{\\text{H}}_{2}\\text{molecules}}\\text{or}\\frac{\\text{2 doz}{\\text{NH}}_{3}\\text{molecules}}{\\text{3 doz}{\\text{H}}_{2}\\text{molecules}}\\text{or}\\frac{\\text{2 mol}{\\text{NH}}_{3}\\text{molecules}}{\\text{3 mol}{\\text{H}}_{2}\\text{molecules}}[\/latex]<\/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 id=\"fs-idp48900096\" data-type=\"example\">\r\n<div class=\"textbox shaded\">\r\n<h3>Example 1<\/h3>\r\n<h4 id=\"fs-idp77269456\"><strong><span data-type=\"title\">Moles of Reactant Required in a Reaction<\/span><\/strong><\/h4>\r\nHow many moles of I<sub>2<\/sub> are required to react with 0.429 mol of Al according to the following equation (see Figure 1)?\r\n<div id=\"fs-idp147980704\" data-type=\"equation\">[latex]2\\text{Al}+3{\\text{I}}_{2}\\rightarrow 2{\\text{AlI}}_{3}[\/latex]<span data-type=\"media\" data-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.\"><span data-type=\"media\" data-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.\">\r\n<\/span><\/span>\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"879\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211233\/CNX_Chem_04_03_iodine1.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=\"879\" height=\"164\" data-media-type=\"image\/jpeg\" \/> Figure 1. 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<\/div>\r\n<h4 id=\"fs-idm34598272\"><span data-type=\"title\">Solution<\/span><\/h4>\r\nReferring to the balanced chemical equation, the stoichiometric factor relating the two substances of interest is [latex]\\frac{\\text{3 mol}{\\text{I}}_{2}}{\\text{2 mol Al.}}[\/latex] The molar amount of iodine is derived by multiplying the provided molar amount of aluminum by this factor:\r\n\r\n<span id=\"fs-idp136092800\" data-type=\"media\" data-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\">\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211234\/CNX_Chem_04_03_moleratio1_img1.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\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n<div id=\"fs-idp102857120\" data-type=\"equation\">[latex]\\begin{array}{ll}\\hfill {\\text{mol I}}_{2}&amp; =0.429\\cancel{\\text{mol Al}}\\times \\frac{\\text{3 mol}{\\text{I}}_{2}}{2\\cancel{\\text{mol Al}}}\\\\ &amp; =\\text{0.644 mol}{\\text{I}}_{2}\\end{array}[\/latex]<\/div>\r\n<h4 id=\"fs-idp30742160\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\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}{\\text{(}\\text{OH}\\text{)}}_{2}+2{\\text{H}}_{3}{\\text{PO}}_{4}\\rightarrow{\\text{Ca}}_{3}{\\text{(}{\\text{PO}}_{4}\\text{)}}_{2}+6{\\text{H}}_{2}\\text{O?}[\/latex]\r\n<div id=\"fs-idp5038208\" data-type=\"note\">\r\n<div data-type=\"title\"><\/div>\r\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer<\/strong>:\u00a02.04 mol<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 2<\/h3>\r\n<div id=\"fs-idp9124448\" data-type=\"example\">\r\n<h4 id=\"fs-idp157170976\"><strong><span data-type=\"title\">Number of Product Molecules Generated by a Reaction<\/span><\/strong><\/h4>\r\nHow many carbon dioxide molecules are produced when 0.75 mol of propane is combusted according to this equation?\r\n<div id=\"fs-idp116053696\" data-type=\"equation\">[latex]{\\text{C}}_{3}{\\text{H}}_{8}+5{\\text{O}}_{2}\\rightarrow 3{\\text{CO}}_{2}+4{\\text{H}}_{2}\\text{O}[\/latex]<\/div>\r\n<h4 id=\"fs-idp121952928\"><span data-type=\"title\">Solution<\/span><\/h4>\r\nThe approach here is the same as for Example 1, 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.\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 id=\"fs-idp114307520\" data-type=\"equation\">[latex]\\frac{\\text{3 mol}{\\text{CO}}_{2}}{\\text{1 mol}{\\text{C}}_{3}{\\text{H}}_{8}}\\text{.}[\/latex]<\/div>\r\n<p id=\"fs-idp83056800\">Using this stoichiometric factor, the provided molar amount of propane, and Avogadro\u2019s number,<\/p>\r\n<span id=\"fs-idp94375456\" data-type=\"media\" data-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\">\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211236\/CNX_Chem_04_03_moleratio2_img1.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\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n<div id=\"fs-idp52373104\" data-type=\"equation\">[latex]0.75\\cancel{\\text{mol}{\\text{C}}_{3}{\\text{H}}_{8}}\\times \\frac{3\\cancel{\\text{mol}{\\text{CO}}_{2}}}{1\\cancel{\\text{mol}{\\text{C}}_{3}{\\text{H}}_{8}}}\\times \\frac{6.022\\times {10}^{23}{\\text{CO}}_{2}\\text{molecules}}{\\cancel{\\text{mol}{\\text{CO}}_{2}}}=1.4\\times {10}^{24}{\\text{CO}}_{2}\\text{molecules}[\/latex]<\/div>\r\n<h4 id=\"fs-idm14771264\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\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:\r\n<div id=\"fs-idp166213360\" data-type=\"equation\">[latex]{\\text{(}{\\text{NH}}_{4}\\text{)}}_{2}{\\text{SO}}_{4}+\\text{Ca}{\\text{(}\\text{OH}\\text{)}}_{2}\\rightarrow 2{\\text{NH}}_{3}+{\\text{CaSO}}_{4}+2{\\text{H}}_{2}\\text{O.}[\/latex]<\/div>\r\n<div id=\"fs-idp15879648\" data-type=\"note\">\r\n<div data-type=\"title\"><\/div>\r\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer<\/strong>:\u00a04.8 \u00d7 10<sup>24<\/sup> NH<sub>3<\/sub> molecules<\/div>\r\n<\/div>\r\n<\/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 id=\"fs-idp113495744\" data-type=\"example\">\r\n<div class=\"textbox shaded\">\r\n<h3>Example\u00a03<\/h3>\r\n<div id=\"fs-idp113495744\" data-type=\"example\">\r\n<h4 id=\"fs-idp56844080\"><strong><span data-type=\"title\">Relating Masses of Reactants and Products<\/span><\/strong><\/h4>\r\nWhat 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?\r\n<div id=\"fs-idp158099664\" data-type=\"equation\">[latex]{\\text{MgCl}}_{2}\\text{(}aq\\text{)}+2\\text{NaOH}\\text{(}aq\\text{)}\\rightarrow\\text{Mg}{\\text{(}\\text{OH}\\text{)}}_{2}\\text{(}s\\text{)}+2\\text{NaCl}\\text{(}aq\\text{)}[\/latex]<\/div>\r\n<h4 id=\"fs-idm3520480\"><span data-type=\"title\">Solution<\/span><\/h4>\r\nThe approach used previously in Example 1 and Example 2 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:<span id=\"fs-idp38697232\" data-type=\"media\" data-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\">\r\n<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211237\/CNX_Chem_04_03_map2_img1.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=\"881\" height=\"391\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n<div id=\"fs-idp122145840\" data-type=\"equation\">[latex]16\\cancel{\\text{g Mg}{\\text{(}\\text{OH}\\text{)}}_{2}}\\times \\frac{1\\cancel{\\text{mol Mg}{\\text{(}\\text{OH}\\text{)}}_{2}}}{58.3\\cancel{\\text{g Mg}{\\text{(}\\text{OH}\\text{)}}_{2}}}\\times \\frac{2\\cancel{\\text{mol NaOH}}}{1\\cancel{\\text{mol Mg}{\\text{(}\\text{OH}\\text{)}}_{2}}}\\times \\frac{\\text{40.0 g NaOH}}{\\cancel{\\text{mol NaOH}}}=\\text{22 g NaOH}[\/latex]<\/div>\r\n<h4 id=\"fs-idp52759328\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\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}\\rightarrow 2{\\text{Ga}}_{2}{\\text{O}}_{3}\\text{.}[\/latex]\r\n<div id=\"fs-idp214975456\" data-type=\"note\">\r\n<div data-type=\"title\"><\/div>\r\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer<\/strong>:\u00a039.0 g<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 4<\/h3>\r\n<h4 id=\"fs-idp64260528\"><strong><span data-type=\"title\">Relating Masses of Reactants<\/span><\/strong><\/h4>\r\nWhat 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?\r\n<div id=\"fs-idp87206624\" data-type=\"equation\">[latex]2{\\text{C}}_{8}{\\text{H}}_{18}+25{\\text{O}}_{2}\\rightarrow 16{\\text{CO}}_{2}+18{\\text{H}}_{2}\\text{O}[\/latex]<\/div>\r\n<h4 id=\"fs-idp35508016\"><span data-type=\"title\">Solution<\/span><\/h4>\r\nThe approach required here is the same as for the Example 3, differing only in that the provided and requested masses are both for reactant species.<span id=\"fs-idp56087040\" data-type=\"media\" data-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\">\r\n<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211238\/CNX_Chem_04_03_map3_img1.