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<title>Mass Relationships in Chemical Equations</title>
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<div class="section" id="averill_1.0-ch03_s04" condition="start-of-chunk" version="5.0" lang="en">
<h2 class="title editable block">
<span class="title-prefix">3.4</span> Mass Relationships in Chemical Equations</h2>
<div class="learning_objectives editable block" id="averill_1.0-ch03_s04_n01">
<h3 class="title">Learning Objective</h3>
<ol class="orderedlist" id="averill_1.0-ch03_s04_l01">
<li>To calculate the quantities of compounds produced or consumed in a chemical reaction.</li>
</ol>
</div>
<p class="para editable block" id="averill_1.0-ch03_s04_p01">A balanced chemical equation gives the identity of the reactants and the products as well as the accurate number of molecules or moles of each that are consumed or produced. <span class="margin_term"><a class="glossterm">Stoichiometry</a><span class="glossdef">A collective term for the quantitative relationships between the masses, the numbers of moles, and the numbers of particles (atoms, molecules, and ions) of the reactants and the products in a balanced chemical equation.</span></span> is a collective term for the quantitative relationships between the masses, the numbers of moles, and the numbers of particles (atoms, molecules, and ions) of the reactants and the products in a balanced chemical equation. A <span class="margin_term"><a class="glossterm">stoichiometric quantity</a><span class="glossdef">The amount of product or reactant specified by the coefficients in a balanced chemical equation.</span></span> is the amount of product or reactant specified by the coefficients in a balanced chemical equation. In <a class="xref" href="averill_1.0-ch03_s03#averill_1.0-ch03_s03">Section 3.3 "Chemical Equations"</a>, for example, you learned how to express the stoichiometry of the reaction for the ammonium dichromate volcano in terms of the atoms, ions, or molecules involved and the numbers of moles, grams, and formula units of each (recognizing, for instance, that 1 mol of ammonium dichromate produces 4 mol of water). This section describes how to use the stoichiometry of a reaction to answer questions like the following: How much oxygen is needed to ensure complete combustion of a given amount of isooctane? (This information is crucial to the design of nonpolluting and efficient automobile engines.) How many grams of pure gold can be obtained from a ton of low-grade gold ore? (The answer determines whether the ore deposit is worth mining.) If an industrial plant must produce a certain number of tons of sulfuric acid per week, how much elemental sulfur must arrive by rail each week?</p>
<p class="para editable block" id="averill_1.0-ch03_s04_p02">All these questions can be answered using the concepts of the mole and molar and formula masses, along with the coefficients in the appropriate balanced chemical equation.</p>
<div class="section" id="averill_1.0-ch03_s04_s01">
<h2 class="title editable block">Stoichiometry Problems</h2>
<p class="para editable block" id="averill_1.0-ch03_s04_s01_p01">When we carry out a reaction in either an industrial setting or a laboratory, it is easier to work with <em class="emphasis">masses</em> of substances than with the numbers of molecules or moles. The general method for converting from the mass of any reactant or product to the mass of any other reactant or product using a balanced chemical equation is outlined in <a class="xref" href="#averill_1.0-ch03_s04_s01_s01_f01">Figure 3.11 "A Flowchart for Stoichiometric Calculations Involving Pure Substances"</a> and described in the following text.</p>
<div class="section" id="averill_1.0-ch03_s04_s01_s01">
<h2 class="title editable block">Steps in Converting between Masses of Reactant and Product</h2>
<ol class="orderedlist editable block" id="averill_1.0-ch03_s04_s01_s01_l01">
<li>Convert the mass of one substance (substance A) to the corresponding number of moles using its molar mass.</li>
<li>From the balanced chemical equation, obtain the number of moles of another substance (B) from the number of moles of substance A using the appropriate mole ratio (the ratio of their coefficients).</li>
<li>Convert the number of moles of substance B to mass using its molar mass. It is important to remember that some species are in excess by virtue of the reaction conditions. For example, if a substance reacts with the oxygen in air, then oxygen is in obvious (but unstated) excess.</li>
</ol>
<p class="para editable block" id="averill_1.0-ch03_s04_s01_s01_p01">Converting amounts of substances to moles—and vice versa—is the key to all stoichiometry problems, whether the amounts are given in units of mass (grams or kilograms), weight (pounds or tons), or volume (liters or gallons).</p>
<div class="figure large editable block" id="averill_1.0-ch03_s04_s01_s01_f01">