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=\"878\" height=\"390\" data-media-type=\"image\/jpeg\" \/><\/span>\r\n<div id=\"fs-idm56417376\" data-type=\"equation\">[latex]702\\cancel{\\text{g}{\\text{C}}_{8}{\\text{H}}_{18}}\\times \\frac{1\\cancel{\\text{mol}{\\text{C}}_{8}{\\text{H}}_{18}}}{114.23\\cancel{\\text{g}{\\text{C}}_{8}{\\text{H}}_{18}}}\\times \\frac{25\\cancel{\\text{mol}{\\text{O}}_{2}}}{2\\cancel{\\text{mol}{\\text{C}}_{8}{\\text{H}}_{18}}}\\times \\frac{\\text{32.00 g}{\\text{O}}_{2}}{\\cancel{\\text{mol}{\\text{O}}_{2}}}=2.46\\times {10}^{3}\\text{g}{\\text{O}}_{2}[\/latex]<\/div>\r\n<h4 id=\"fs-idp24638432\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\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 [latex]{\\text{Fe}}_{2}{\\text{O}}_{3}+3\\text{CO}\\rightarrow 2\\text{Fe}+3{\\text{CO}}_{2}?[\/latex]\r\n<div id=\"fs-idp58164000\" data-type=\"note\">\r\n<div data-type=\"title\"><\/div>\r\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer<\/strong>:\u00a013.22 g<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\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. Figure 2 provides a general outline of the various computational steps associated with many reaction stoichiometry calculations.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"880\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211239\/CNX_Chem_04_03_flowchart1.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=\"880\" height=\"539\" data-media-type=\"image\/jpeg\" \/> Figure 2. The flow chart depicts the various computational steps involved in most reaction stoichiometry calculations.[\/caption]\r\n\r\n<div id=\"fs-idp84121360\" class=\"chemistry everyday-life\" data-type=\"note\">\r\n<div class=\"textbox shaded\" data-type=\"title\">\r\n<h3 style=\"text-align: left;\" data-type=\"title\">Airbags<\/h3>\r\n<p id=\"fs-idp124943600\">Airbags (Figure 3) 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 id=\"fs-idp113504400\" data-type=\"equation\">[latex]2{\\text{NaN}}_{3}\\text{(}s\\text{)}\\rightarrow 3{\\text{N}}_{2}\\text{(}g\\text{)}+2\\text{Na}\\text{(}s\\text{)}[\/latex]<\/div>\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). 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. For example, a small mass (~100 g) of NaN<sub>3<\/sub> will generate approximately 50 L of N<sub>2<\/sub>.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"501\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211241\/CNX_Chem_04_03_airbag1.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=\"501\" height=\"376\" data-media-type=\"image\/jpeg\" \/> Figure 3. Airbags deploy upon impact to minimize serious injuries to passengers. (credit: Jon Seidman)[\/caption]\r\n<p id=\"fs-idp15096240\">For more information about the chemistry and physics behind airbags and for helpful diagrams on how airbags work, <a href=\"http:\/\/auto.howstuffworks.com\/car-driving-safety\/safety-regulatory-devices\/airbag.htm\" target=\"_blank\">go to How Stuff Works' \"How Airbags Work\" article<\/a>.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<section id=\"fs-idp102563712\" class=\"summary\" data-depth=\"1\"><section id=\"fs-idm40478192\" class=\"summary\" data-depth=\"1\">\r\n<div id=\"post-355\" class=\"post-355 chapter type-chapter status-publish hentry type-1\">\r\n<div class=\"entry-content\">\r\n<div class=\"im_section\">\r\n<div class=\"im_section\">\r\n<div id=\"mclean-ch03_s01_s02_n01\" class=\"im_key_takeaways im_editable im_block\">\r\n<div class=\"bcc-box bcc-success\">\r\n<h2>Key Concepts and Summary<\/h2>\r\n<section>\r\n<div data-type=\"note\">\r\n<p id=\"fs-idp236281408\">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<\/div>\r\n<\/section><\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Chemistry End of Chapter Exercises<\/h3>\r\n<div id=\"fs-idp166618800\" data-type=\"exercise\">\r\n<div id=\"fs-idp148234256\" data-type=\"problem\">\r\n<ol>\r\n\t<li id=\"fs-idp87549088\">Write the balanced equation, then outline the steps necessary to determine the information requested in each of the following:\r\n<ol>\r\n\t<li>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.<\/li>\r\n\t<li>The number of moles and the mass of oxygen formed by the decomposition of 1.252 g of mercury(II) oxide.<\/li>\r\n\t<li>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.)<\/li>\r\n\t<li>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.<\/li>\r\n\t<li>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.)<\/li>\r\n\t<li><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211242\/CNX_Chem_04_03_etheneBr_img1.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\" data-media-type=\"image\/jpeg\" \/><\/li>\r\n<\/ol>\r\n<\/li>\r\n\t<li>Determine the number of moles and the mass requested for each reaction in Exercise 1.<\/li>\r\n\t<li>Write the balanced equation, then outline the steps necessary to determine the information requested in each of the following:\r\n<ol>\r\n\t<li>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>.<\/li>\r\n\t<li>The number of moles and the mass of oxygen formed by the decomposition of 1.252 g of silver(I) oxide.<\/li>\r\n\t<li>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.)<\/li>\r\n\t<li>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.<\/li>\r\n\t<li>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.)<\/li>\r\n\t<li><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211243\/CNX_Chem_04_03_ethene_img1.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.\" data-media-type=\"image\/jpeg\" \/><\/li>\r\n<\/ol>\r\n<\/li>\r\n\t<li>Determine the number of moles and the mass requested for each reaction in Exercise 3.<\/li>\r\n\t<li>H<sub>2<\/sub> is produced by the reaction of 118.5 mL of a 0.8775-M solution of H<sub>3<\/sub>PO<sub>4<\/sub> according to the following equation: [latex]2\\text{Cr}+2{\\text{H}}_{3}{\\text{PO}}_{4}\\rightarrow 3{\\text{H}}_{2}+2{\\text{CrPO}}_{4}\\text{.}[\/latex]\r\n<ol>\r\n\t<li>Outline the steps necessary to determine the number of moles and mass of H<sub>2<\/sub>.<\/li>\r\n\t<li>Perform the calculations outlined.<\/li>\r\n<\/ol>\r\n<\/li>\r\n\t<li>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}\\rightarrow 2{\\text{GaCl}}_{3}+3{\\text{H}}_{2}\\text{.}[\/latex]\r\n<ol>\r\n\t<li>Outline the steps necessary to determine the number of moles and mass of gallium chloride.<\/li>\r\n\t<li>Perform the calculations outlined.<\/li>\r\n<\/ol>\r\n<\/li>\r\n\t<li>I<sub>2<\/sub> is produced by the reaction of 0.4235 mol of CuCl<sub>2<\/sub> according to the following equation: [latex]2{\\text{CuCl}}_{2}+4\\text{KI}\\rightarrow 2\\text{CuI}+4\\text{KCl}+{\\text{I}}_{2}\\text{.}[\/latex]\r\n<ol>\r\n\t<li>How many molecules of I<sub>2<\/sub> are produced?<\/li>\r\n\t<li>What mass of I<sub>2<\/sub> is produced?<\/li>\r\n<\/ol>\r\n<\/li>\r\n\t<li>Silver is often extracted from ores as K[Ag(CN)<sub>2<\/sub>] and then recovered by the reaction [latex]2\\text{K}\\left[\\text{Ag}{\\text{(}\\text{CN}\\text{)}}_{2}\\right]\\text{(}aq\\text{)}+\\text{Zn}\\text{(}s\\text{)}\\rightarrow 2Ag\\text{(}s\\text{)}+\\text{Zn}{\\text{(}\\text{CN}\\text{)}}_{2}\\text{(}aq\\text{)}+2\\text{KCN}\\text{(}aq\\text{)}[\/latex] \u00a0(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>]? (b) What mass of Zn(CN)<sub>2<\/sub> is produced?<\/li>\r\n\t<li>What mass of silver oxide, Ag<sub>2<\/sub>O, is required to produce 25.0 g of silver sulfadiazine, AgC<sub>10<\/sub>H<sub>9<\/sub>N<sub>4<\/sub>SO<sub>2<\/sub>, from the reaction of silver oxide and sulfadiazine? [latex]2{\\text{C}}_{10}{\\text{H}}_{10}{\\text{N}}_{4}{\\text{SO}}_{2}+{\\text{Ag}}_{2}\\text{O}\\rightarrow 2{\\text{AgC}}_{10}{\\text{H}}_{9}{\\text{N}}_{4}{\\text{SO}}_{2}+{\\text{H}}_{2}\\text{O}[\/latex]<\/li>\r\n\t<li>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.<\/li>\r\n\t<li>Automotive air bags inflate when a sample of sodium azide, NaN<sub>3<\/sub>, is very rapidly decomposed.[latex]2{\\text{NaN}}_{3}\\text{(}s\\text{)}\\rightarrow 2\\text{Na}\\text{(}s\\text{)}+3{\\text{N}}_{2}\\text{(}g\\text{)}[\/latex]\u00a0What mass of sodium azide is required to produce 2.6 ft<sup>3<\/sup> (73.6 L) of nitrogen gas with a density of 1.25 g\/L?<\/li>\r\n\t<li>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 [latex]1.00\\times {10}^{3}\\text{kg}[\/latex] of carbon followed by the reaction? [latex]{\\text{CO}}_{2}\\text{(}g\\text{)}+2{\\text{NH}}_{3}\\text{(}g\\text{)}\\rightarrow\\text{CO}{\\text{(}{\\text{NH}}_{2}\\text{)}}_{2}\\text{(}s\\text{)}+{\\text{H}}_{2}\\text{O}\\text{(}l\\text{)}[\/latex]<\/li>\r\n\t<li>In an accident, a solution containing 2.5 kg of nitric acid was spilled. Two kilograms of Na<sub>2<\/sub>CO<sub>3<\/sub> was quickly spread on the area and CO<sub>2<\/sub> was released by the reaction. Was sufficient Na<sub>2<\/sub>CO<sub>3<\/sub> used to neutralize all of the acid?<\/li>\r\n\t<li>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).<\/li>\r\n\t<li>What volume of a 0.750 M solution of hydrochloric acid, a solution of HCl, can be prepared from the HCl produced by the reaction of 25.0 g of NaCl with an excess of sulfuric acid? [latex]\\text{NaCl}\\text{(}s\\text{)}+{\\text{H}}_{2}{\\text{SO}}_{4}\\text{(}l\\text{)}\\rightarrow\\text{HCl}\\text{(}g\\text{)}+{\\text{NaHSO}}_{4}\\text{(}s\\text{)}[\/latex]<\/li>\r\n\t<li>What volume of a 0.2089 M 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 M solution of Cu(NO<sub>3<\/sub>)<sub>2<\/sub>? [latex]2\\text{Cu}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}+4\\text{KI}\\rightarrow 2\\text{CuI}+{\\text{I}}_{2}+4{\\text{KNO}}_{3}[\/latex]<\/li>\r\n\t<li>A mordant is a substance that combines with a dye to produce a stable fixed color in a dyed fabric. Calcium acetate is used as a mordant. It is prepared by the reaction of acetic acid with calcium hydroxide.[latex]2{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}+\\text{Ca}{\\text{(}\\text{OH}\\text{)}}_{2}\\rightarrow\\text{Ca}{\\text{(}{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{)}}_{2}+2{\\text{H}}_{2}\\text{O}[\/latex] What mass of Ca(OH)<sub>2<\/sub> is required to react with the acetic acid in 25.0 mL of a solution having a density of 1.065 g\/mL and containing 58.0% acetic acid by mass?<\/li>\r\n\t<li>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}\\rightarrow 2{\\text{FeCl}}_{3}+2{\\text{TiO}}_{2}+2{\\text{H}}_{2}\\text{O}[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"post-355\" class=\"post-355 chapter type-chapter status-publish hentry type-1\">\r\n<div class=\"entry-content\">\r\n<div class=\"im_section\">\r\n<div class=\"im_section\">\r\n<div id=\"mclean-ch03_s01_s02_n01\" class=\"im_key_takeaways im_editable im_block\">\r\n<div class=\"bcc-box bcc-info\">\r\n<h4>Selected Answers<\/h4>\r\n2. (a)\u00a00.435 mol Na, 0.271 mol Cl<sub>2<\/sub>, 15.4 g Cl<sub>2<\/sub>;\r\n\r\n(b) 0.005780 mol HgO, 2.890 <span id=\"MathJax-Element-756-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-20861\" class=\"math\"><span id=\"MathJax-Span-20862\" class=\"mrow\"><span id=\"MathJax-Span-20863\" class=\"semantics\"><span id=\"MathJax-Span-20864\" class=\"mrow\"><span id=\"MathJax-Span-20865\" class=\"mrow\"><span id=\"MathJax-Span-20866\" class=\"mo\">\u00d7<\/span><\/span><\/span><\/span><\/span><\/span><\/span> 10<sup>\u22123<\/sup> mol O<sub>2<\/sub>, 9.248 <span id=\"MathJax-Element-757-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-20867\" class=\"math\"><span id=\"MathJax-Span-20868\" class=\"mrow\"><span id=\"MathJax-Span-20869\" class=\"semantics\"><span id=\"MathJax-Span-20870\" class=\"mrow\"><span id=\"MathJax-Span-20871\" class=\"mrow\"><span id=\"MathJax-Span-20872\" class=\"mo\">\u00d7<\/span><\/span><\/span><\/span><\/span><\/span><\/span> 10<sup>\u22122<\/sup> g O<sub>2<\/sub>;\r\n\r\n(c) 8.00 mol NaNO<sub>3<\/sub>, 6.8 <span id=\"MathJax-Element-758-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-20873\" class=\"math\"><span id=\"MathJax-Span-20874\" class=\"mrow\"><span id=\"MathJax-Span-20875\" class=\"semantics\"><span id=\"MathJax-Span-20876\" class=\"mrow\"><span id=\"MathJax-Span-20877\" class=\"mrow\"><span id=\"MathJax-Span-20878\" class=\"mo\">\u00d7<\/span><\/span><\/span><\/span><\/span><\/span><\/span> 10<sup>2<\/sup> g NaNO<sub>3<\/sub>;\r\n\r\n(d) 1665 mol CO<sub>2<\/sub>, 73.3 kg CO<sub>2<\/sub>;\r\n\r\n(e) 18.86 mol CuO, 2.330 kg CuCO<sub>3<\/sub>;\r\n\r\n(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>\r\n<div data-type=\"newline\">\r\n\r\n4. \u00a0(a) [latex]\\text{mol Mg}=5.00\\cancel{\\text{g HCl}}\\times \\frac{1\\cancel{\\text{mol HCl}}}{36.4606\\cancel{\\text{g}}}\\times \\frac{\\text{1 mol Mg}}{2\\cancel{\\text{mol HCl}}}=0.0686\\text{mol,}[\/latex] [latex]\\text{g Mg}=0.0686\\cancel{\\text{mol Mg}}\\times \\frac{\\text{24.305 g}}{1\\cancel{\\text{mol Mg}}}=\\text{1.67 g};[\/latex]\r\n<div data-type=\"newline\">(b) [latex]{\\text{mol O}}_{2}=1.252\\cancel{\\text{g}{\\text{Ag}}_{2}\\text{O}}\\times \\frac{1\\cancel{\\text{mol}{\\text{Ag}}_{2}\\text{O}}}{231.7358\\cancel{\\text{g}}}\\times \\frac{\\text{1 mol}{\\text{O}}_{2}}{2\\cancel{\\text{mol}{\\text{Ag}}_{2}\\text{O}}}=2.701\\times {10}^{-3}[\/latex], [latex]{\\text{g O}}_{2}=2.701\\times {10}^{-3}\\cancel{{\\text{mol O}}_{2}}\\times \\frac{\\text{31.9988 g}}{1\\cancel{\\text{mol}{\\text{O}}_{2}}}=0.08644\\text{g;}[\/latex]<\/div>\r\n<div data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">(c) [latex]{\\text{mol MgCO}}_{3}=283\\cancel{\\text{g}{\\text{CO}}_{2}}\\times \\frac{1\\cancel{\\text{mol}{\\text{CO}}_{2}}}{44.010\\cancel{\\text{g}}}\\times \\frac{\\text{1 mol}{\\text{MgCO}}_{3}}{1\\cancel{\\text{mol}{\\text{CO}}_{2}}}=6.43\\text{mol,}[\/latex] [latex]{\\text{g MgCO}}_{3}=6.43\\cancel{{\\text{mol MgCO}}_{3}}\\times \\frac{\\text{84.314 g}}{1\\cancel{{\\text{mol MgCO}}_{3}}}=542\\text{g;}[\/latex]<\/div>\r\n<div data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">(d) [latex]{\\text{mol H}}_{2}\\text{O}=2.00\\times {10}^{4}\\cancel{\\text{g}{\\text{C}}_{2}{\\text{H}}_{4}}\\times \\frac{1\\cancel{\\text{mol}{\\text{C}}_{2}{\\text{H}}_{2}}}{28.054\\cancel{\\text{g}}}\\times \\frac{\\text{1 mol}{\\text{H}}_{2}\\text{O}}{1\\cancel{\\text{mol}{\\text{C}}_{2}{\\text{H}}_{2}}}=713\\text{mol,}[\/latex] [latex]{\\text{g H}}_{2}\\text{O}=713\\cancel{\\text{mol}{\\text{H}}_{2}\\text{O}}\\times \\frac{18.01528\\cancel{\\text{g}}}{1\\cancel{\\text{mol}{\\text{H}}_{2}\\text{O}}}\\times \\frac{\\text{1 kg}}{1000\\cancel{\\text{g}}}=\\text{12.8 kg}[\/latex]<\/div>\r\n<div data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">(e) [latex]2.500\\cancel{\\text{kg BaO}}\\times \\frac{1000\\cancel{\\text{g BaO}}}{1\\cancel{\\text{kg BaO}}}\\times \\frac{1\\cancel{\\text{mol BaO}}}{153.326\\cancel{\\text{g BaO}}}\\times \\frac{\\text{2 mol}{\\text{BaO}}_{2}}{2\\cancel{\\text{mol BaO}}}=\\text{16.31 mol}{\\text{BaO}}_{2}[\/latex] [latex]16.31\\cancel{\\text{mol}{\\text{BaO}}_{2}}\\times \\frac{\\text{169.326 g}{\\text{BaO}}_{2}}{1\\cancel{\\text{mol}{\\text{BaO}}_{2}}}=\\text{2762 g}{\\text{BaO}}_{2}[\/latex]<\/div>\r\n<div data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">(f) [latex]9.55\\cancel{\\text{g}{\\text{C}}_{2}{\\text{H}}_{6}\\text{O}}\\times \\frac{1\\cancel{\\text{mol}{\\text{C}}_{2}{\\text{H}}_{6}\\text{O}}}{46.068\\cancel{\\text{g}{\\text{C}}_{2}{\\text{H}}_{6}\\text{O}}}\\times \\frac{\\text{1 mol}{\\text{C}}_{2}{\\text{H}}_{4}}{1\\cancel{\\text{mol}{\\text{C}}_{2}{\\text{H}}_{6}\\text{O}}}=\\text{0.207 mol}{\\text{C}}_{2}{\\text{H}}_{4}[\/latex] [latex]0.207\\cancel{\\text{mol}{\\text{C}}_{2}{\\text{H}}_{4}}\\times \\frac{\\text{28.053 g}{\\text{C}}_{2}{\\text{H}}_{4}}{1\\cancel{\\text{mol}{\\text{C}}_{2}{\\text{H}}_{4}}}=\\text{5.81 g}{\\text{C}}_{2}{\\text{H}}_{4}[\/latex]<\/div>\r\n<div data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">6. (a) [latex]\\text{volume HCl solution}\\rightarrow\\text{mol HCl}\\rightarrow{\\text{mol GaCl}}_{3};[\/latex]<\/div>\r\n<div data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">(b) [latex]2.6\\cancel{\\text{L HCl}}\\times \\frac{1.44\\cancel{\\text{mol HCl}}}{1\\cancel{\\text{L HCl}}}\\times \\frac{2\\cancel{\\text{mol}{\\text{GaCl}}_{3}}}{6\\cancel{\\text{mol HCl}}}\\times \\frac{\\text{180.079 g}{\\text{GaCl}}_{3}}{1\\cancel{\\text{mol}{\\text{GaCl}}_{3}}}=2.3\\times {10}^{2}\\text{g}{\\text{GaCl}}_{3}[\/latex]<\/div>\r\n<div data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">8.\u00a0The development requires the following: [latex]\\text{mass K}\\left[\\text{Ag}{\\text{(}\\text{CN}\\text{)}}_{2}\\right]\\rightarrow\\text{mol K}\\left[\\text{Ag}{\\text{(}\\text{CN}\\text{)}}_{2}\\right]\\rightarrow\\text{mol Zn}{\\text{(}\\text{CN}\\text{)}}_{2}\\rightarrow\\text{molecules of Zn}{\\text{(}\\text{CN}\\text{)}}_{2}\\text{g Zn}{\\text{(}\\text{CN}\\text{)}}_{2};[\/latex]<\/div>\r\n<div data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">(a) [latex]35.27\\cancel{\\text{g K}\\left[\\text{Ag}{\\text{(}\\text{CN}\\text{)}}_{2}\\right]}\\times \\frac{1\\cancel{\\text{mol K}\\left[\\text{Ag}{\\text{(}\\text{CN}\\text{)}}_{2}\\right]}}{199.002\\cancel{\\text{g K}\\left[\\text{Ag}{\\text{(}\\text{CN}\\text{)}}_{2}\\right]}}\\times \\frac{1\\cancel{\\text{mol Zn}{\\text{(}\\text{CN}\\text{)}}_{2}}}{2\\cancel{\\text{mol K}\\left[\\text{Ag}{\\text{(}\\text{CN}\\text{)}}_{2}\\right]}}\\times \\frac{6.022\\times {10}^{23}}{1\\cancel{\\text{mol Zn}{\\text{(}\\text{CN}\\text{)}}_{2}}}=5.337\\times {10}^{22}\\text{molecules}[\/latex]<\/div>\r\n<div data-type=\"newline\"><\/div>\r\n<div data-type=\"newline\">(b) [latex]5.337\\times {10}^{22}\\cancel{\\text{molecules}}\\times \\frac{1\\cancel{\\text{mol Zn}{\\text{(}\\text{CN}\\text{)}}_{2}}}{6.022\\times {10}^{23}\\cancel{\\text{molecules}}}\\times \\frac{\\text{117.43 g Zn}{\\text{(}\\text{CN}\\text{)}}_{2}}{1\\cancel{\\text{mol Zn}{\\text{(}\\text{CN}\\text{)}}_{2}}}=\\text{10.41 g Zn}{\\text{(}\\text{CN}\\text{)}}_{2}[\/latex]<\/div>\r\n<div data-type=\"newline\">\r\n\r\n&nbsp;\r\n<div data-type=\"newline\">\r\n\r\n10. [latex]{\\text{SiO}}_{2}+3\\text{C}\\rightarrow\\text{SiC}+2\\text{CO.}[\/latex] From the balanced equation, 1 mol of SiO<sub>2<\/sub> produces 1 mol of SiC. The unknown is the mass of SiO<sub>2<\/sub> required to produce 3.00 kg (3000 g) of SiC. To calculate the mass of SiO<sub>2<\/sub> required, determine the molar masses of SiO<sub>2<\/sub> and SiC. Then calculate the number of moles of SiC required, and through the mole relation of SiO<sub>2<\/sub> to SiC, find the mass of SiO<sub>2<\/sub> required. The conversions required are: [latex]\\text{g SiC}\\rightarrow\\text{mol SiC}\\rightarrow\\text{mol}{\\text{SiO}}_{2}\\rightarrow\\text{g}{\\text{SiO}}_{2}[\/latex]\r\n<div data-type=\"newline\">Molar masses: SiO<sub>2<\/sub> = 60.0843 g mol<sup>\u20131<\/sup>; SiC = 40.0955 g mol<sup>\u20131<\/sup><\/div>\r\n<div data-type=\"newline\">[latex]\\text{mass SiO}2=3000\\cancel{\\text{g SiC}}\\times \\frac{1\\cancel{\\text{mol SiC}}}{40.955\\cancel{\\text{g SiC}}}\\times \\frac{1\\cancel{\\text{mol}{\\text{SiO}}_{2}}}{1\\cancel{\\text{mol SiC}}}\\times \\frac{\\text{60.843 g}{\\text{SiO}}_{2}}{1\\cancel{\\text{mol}{\\text{SiO}}_{2}}}=\\text{4496 g}{\\text{SiO}}_{2}=\\text{4.50 kg}{\\text{SiO}}_{2}[\/latex]<\/div>\r\n<div data-type=\"newline\">\r\n\r\n&nbsp;\r\n<div data-type=\"newline\">\r\n\r\n12. Molar mass urea = 12.011 + 15.9994 + 2(14.0067) + 4(1.0079) = 60.054 g mol<sup>\u20131<\/sup>\r\n<div data-type=\"newline\">[latex]\\text{1 mol C}\\rightarrow 1{\\text{mol CO}}_{2}\\rightarrow 1\\text{mol urea}[\/latex]<\/div>\r\n<div data-type=\"newline\">[latex]\\begin{array}{ll}\\hfill \\text{mass urea}&amp; =1.00\\times {10}^{3}\\cancel{\\text{kg}}\\times \\frac{1000\\cancel{\\text{g}}}{\\cancel{\\text{kg}}}\\times \\frac{1\\cancel{\\text{mol C}}}{12.0\\cancel{\\text{g C}}}\\times \\frac{1\\cancel{\\text{mol urea}}}{1\\cancel{\\text{mol C}}}\\times \\frac{\\text{60.054 g urea}}{1\\cancel{\\text{mol urea}}}\\\\ &amp; =5.00\\times {10}^{6}\\text{g or}5.00\\times {10}^{3}\\text{kg}\\end{array}[\/latex]<\/div>\r\n<div data-type=\"newline\">\r\n\r\n&nbsp;\r\n<div data-type=\"newline\">\r\n\r\n14. \u00a0The balanced chemical equation is [latex]\\text{C}\\text{(}s\\text{)}+{\\text{O}}_{2}\\text{(}g\\text{)}\\rightarrow{\\text{CO}}_{2}\\text{(}g\\text{)}[\/latex]\r\n<div data-type=\"newline\">[latex]\\begin{array}{l}\\\\ \\\\ \\\\ 500\\cancel{\\text{miles}}\\times \\frac{1\\cancel{\\text{gallon}}}{37.5\\cancel{\\text{miles}}}\\times \\frac{3.785\\cancel{\\text{L}}}{1\\cancel{\\text{gallon}}}\\times \\frac{1000\\cancel{\\text{mL}}}{1\\cancel{\\text{L}}}\\times \\frac{0.8205\\cancel{\\text{g gas}}}{1\\cancel{\\text{mL}}\\text{gas}}\\times \\frac{84.2\\cancel{\\text{g C}}}{100\\cancel{\\text{g gas}}}\\times \\frac{1\\cancel{\\text{mol C}}}{12.01\\cancel{\\text{g C}}}\\\\ \\times \\frac{1\\cancel{\\text{mol}{\\text{CO}}_{2}}}{1\\cancel{\\text{mol C}}}\\times \\frac{\\text{44.01 g}{\\text{CO}}_{2}}{1\\cancel{\\text{mol}{\\text{CO}}_{2}}}=1.28\\times {10}^{5}\\text{g}{\\text{CO}}_{2}\\end{array}[\/latex]<\/div>\r\n<div data-type=\"newline\">\r\n\r\n&nbsp;\r\n<div data-type=\"newline\">\r\n\r\n16. Use molarity to convert. This solution involves the following steps:\r\n\r\nConverting the volume of KI to moles of KI\r\n\r\nConverting the moles of KI to moles of Cu(NO<sub>3<\/sub>)<sub>2<\/sub>\r\n\r\nConverting the moles of [latex]\\text{K}\\rightarrow\\text{Cu}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}[\/latex] to a volume of KI. Cu(NO<sub>3<\/sub>)<sub>2<\/sub> solution\r\n<div data-type=\"newline\">[latex]\\text{43.88 mL}\\times \\frac{\\text{1 L}}{\\text{1000 mL}}\\times \\frac{0.3842\\cancel{\\text{mol Cu}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}}}{\\text{1 L}}\\times \\frac{4\\cancel{\\text{mol KI}}}{2\\cancel{\\text{mol Cu}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}}}\\times \\frac{\\text{1 L KI}}{0.2089\\cancel{\\text{mol KI}}}=\\text{161.4 mL}[\/latex]<\/div>\r\n<div data-type=\"newline\">All of these steps can be shown together, as follows:<\/div>\r\n<div data-type=\"newline\">[latex]\\frac{\\text{43.88 mL Cu}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}}{1}\\times \\frac{\\text{0.3842 mol Cu}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}}{\\text{1000 mL Cu}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}}\\times \\frac{\\text{4 mol KI}}{\\text{2 mol Cu}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}}\\times \\frac{\\text{1000 mL KI}}{\\text{0.2089 mol KI}}=\\text{161.40 mL KI solution}[\/latex]<\/div>\r\n<div data-type=\"newline\">\r\n\r\n&nbsp;\r\n<div data-type=\"newline\">\r\n\r\n18. Find from worked example, check your learning problem\r\n<div data-type=\"newline\">[latex]\\text{mass of ilmenite}=379\\cancel{\\text{g ore}}\\times \\frac{\\text{0.883 g}{\\text{FeTiO}}_{3}}{1\\cancel{\\text{g ore}}}=\\text{334.6 g}{\\text{FeTiO}}_{3}[\/latex]<\/div>\r\n<div data-type=\"newline\">[latex]\\text{mass of rutile}=334.6\\cancel{\\text{g}{\\text{FeTiO}}_{3}}\\times \\frac{1\\cancel{\\text{mol}{\\text{FeTiO}}_{3}}}{151.7\\cancel{\\text{g}{\\text{FeTiO}}_{3}}}\\times \\frac{2\\cancel{\\text{mol}{\\text{TiO}}_{2}}}{2\\cancel{\\text{mol}{\\text{FeTiO}}_{3}}}\\times \\frac{\\text{79.88 g}{\\text{TiO}}_{2}}{1\\cancel{\\text{mol}{\\text{TiO}}_{2}}}=\\text{176 g}{\\text{TiO}}_{2}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section>\r\n<div class=\"bcc-box bcc-success\"><section id=\"glossary\">\r\n<h3>Glossary<\/h3>\r\n<div data-type=\"definition\">\r\n<div id=\"fs-idm8143856\" data-type=\"definition\">\r\n<p data-type=\"definition\"><strong><span data-type=\"term\">stoichiometric factor\r\n<\/span><\/strong>ratio of coefficients in a balanced chemical equation, used in computations relating amounts of reactants and products<\/p>\r\n\r\n<div id=\"fs-idp127462672\" data-type=\"definition\">\r\n\r\n<strong><span data-type=\"term\">stoichiometry\r\n<\/span><\/strong>relationships between the amounts of reactants and products of a chemical reaction\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section>","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>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><span data-type=\"term\">stoichiometry<\/span><\/strong>, a term derived from the Greek words <em data-effect=\"italics\">stoicheion<\/em> (meaning \u201celement\u201d) and <em data-effect=\"italics\">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 id=\"fs-idp116126928\" data-type=\"equation\">[latex]1\\text{cup mix}+\\frac{3}{4}\\text{cup milk}+1\\text{egg}\\rightarrow 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 id=\"fs-idp58586688\" data-type=\"equation\">[latex]24\\cancel{\\text{pancakes}}\\times \\frac{1\\text{egg}}{8\\cancel{\\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><span data-type=\"term\">stoichiometric factors<\/span><\/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 id=\"fs-idp124603552\" data-type=\"equation\">[latex]{\\text{N}}_{2}\\text{(}g\\text{)}+3{\\text{H}}_{2}\\text{(}g\\text{)}\\rightarrow 2{\\text{NH}}_{3}\\text{(}g\\text{)}[\/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 id=\"fs-idp64965808\" data-type=\"equation\">[latex]\\frac{2{\\text{NH}}_{3}\\text{molecules}}{3{\\text{H}}_{2}\\text{molecules}}\\text{or}\\frac{\\text{2 doz}{\\text{NH}}_{3}\\text{molecules}}{\\text{3 doz}{\\text{H}}_{2}\\text{molecules}}\\text{or}\\frac{\\text{2 mol}{\\text{NH}}_{3}\\text{molecules}}{\\text{3 mol}{\\text{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 id=\"fs-idp48900096\" data-type=\"example\">\n<div class=\"textbox shaded\">\n<h3>Example 1<\/h3>\n<h4 id=\"fs-idp77269456\"><strong><span data-type=\"title\">Moles of Reactant Required in a Reaction<\/span><\/strong><\/h4>\n<p>How many moles of I<sub>2<\/sub> are required to react with 0.429 mol of Al according to the following equation (see Figure 1)?<\/p>\n<div id=\"fs-idp147980704\" data-type=\"equation\">[latex]2\\text{Al}+3{\\text{I}}_{2}\\rightarrow 2{\\text{AlI}}_{3}[\/latex]<span data-type=\"media\" data-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.\"><span data-type=\"media\" data-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.\"><br \/>\n<\/span><\/span><\/p>\n<div style=\"width: 889px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211233\/CNX_Chem_04_03_iodine1.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=\"879\" height=\"164\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. 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)<\/p>\n<\/div>\n<\/div>\n<h4 id=\"fs-idm34598272\"><span data-type=\"title\">Solution<\/span><\/h4>\n<p>Referring to the balanced chemical equation, the stoichiometric factor relating the two substances of interest is [latex]\\frac{\\text{3 mol}{\\text{I}}_{2}}{\\text{2 mol Al.}}[\/latex] The molar amount of iodine is derived by multiplying the provided molar amount of aluminum by this factor:<\/p>\n<p><span id=\"fs-idp136092800\" data-type=\"media\" data-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\"><br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211234\/CNX_Chem_04_03_moleratio1_img1.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\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<div id=\"fs-idp102857120\" data-type=\"equation\">[latex]\\begin{array}{ll}\\hfill {\\text{mol I}}_{2}& =0.429\\cancel{\\text{mol Al}}\\times \\frac{\\text{3 mol}{\\text{I}}_{2}}{2\\cancel{\\text{mol Al}}}\\\\ & =\\text{0.644 mol}{\\text{I}}_{2}\\end{array}[\/latex]<\/div>\n<h4 id=\"fs-idp30742160\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\n<p>How 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}{\\text{(}\\text{OH}\\text{)}}_{2}+2{\\text{H}}_{3}{\\text{PO}}_{4}\\rightarrow{\\text{Ca}}_{3}{\\text{(}{\\text{PO}}_{4}\\text{)}}_{2}+6{\\text{H}}_{2}\\text{O?}[\/latex]<\/p>\n<div id=\"fs-idp5038208\" data-type=\"note\">\n<div data-type=\"title\"><\/div>\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer<\/strong>:\u00a02.04 mol<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 2<\/h3>\n<div id=\"fs-idp9124448\" data-type=\"example\">\n<h4 id=\"fs-idp157170976\"><strong><span data-type=\"title\">Number of Product Molecules Generated by a Reaction<\/span><\/strong><\/h4>\n<p>How many carbon dioxide molecules are produced when 0.75 mol of propane is combusted according to this equation?<\/p>\n<div id=\"fs-idp116053696\" data-type=\"equation\">[latex]{\\text{C}}_{3}{\\text{H}}_{8}+5{\\text{O}}_{2}\\rightarrow 3{\\text{CO}}_{2}+4{\\text{H}}_{2}\\text{O}[\/latex]<\/div>\n<h4 id=\"fs-idp121952928\"><span data-type=\"title\">Solution<\/span><\/h4>\n<p>The approach here is the same as for Example 1, 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 id=\"fs-idp114307520\" data-type=\"equation\">[latex]\\frac{\\text{3 mol}{\\text{CO}}_{2}}{\\text{1 mol}{\\text{C}}_{3}{\\text{H}}_{8}}\\text{.}[\/latex]<\/div>\n<p id=\"fs-idp83056800\">Using this stoichiometric factor, the provided molar amount of propane, and Avogadro\u2019s number,<\/p>\n<p><span id=\"fs-idp94375456\" data-type=\"media\" data-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\"><br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211236\/CNX_Chem_04_03_moleratio2_img1.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\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<div id=\"fs-idp52373104\" data-type=\"equation\">[latex]0.75\\cancel{\\text{mol}{\\text{C}}_{3}{\\text{H}}_{8}}\\times \\frac{3\\cancel{\\text{mol}{\\text{CO}}_{2}}}{1\\cancel{\\text{mol}{\\text{C}}_{3}{\\text{H}}_{8}}}\\times \\frac{6.022\\times {10}^{23}{\\text{CO}}_{2}\\text{molecules}}{\\cancel{\\text{mol}{\\text{CO}}_{2}}}=1.4\\times {10}^{24}{\\text{CO}}_{2}\\text{molecules}[\/latex]<\/div>\n<h4 id=\"fs-idm14771264\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\n<p>How 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 id=\"fs-idp166213360\" data-type=\"equation\">[latex]{\\text{(}{\\text{NH}}_{4}\\text{)}}_{2}{\\text{SO}}_{4}+\\text{Ca}{\\text{(}\\text{OH}\\text{)}}_{2}\\rightarrow 2{\\text{NH}}_{3}+{\\text{CaSO}}_{4}+2{\\text{H}}_{2}\\text{O.}[\/latex]<\/div>\n<div id=\"fs-idp15879648\" data-type=\"note\">\n<div data-type=\"title\"><\/div>\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer<\/strong>:\u00a04.8 \u00d7 10<sup>24<\/sup> NH<sub>3<\/sub> molecules<\/div>\n<\/div>\n<\/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 id=\"fs-idp113495744\" data-type=\"example\">\n<div class=\"textbox shaded\">\n<h3>Example\u00a03<\/h3>\n<div id=\"fs-idp113495744\" data-type=\"example\">\n<h4 id=\"fs-idp56844080\"><strong><span data-type=\"title\">Relating Masses of Reactants and Products<\/span><\/strong><\/h4>\n<p>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 id=\"fs-idp158099664\" data-type=\"equation\">[latex]{\\text{MgCl}}_{2}\\text{(}aq\\text{)}+2\\text{NaOH}\\text{(}aq\\text{)}\\rightarrow\\text{Mg}{\\text{(}\\text{OH}\\text{)}}_{2}\\text{(}s\\text{)}+2\\text{NaCl}\\text{(}aq\\text{)}[\/latex]<\/div>\n<h4 id=\"fs-idm3520480\"><span data-type=\"title\">Solution<\/span><\/h4>\n<p>The approach used previously in Example 1 and Example 2 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:<span id=\"fs-idp38697232\" data-type=\"media\" data-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\"><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211237\/CNX_Chem_04_03_map2_img1.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=\"881\" height=\"391\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<div id=\"fs-idp122145840\" data-type=\"equation\">[latex]16\\cancel{\\text{g Mg}{\\text{(}\\text{OH}\\text{)}}_{2}}\\times \\frac{1\\cancel{\\text{mol Mg}{\\text{(}\\text{OH}\\text{)}}_{2}}}{58.3\\cancel{\\text{g Mg}{\\text{(}\\text{OH}\\text{)}}_{2}}}\\times \\frac{2\\cancel{\\text{mol NaOH}}}{1\\cancel{\\text{mol Mg}{\\text{(}\\text{OH}\\text{)}}_{2}}}\\times \\frac{\\text{40.0 g NaOH}}{\\cancel{\\text{mol NaOH}}}=\\text{22 g NaOH}[\/latex]<\/div>\n<h4 id=\"fs-idp52759328\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\n<p>What 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}\\rightarrow 2{\\text{Ga}}_{2}{\\text{O}}_{3}\\text{.}[\/latex]<\/p>\n<div id=\"fs-idp214975456\" data-type=\"note\">\n<div data-type=\"title\"><\/div>\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer<\/strong>:\u00a039.0 g<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 4<\/h3>\n<h4 id=\"fs-idp64260528\"><strong><span data-type=\"title\">Relating Masses of Reactants<\/span><\/strong><\/h4>\n<p>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 id=\"fs-idp87206624\" data-type=\"equation\">[latex]2{\\text{C}}_{8}{\\text{H}}_{18}+25{\\text{O}}_{2}\\rightarrow 16{\\text{CO}}_{2}+18{\\text{H}}_{2}\\text{O}[\/latex]<\/div>\n<h4 id=\"fs-idp35508016\"><span data-type=\"title\">Solution<\/span><\/h4>\n<p>The approach required here is the same as for the Example 3, differing only in that the provided and requested masses are both for reactant species.<span id=\"fs-idp56087040\" data-type=\"media\" data-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\"><br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211238\/CNX_Chem_04_03_map3_img1.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=\"878\" height=\"390\" data-media-type=\"image\/jpeg\" \/><\/span><\/p>\n<div id=\"fs-idm56417376\" data-type=\"equation\">[latex]702\\cancel{\\text{g}{\\text{C}}_{8}{\\text{H}}_{18}}\\times \\frac{1\\cancel{\\text{mol}{\\text{C}}_{8}{\\text{H}}_{18}}}{114.23\\cancel{\\text{g}{\\text{C}}_{8}{\\text{H}}_{18}}}\\times \\frac{25\\cancel{\\text{mol}{\\text{O}}_{2}}}{2\\cancel{\\text{mol}{\\text{C}}_{8}{\\text{H}}_{18}}}\\times \\frac{\\text{32.00 g}{\\text{O}}_{2}}{\\cancel{\\text{mol}{\\text{O}}_{2}}}=2.46\\times {10}^{3}\\text{g}{\\text{O}}_{2}[\/latex]<\/div>\n<h4 id=\"fs-idp24638432\"><strong><span data-type=\"title\">Check Your Learning<\/span><\/strong><\/h4>\n<p>What mass of CO is required to react with 25.13 g of Fe<sub>2<\/sub>O<sub>3<\/sub> according to the equation [latex]{\\text{Fe}}_{2}{\\text{O}}_{3}+3\\text{CO}\\rightarrow 2\\text{Fe}+3{\\text{CO}}_{2}?[\/latex]<\/p>\n<div id=\"fs-idp58164000\" data-type=\"note\">\n<div data-type=\"title\"><\/div>\n<div style=\"text-align: right;\" data-type=\"title\"><strong>Answer<\/strong>:\u00a013.22 g<\/div>\n<\/div>\n<\/div>\n<\/div>\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. Figure 2 provides a general outline of the various computational steps associated with many reaction stoichiometry calculations.<\/p>\n<div style=\"width: 890px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211239\/CNX_Chem_04_03_flowchart1.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=\"880\" height=\"539\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. The flow chart depicts the various computational steps involved in most reaction stoichiometry calculations.<\/p>\n<\/div>\n<div id=\"fs-idp84121360\" class=\"chemistry everyday-life\" data-type=\"note\">\n<div class=\"textbox shaded\" data-type=\"title\">\n<h3 style=\"text-align: left;\" data-type=\"title\">Airbags<\/h3>\n<p id=\"fs-idp124943600\">Airbags (Figure 3) 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 id=\"fs-idp113504400\" data-type=\"equation\">[latex]2{\\text{NaN}}_{3}\\text{(}s\\text{)}\\rightarrow 3{\\text{N}}_{2}\\text{(}g\\text{)}+2\\text{Na}\\text{(}s\\text{)}[\/latex]<\/div>\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). 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. For example, a small mass (~100 g) of NaN<sub>3<\/sub> will generate approximately 50 L of N<sub>2<\/sub>.<\/p>\n<div style=\"width: 511px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211241\/CNX_Chem_04_03_airbag1.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=\"501\" height=\"376\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3. Airbags deploy upon impact to minimize serious injuries to passengers. (credit: Jon Seidman)<\/p>\n<\/div>\n<p id=\"fs-idp15096240\">For more information about the chemistry and physics behind airbags and for helpful diagrams on how airbags work, <a href=\"http:\/\/auto.howstuffworks.com\/car-driving-safety\/safety-regulatory-devices\/airbag.htm\" target=\"_blank\">go to How Stuff Works&#8217; &#8220;How Airbags Work&#8221; article<\/a>.<\/p>\n<\/div>\n<\/div>\n<section id=\"fs-idp102563712\" class=\"summary\" data-depth=\"1\">\n<section id=\"fs-idm40478192\" class=\"summary\" data-depth=\"1\">\n<div id=\"post-355\" class=\"post-355 chapter type-chapter status-publish hentry type-1\">\n<div class=\"entry-content\">\n<div class=\"im_section\">\n<div class=\"im_section\">\n<div id=\"mclean-ch03_s01_s02_n01\" class=\"im_key_takeaways im_editable im_block\">\n<div class=\"bcc-box bcc-success\">\n<h2>Key Concepts and Summary<\/h2>\n<section>\n<div data-type=\"note\">\n<p id=\"fs-idp236281408\">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<\/div>\n<\/section>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Chemistry End of Chapter Exercises<\/h3>\n<div id=\"fs-idp166618800\" data-type=\"exercise\">\n<div id=\"fs-idp148234256\" data-type=\"problem\">\n<ol>\n<li id=\"fs-idp87549088\">Write the balanced equation, then outline the steps necessary to determine the information requested in each of the following:\n<ol>\n<li>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.<\/li>\n<li>The number of moles and the mass of oxygen formed by the decomposition of 1.252 g of mercury(II) oxide.<\/li>\n<li>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.)<\/li>\n<li>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.<\/li>\n<li>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.)<\/li>\n<li><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211242\/CNX_Chem_04_03_etheneBr_img1.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\" data-media-type=\"image\/jpeg\" \/><\/li>\n<\/ol>\n<\/li>\n<li>Determine the number of moles and the mass requested for each reaction in Exercise 1.<\/li>\n<li>Write the balanced equation, then outline the steps necessary to determine the information requested in each of the following:\n<ol>\n<li>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>.<\/li>\n<li>The number of moles and the mass of oxygen formed by the decomposition of 1.252 g of silver(I) oxide.<\/li>\n<li>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.)<\/li>\n<li>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.<\/li>\n<li>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.)<\/li>\n<li><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/887\/2015\/04\/23211243\/CNX_Chem_04_03_ethene_img1.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.\" data-media-type=\"image\/jpeg\" \/><\/li>\n<\/ol>\n<\/li>\n<li>Determine the number of moles and the mass requested for each reaction in Exercise 3.<\/li>\n<li>H<sub>2<\/sub> is produced by the reaction of 118.5 mL of a 0.8775-M solution of H<sub>3<\/sub>PO<sub>4<\/sub> according to the following equation: [latex]2\\text{Cr}+2{\\text{H}}_{3}{\\text{PO}}_{4}\\rightarrow 3{\\text{H}}_{2}+2{\\text{CrPO}}_{4}\\text{.}[\/latex]\n<ol>\n<li>Outline the steps necessary to determine the number of moles and mass of H<sub>2<\/sub>.<\/li>\n<li>Perform the calculations outlined.<\/li>\n<\/ol>\n<\/li>\n<li>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}\\rightarrow 2{\\text{GaCl}}_{3}+3{\\text{H}}_{2}\\text{.}[\/latex]\n<ol>\n<li>Outline the steps necessary to determine the number of moles and mass of gallium chloride.<\/li>\n<li>Perform the calculations outlined.<\/li>\n<\/ol>\n<\/li>\n<li>I<sub>2<\/sub> is produced by the reaction of 0.4235 mol of CuCl<sub>2<\/sub> according to the following equation: [latex]2{\\text{CuCl}}_{2}+4\\text{KI}\\rightarrow 2\\text{CuI}+4\\text{KCl}+{\\text{I}}_{2}\\text{.}[\/latex]\n<ol>\n<li>How many molecules of I<sub>2<\/sub> are produced?<\/li>\n<li>What mass of I<sub>2<\/sub> is produced?<\/li>\n<\/ol>\n<\/li>\n<li>Silver is often extracted from ores as K[Ag(CN)<sub>2<\/sub>] and then recovered by the reaction [latex]2\\text{K}\\left[\\text{Ag}{\\text{(}\\text{CN}\\text{)}}_{2}\\right]\\text{(}aq\\text{)}+\\text{Zn}\\text{(}s\\text{)}\\rightarrow 2Ag\\text{(}s\\text{)}+\\text{Zn}{\\text{(}\\text{CN}\\text{)}}_{2}\\text{(}aq\\text{)}+2\\text{KCN}\\text{(}aq\\text{)}[\/latex] \u00a0(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>]? (b) What mass of Zn(CN)<sub>2<\/sub> is produced?<\/li>\n<li>What mass of silver oxide, Ag<sub>2<\/sub>O, is required to produce 25.0 g of silver sulfadiazine, AgC<sub>10<\/sub>H<sub>9<\/sub>N<sub>4<\/sub>SO<sub>2<\/sub>, from the reaction of silver oxide and sulfadiazine? [latex]2{\\text{C}}_{10}{\\text{H}}_{10}{\\text{N}}_{4}{\\text{SO}}_{2}+{\\text{Ag}}_{2}\\text{O}\\rightarrow 2{\\text{AgC}}_{10}{\\text{H}}_{9}{\\text{N}}_{4}{\\text{SO}}_{2}+{\\text{H}}_{2}\\text{O}[\/latex]<\/li>\n<li>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.<\/li>\n<li>Automotive air bags inflate when a sample of sodium azide, NaN<sub>3<\/sub>, is very rapidly decomposed.[latex]2{\\text{NaN}}_{3}\\text{(}s\\text{)}\\rightarrow 2\\text{Na}\\text{(}s\\text{)}+3{\\text{N}}_{2}\\text{(}g\\text{)}[\/latex]\u00a0What mass of sodium azide is required to produce 2.6 ft<sup>3<\/sup> (73.6 L) of nitrogen gas with a density of 1.25 g\/L?<\/li>\n<li>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 [latex]1.00\\times {10}^{3}\\text{kg}[\/latex] of carbon followed by the reaction? [latex]{\\text{CO}}_{2}\\text{(}g\\text{)}+2{\\text{NH}}_{3}\\text{(}g\\text{)}\\rightarrow\\text{CO}{\\text{(}{\\text{NH}}_{2}\\text{)}}_{2}\\text{(}s\\text{)}+{\\text{H}}_{2}\\text{O}\\text{(}l\\text{)}[\/latex]<\/li>\n<li>In an accident, a solution containing 2.5 kg of nitric acid was spilled. Two kilograms of Na<sub>2<\/sub>CO<sub>3<\/sub> was quickly spread on the area and CO<sub>2<\/sub> was released by the reaction. Was sufficient Na<sub>2<\/sub>CO<sub>3<\/sub> used to neutralize all of the acid?<\/li>\n<li>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).<\/li>\n<li>What volume of a 0.750 M solution of hydrochloric acid, a solution of HCl, can be prepared from the HCl produced by the reaction of 25.0 g of NaCl with an excess of sulfuric acid? [latex]\\text{NaCl}\\text{(}s\\text{)}+{\\text{H}}_{2}{\\text{SO}}_{4}\\text{(}l\\text{)}\\rightarrow\\text{HCl}\\text{(}g\\text{)}+{\\text{NaHSO}}_{4}\\text{(}s\\text{)}[\/latex]<\/li>\n<li>What volume of a 0.2089 M 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 M solution of Cu(NO<sub>3<\/sub>)<sub>2<\/sub>? [latex]2\\text{Cu}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}+4\\text{KI}\\rightarrow 2\\text{CuI}+{\\text{I}}_{2}+4{\\text{KNO}}_{3}[\/latex]<\/li>\n<li>A mordant is a substance that combines with a dye to produce a stable fixed color in a dyed fabric. Calcium acetate is used as a mordant. It is prepared by the reaction of acetic acid with calcium hydroxide.[latex]2{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}+\\text{Ca}{\\text{(}\\text{OH}\\text{)}}_{2}\\rightarrow\\text{Ca}{\\text{(}{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{)}}_{2}+2{\\text{H}}_{2}\\text{O}[\/latex] What mass of Ca(OH)<sub>2<\/sub> is required to react with the acetic acid in 25.0 mL of a solution having a density of 1.065 g\/mL and containing 58.0% acetic acid by mass?<\/li>\n<li>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}\\rightarrow 2{\\text{FeCl}}_{3}+2{\\text{TiO}}_{2}+2{\\text{H}}_{2}\\text{O}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"post-355\" class=\"post-355 chapter type-chapter status-publish hentry type-1\">\n<div class=\"entry-content\">\n<div class=\"im_section\">\n<div class=\"im_section\">\n<div id=\"mclean-ch03_s01_s02_n01\" class=\"im_key_takeaways im_editable im_block\">\n<div class=\"bcc-box bcc-info\">\n<h4>Selected Answers<\/h4>\n<p>2. (a)\u00a00.435 mol Na, 0.271 mol Cl<sub>2<\/sub>, 15.4 g Cl<sub>2<\/sub>;<\/p>\n<p>(b) 0.005780 mol HgO, 2.890 <span id=\"MathJax-Element-756-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-20861\" class=\"math\"><span id=\"MathJax-Span-20862\" class=\"mrow\"><span id=\"MathJax-Span-20863\" class=\"semantics\"><span id=\"MathJax-Span-20864\" class=\"mrow\"><span id=\"MathJax-Span-20865\" class=\"mrow\"><span id=\"MathJax-Span-20866\" class=\"mo\">\u00d7<\/span><\/span><\/span><\/span><\/span><\/span><\/span> 10<sup>\u22123<\/sup> mol O<sub>2<\/sub>, 9.248 <span id=\"MathJax-Element-757-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-20867\" class=\"math\"><span id=\"MathJax-Span-20868\" class=\"mrow\"><span id=\"MathJax-Span-20869\" class=\"semantics\"><span id=\"MathJax-Span-20870\" class=\"mrow\"><span id=\"MathJax-Span-20871\" class=\"mrow\"><span id=\"MathJax-Span-20872\" class=\"mo\">\u00d7<\/span><\/span><\/span><\/span><\/span><\/span><\/span> 10<sup>\u22122<\/sup> g O<sub>2<\/sub>;<\/p>\n<p>(c) 8.00 mol NaNO<sub>3<\/sub>, 6.8 <span id=\"MathJax-Element-758-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-20873\" class=\"math\"><span id=\"MathJax-Span-20874\" class=\"mrow\"><span id=\"MathJax-Span-20875\" class=\"semantics\"><span id=\"MathJax-Span-20876\" class=\"mrow\"><span id=\"MathJax-Span-20877\" class=\"mrow\"><span id=\"MathJax-Span-20878\" class=\"mo\">\u00d7<\/span><\/span><\/span><\/span><\/span><\/span><\/span> 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<div data-type=\"newline\">\n<p>4. \u00a0(a) [latex]\\text{mol Mg}=5.00\\cancel{\\text{g HCl}}\\times \\frac{1\\cancel{\\text{mol HCl}}}{36.4606\\cancel{\\text{g}}}\\times \\frac{\\text{1 mol Mg}}{2\\cancel{\\text{mol HCl}}}=0.0686\\text{mol,}[\/latex] [latex]\\text{g Mg}=0.0686\\cancel{\\text{mol Mg}}\\times \\frac{\\text{24.305 g}}{1\\cancel{\\text{mol Mg}}}=\\text{1.67 g};[\/latex]<\/p>\n<div data-type=\"newline\">(b) [latex]{\\text{mol O}}_{2}=1.252\\cancel{\\text{g}{\\text{Ag}}_{2}\\text{O}}\\times \\frac{1\\cancel{\\text{mol}{\\text{Ag}}_{2}\\text{O}}}{231.7358\\cancel{\\text{g}}}\\times \\frac{\\text{1 mol}{\\text{O}}_{2}}{2\\cancel{\\text{mol}{\\text{Ag}}_{2}\\text{O}}}=2.701\\times {10}^{-3}[\/latex], [latex]{\\text{g O}}_{2}=2.701\\times {10}^{-3}\\cancel{{\\text{mol O}}_{2}}\\times \\frac{\\text{31.9988 g}}{1\\cancel{\\text{mol}{\\text{O}}_{2}}}=0.08644\\text{g;}[\/latex]<\/div>\n<div data-type=\"newline\"><\/div>\n<div data-type=\"newline\">(c) [latex]{\\text{mol MgCO}}_{3}=283\\cancel{\\text{g}{\\text{CO}}_{2}}\\times \\frac{1\\cancel{\\text{mol}{\\text{CO}}_{2}}}{44.010\\cancel{\\text{g}}}\\times \\frac{\\text{1 mol}{\\text{MgCO}}_{3}}{1\\cancel{\\text{mol}{\\text{CO}}_{2}}}=6.43\\text{mol,}[\/latex] [latex]{\\text{g MgCO}}_{3}=6.43\\cancel{{\\text{mol MgCO}}_{3}}\\times \\frac{\\text{84.314 g}}{1\\cancel{{\\text{mol MgCO}}_{3}}}=542\\text{g;}[\/latex]<\/div>\n<div data-type=\"newline\"><\/div>\n<div data-type=\"newline\">(d) [latex]{\\text{mol H}}_{2}\\text{O}=2.00\\times {10}^{4}\\cancel{\\text{g}{\\text{C}}_{2}{\\text{H}}_{4}}\\times \\frac{1\\cancel{\\text{mol}{\\text{C}}_{2}{\\text{H}}_{2}}}{28.054\\cancel{\\text{g}}}\\times \\frac{\\text{1 mol}{\\text{H}}_{2}\\text{O}}{1\\cancel{\\text{mol}{\\text{C}}_{2}{\\text{H}}_{2}}}=713\\text{mol,}[\/latex] [latex]{\\text{g H}}_{2}\\text{O}=713\\cancel{\\text{mol}{\\text{H}}_{2}\\text{O}}\\times \\frac{18.01528\\cancel{\\text{g}}}{1\\cancel{\\text{mol}{\\text{H}}_{2}\\text{O}}}\\times \\frac{\\text{1 kg}}{1000\\cancel{\\text{g}}}=\\text{12.8 kg}[\/latex]<\/div>\n<div data-type=\"newline\"><\/div>\n<div data-type=\"newline\">(e) [latex]2.500\\cancel{\\text{kg BaO}}\\times \\frac{1000\\cancel{\\text{g BaO}}}{1\\cancel{\\text{kg BaO}}}\\times \\frac{1\\cancel{\\text{mol BaO}}}{153.326\\cancel{\\text{g BaO}}}\\times \\frac{\\text{2 mol}{\\text{BaO}}_{2}}{2\\cancel{\\text{mol BaO}}}=\\text{16.31 mol}{\\text{BaO}}_{2}[\/latex] [latex]16.31\\cancel{\\text{mol}{\\text{BaO}}_{2}}\\times \\frac{\\text{169.326 g}{\\text{BaO}}_{2}}{1\\cancel{\\text{mol}{\\text{BaO}}_{2}}}=\\text{2762 g}{\\text{BaO}}_{2}[\/latex]<\/div>\n<div data-type=\"newline\"><\/div>\n<div data-type=\"newline\">(f) [latex]9.55\\cancel{\\text{g}{\\text{C}}_{2}{\\text{H}}_{6}\\text{O}}\\times \\frac{1\\cancel{\\text{mol}{\\text{C}}_{2}{\\text{H}}_{6}\\text{O}}}{46.068\\cancel{\\text{g}{\\text{C}}_{2}{\\text{H}}_{6}\\text{O}}}\\times \\frac{\\text{1 mol}{\\text{C}}_{2}{\\text{H}}_{4}}{1\\cancel{\\text{mol}{\\text{C}}_{2}{\\text{H}}_{6}\\text{O}}}=\\text{0.207 mol}{\\text{C}}_{2}{\\text{H}}_{4}[\/latex] [latex]0.207\\cancel{\\text{mol}{\\text{C}}_{2}{\\text{H}}_{4}}\\times \\frac{\\text{28.053 g}{\\text{C}}_{2}{\\text{H}}_{4}}{1\\cancel{\\text{mol}{\\text{C}}_{2}{\\text{H}}_{4}}}=\\text{5.81 g}{\\text{C}}_{2}{\\text{H}}_{4}[\/latex]<\/div>\n<div data-type=\"newline\"><\/div>\n<div data-type=\"newline\"><\/div>\n<div data-type=\"newline\">6. (a) [latex]\\text{volume HCl solution}\\rightarrow\\text{mol HCl}\\rightarrow{\\text{mol GaCl}}_{3};[\/latex]<\/div>\n<div data-type=\"newline\"><\/div>\n<div data-type=\"newline\">(b) [latex]2.6\\cancel{\\text{L HCl}}\\times \\frac{1.44\\cancel{\\text{mol HCl}}}{1\\cancel{\\text{L HCl}}}\\times \\frac{2\\cancel{\\text{mol}{\\text{GaCl}}_{3}}}{6\\cancel{\\text{mol HCl}}}\\times \\frac{\\text{180.079 g}{\\text{GaCl}}_{3}}{1\\cancel{\\text{mol}{\\text{GaCl}}_{3}}}=2.3\\times {10}^{2}\\text{g}{\\text{GaCl}}_{3}[\/latex]<\/div>\n<div data-type=\"newline\"><\/div>\n<div data-type=\"newline\"><\/div>\n<div data-type=\"newline\">8.\u00a0The development requires the following: [latex]\\text{mass K}\\left[\\text{Ag}{\\text{(}\\text{CN}\\text{)}}_{2}\\right]\\rightarrow\\text{mol K}\\left[\\text{Ag}{\\text{(}\\text{CN}\\text{)}}_{2}\\right]\\rightarrow\\text{mol Zn}{\\text{(}\\text{CN}\\text{)}}_{2}\\rightarrow\\text{molecules of Zn}{\\text{(}\\text{CN}\\text{)}}_{2}\\text{g Zn}{\\text{(}\\text{CN}\\text{)}}_{2};[\/latex]<\/div>\n<div data-type=\"newline\"><\/div>\n<div data-type=\"newline\">(a) [latex]35.27\\cancel{\\text{g K}\\left[\\text{Ag}{\\text{(}\\text{CN}\\text{)}}_{2}\\right]}\\times \\frac{1\\cancel{\\text{mol K}\\left[\\text{Ag}{\\text{(}\\text{CN}\\text{)}}_{2}\\right]}}{199.002\\cancel{\\text{g K}\\left[\\text{Ag}{\\text{(}\\text{CN}\\text{)}}_{2}\\right]}}\\times \\frac{1\\cancel{\\text{mol Zn}{\\text{(}\\text{CN}\\text{)}}_{2}}}{2\\cancel{\\text{mol K}\\left[\\text{Ag}{\\text{(}\\text{CN}\\text{)}}_{2}\\right]}}\\times \\frac{6.022\\times {10}^{23}}{1\\cancel{\\text{mol Zn}{\\text{(}\\text{CN}\\text{)}}_{2}}}=5.337\\times {10}^{22}\\text{molecules}[\/latex]<\/div>\n<div data-type=\"newline\"><\/div>\n<div data-type=\"newline\">(b) [latex]5.337\\times {10}^{22}\\cancel{\\text{molecules}}\\times \\frac{1\\cancel{\\text{mol Zn}{\\text{(}\\text{CN}\\text{)}}_{2}}}{6.022\\times {10}^{23}\\cancel{\\text{molecules}}}\\times \\frac{\\text{117.43 g Zn}{\\text{(}\\text{CN}\\text{)}}_{2}}{1\\cancel{\\text{mol Zn}{\\text{(}\\text{CN}\\text{)}}_{2}}}=\\text{10.41 g Zn}{\\text{(}\\text{CN}\\text{)}}_{2}[\/latex]<\/div>\n<div data-type=\"newline\">\n<p>&nbsp;<\/p>\n<div data-type=\"newline\">\n<p>10. [latex]{\\text{SiO}}_{2}+3\\text{C}\\rightarrow\\text{SiC}+2\\text{CO.}[\/latex] From the balanced equation, 1 mol of SiO<sub>2<\/sub> produces 1 mol of SiC. The unknown is the mass of SiO<sub>2<\/sub> required to produce 3.00 kg (3000 g) of SiC. To calculate the mass of SiO<sub>2<\/sub> required, determine the molar masses of SiO<sub>2<\/sub> and SiC. Then calculate the number of moles of SiC required, and through the mole relation of SiO<sub>2<\/sub> to SiC, find the mass of SiO<sub>2<\/sub> required. The conversions required are: [latex]\\text{g SiC}\\rightarrow\\text{mol SiC}\\rightarrow\\text{mol}{\\text{SiO}}_{2}\\rightarrow\\text{g}{\\text{SiO}}_{2}[\/latex]<\/p>\n<div data-type=\"newline\">Molar masses: SiO<sub>2<\/sub> = 60.0843 g mol<sup>\u20131<\/sup>; SiC = 40.0955 g mol<sup>\u20131<\/sup><\/div>\n<div data-type=\"newline\">[latex]\\text{mass SiO}2=3000\\cancel{\\text{g SiC}}\\times \\frac{1\\cancel{\\text{mol SiC}}}{40.955\\cancel{\\text{g SiC}}}\\times \\frac{1\\cancel{\\text{mol}{\\text{SiO}}_{2}}}{1\\cancel{\\text{mol SiC}}}\\times \\frac{\\text{60.843 g}{\\text{SiO}}_{2}}{1\\cancel{\\text{mol}{\\text{SiO}}_{2}}}=\\text{4496 g}{\\text{SiO}}_{2}=\\text{4.50 kg}{\\text{SiO}}_{2}[\/latex]<\/div>\n<div data-type=\"newline\">\n<p>&nbsp;<\/p>\n<div data-type=\"newline\">\n<p>12. Molar mass urea = 12.011 + 15.9994 + 2(14.0067) + 4(1.0079) = 60.054 g mol<sup>\u20131<\/sup><\/p>\n<div data-type=\"newline\">[latex]\\text{1 mol C}\\rightarrow 1{\\text{mol CO}}_{2}\\rightarrow 1\\text{mol urea}[\/latex]<\/div>\n<div data-type=\"newline\">[latex]\\begin{array}{ll}\\hfill \\text{mass urea}& =1.00\\times {10}^{3}\\cancel{\\text{kg}}\\times \\frac{1000\\cancel{\\text{g}}}{\\cancel{\\text{kg}}}\\times \\frac{1\\cancel{\\text{mol C}}}{12.0\\cancel{\\text{g C}}}\\times \\frac{1\\cancel{\\text{mol urea}}}{1\\cancel{\\text{mol C}}}\\times \\frac{\\text{60.054 g urea}}{1\\cancel{\\text{mol urea}}}\\\\ & =5.00\\times {10}^{6}\\text{g or}5.00\\times {10}^{3}\\text{kg}\\end{array}[\/latex]<\/div>\n<div data-type=\"newline\">\n<p>&nbsp;<\/p>\n<div data-type=\"newline\">\n<p>14. \u00a0The balanced chemical equation is [latex]\\text{C}\\text{(}s\\text{)}+{\\text{O}}_{2}\\text{(}g\\text{)}\\rightarrow{\\text{CO}}_{2}\\text{(}g\\text{)}[\/latex]<\/p>\n<div data-type=\"newline\">[latex]\\begin{array}{l}\\\\ \\\\ \\\\ 500\\cancel{\\text{miles}}\\times \\frac{1\\cancel{\\text{gallon}}}{37.5\\cancel{\\text{miles}}}\\times \\frac{3.785\\cancel{\\text{L}}}{1\\cancel{\\text{gallon}}}\\times \\frac{1000\\cancel{\\text{mL}}}{1\\cancel{\\text{L}}}\\times \\frac{0.8205\\cancel{\\text{g gas}}}{1\\cancel{\\text{mL}}\\text{gas}}\\times \\frac{84.2\\cancel{\\text{g C}}}{100\\cancel{\\text{g gas}}}\\times \\frac{1\\cancel{\\text{mol C}}}{12.01\\cancel{\\text{g C}}}\\\\ \\times \\frac{1\\cancel{\\text{mol}{\\text{CO}}_{2}}}{1\\cancel{\\text{mol C}}}\\times \\frac{\\text{44.01 g}{\\text{CO}}_{2}}{1\\cancel{\\text{mol}{\\text{CO}}_{2}}}=1.28\\times {10}^{5}\\text{g}{\\text{CO}}_{2}\\end{array}[\/latex]<\/div>\n<div data-type=\"newline\">\n<p>&nbsp;<\/p>\n<div data-type=\"newline\">\n<p>16. Use molarity to convert. This solution involves the following steps:<\/p>\n<p>Converting the volume of KI to moles of KI<\/p>\n<p>Converting the moles of KI to moles of Cu(NO<sub>3<\/sub>)<sub>2<\/sub><\/p>\n<p>Converting the moles of [latex]\\text{K}\\rightarrow\\text{Cu}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}[\/latex] to a volume of KI. Cu(NO<sub>3<\/sub>)<sub>2<\/sub> solution<\/p>\n<div data-type=\"newline\">[latex]\\text{43.88 mL}\\times \\frac{\\text{1 L}}{\\text{1000 mL}}\\times \\frac{0.3842\\cancel{\\text{mol Cu}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}}}{\\text{1 L}}\\times \\frac{4\\cancel{\\text{mol KI}}}{2\\cancel{\\text{mol Cu}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}}}\\times \\frac{\\text{1 L KI}}{0.2089\\cancel{\\text{mol KI}}}=\\text{161.4 mL}[\/latex]<\/div>\n<div data-type=\"newline\">All of these steps can be shown together, as follows:<\/div>\n<div data-type=\"newline\">[latex]\\frac{\\text{43.88 mL Cu}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}}{1}\\times \\frac{\\text{0.3842 mol Cu}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}}{\\text{1000 mL Cu}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}}\\times \\frac{\\text{4 mol KI}}{\\text{2 mol Cu}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}}\\times \\frac{\\text{1000 mL KI}}{\\text{0.2089 mol KI}}=\\text{161.40 mL KI solution}[\/latex]<\/div>\n<div data-type=\"newline\">\n<p>&nbsp;<\/p>\n<div data-type=\"newline\">\n<p>18. Find from worked example, check your learning problem<\/p>\n<div data-type=\"newline\">[latex]\\text{mass of ilmenite}=379\\cancel{\\text{g ore}}\\times \\frac{\\text{0.883 g}{\\text{FeTiO}}_{3}}{1\\cancel{\\text{g ore}}}=\\text{334.6 g}{\\text{FeTiO}}_{3}[\/latex]<\/div>\n<div data-type=\"newline\">[latex]\\text{mass of rutile}=334.6\\cancel{\\text{g}{\\text{FeTiO}}_{3}}\\times \\frac{1\\cancel{\\text{mol}{\\text{FeTiO}}_{3}}}{151.7\\cancel{\\text{g}{\\text{FeTiO}}_{3}}}\\times \\frac{2\\cancel{\\text{mol}{\\text{TiO}}_{2}}}{2\\cancel{\\text{mol}{\\text{FeTiO}}_{3}}}\\times \\frac{\\text{79.88 g}{\\text{TiO}}_{2}}{1\\cancel{\\text{mol}{\\text{TiO}}_{2}}}=\\text{176 g}{\\text{TiO}}_{2}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<div class=\"bcc-box bcc-success\">\n<section id=\"glossary\">\n<h3>Glossary<\/h3>\n<div data-type=\"definition\">\n<div id=\"fs-idm8143856\" data-type=\"definition\">\n<p data-type=\"definition\"><strong><span data-type=\"term\">stoichiometric factor<br \/>\n<\/span><\/strong>ratio of coefficients in a balanced chemical equation, used in computations relating amounts of reactants and products<\/p>\n<div id=\"fs-idp127462672\" data-type=\"definition\">\n<p><strong><span data-type=\"term\">stoichiometry<br \/>\n<\/span><\/strong>relationships between the amounts of reactants and products of a chemical reaction<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<\/section>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1685\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Chemistry. <strong>Provided by<\/strong>: OpenStax College. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/openstaxcollege.org\">http:\/\/openstaxcollege.org<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at https:\/\/openstaxcollege.org\/textbooks\/chemistry\/get<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":5,"menu_order":7,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Chemistry\",\"author\":\"\",\"organization\":\"OpenStax College\",\"url\":\"http:\/\/openstaxcollege.org\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at 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