<p class="title"><span class="title-prefix">Figure 3.11</span> A Flowchart for Stoichiometric Calculations Involving Pure Substances</p>
<img src="section_07/ad5779e7879742495eeb43ea41506295.jpg">
<p class="para">The molar masses of the reactants and the products are used as conversion factors so that you can calculate the mass of product from the mass of reactant and vice versa.</p>
</div>
<p class="para editable block" id="averill_1.0-ch03_s04_s01_s01_p02">To illustrate this procedure, let’s return to the combustion of glucose. We saw earlier that glucose reacts with oxygen to produce carbon dioxide and water:</p>
<div class="equation block" id="averill_1.0-ch03_s04_s01_s01_eq01">
<p class="title editable"><span class="title-prefix">Equation 3.20</span> </p>
<span class="mathphrase">C<sub class="subscript">6</sub>H<sub class="subscript">12</sub>O<sub class="subscript">6</sub>(s) + 6O<sub class="subscript">2</sub>(g) → 6CO<sub class="subscript">2</sub>(g) + 6H<sub class="subscript">2</sub>O(l)</span>
</div>
<p class="para editable block" id="averill_1.0-ch03_s04_s01_s01_p03">Just before a chemistry exam, suppose a friend reminds you that glucose is the major fuel used by the human brain. You therefore decide to eat a candy bar to make sure that your brain doesn’t run out of energy during the exam (even though there is no direct evidence that consumption of candy bars improves performance on chemistry exams). If a typical 2 oz candy bar contains the equivalent of 45.3 g of glucose and the glucose is completely converted to carbon dioxide during the exam, how many grams of carbon dioxide will you produce and exhale into the exam room?</p>
<p class="para editable block" id="averill_1.0-ch03_s04_s01_s01_p04">The initial step in solving a problem of this type must be to write the balanced chemical equation for the reaction. Inspection of <a class="xref" href="#averill_1.0-ch03_s04_s01_s01_eq01" xrefstyle="select:label">Equation 3.20</a> shows that it is balanced as written, so we can proceed to the strategy outlined in <a class="xref" href="#averill_1.0-ch03_s04_s01_s01_f01">Figure 3.11 "A Flowchart for Stoichiometric Calculations Involving Pure Substances"</a>, adapting it as follows:</p>
<ol class="orderedlist block" id="averill_1.0-ch03_s04_s01_s01_l02">
<li>
<p class="para">Use the molar mass of glucose (to one decimal place, 180.2 g/mol) to determine the number of moles of glucose in the candy bar:</p>
<span class="informalequation">
<math xml:id="averill_1.0-ch03_m042" display="block">
<semantics>
<mrow>
<mtext>moles glucose </mtext>
<mo>=</mo>
<mtext> 45</mtext>
<mtext>.3 </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<mtext>g glucose</mtext>
</mrow>
</menclose>
<mo>×</mo>
<mfrac>
<mrow>
<mtext>1 mol glucose</mtext>
</mrow>
<mrow>
<mtext>180</mtext>
<mtext>.2 </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<mtext>g glucose</mtext>
</mrow>
</menclose>
</mrow>
</mfrac>
<mo>=</mo>
<mtext> 0</mtext>
<mtext>.251 mol glucose</mtext>
</mrow>
</semantics>
</math>
</span>
</li>
<li>
<p class="para">According to the balanced chemical equation, 6 mol of CO<sub class="subscript">2</sub> is produced per mole of glucose; the mole ratio of CO<sub class="subscript">2</sub> to glucose is therefore 6:1. The number of moles of CO<sub class="subscript">2</sub> produced is thus</p>
<span class="informalequation">
<math xml:id="averill_1.0-ch03_m043" display="block">
<semantics>
<mtable columnalign="left">
<mtr columnalign="left">
<mtd columnalign="left">
<msub>
<mtext>moles CO</mtext>
<mtext>2</mtext>
</msub>
</mtd>
<mtd columnalign="left">
<mo>=</mo>
<mtext> mol glucose</mtext>
<mo>×</mo>
<mfrac>
<mrow>
<msub>
<mrow>
<mtext>6 mol CO</mtext>
</mrow>
<mtext>2</mtext>
</msub>
</mrow>
<mrow>
<mtext>1 mol glucose</mtext>
</mrow>
</mfrac>
</mtd>
</mtr>
<mtr columnalign="left">
<mtd columnalign="left">
<mrow></mrow>
</mtd>
<mtd columnalign="left">
<mo>=</mo>
<mtext> 0</mtext>
<mtext>.251 </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<mtext>mol glucose</mtext>
</mrow>
</menclose>
<mo>×</mo>
<mfrac>
<mrow>
<msub>
<mrow>
<mtext>6 mol CO</mtext>
</mrow>
<mtext>2</mtext>
</msub>
</mrow>
<mrow>
<mtext>1 </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<mtext>mol glucose</mtext>
</mrow>
</menclose>
</mrow>
</mfrac>
</mtd>
</mtr>
<mtr columnalign="left">
<mtd columnalign="left">
<mrow></mrow>
</mtd>
<mtd columnalign="left">
<mo>=</mo>
<mtext> 1</mtext>
<msub>
<mtext>.51 mol CO</mtext>
<mtext>2</mtext>
</msub>
</mtd>
</mtr>
</mtable>
</semantics>
</math>
</span>
</li>
<li>
<p class="para">Use the molar mass of CO<sub class="subscript">2</sub> (44.010 g/mol) to calculate the mass of CO<sub class="subscript">2</sub> corresponding to 1.51 mol of CO<sub class="subscript">2</sub>:</p>
<span class="informalequation">
<math xml:id="averill_1.0-ch03_m044" display="block">
<semantics>
<mrow>
<msub>
<mrow>
<mtext>mass of CO</mtext>
</mrow>
<mtext>2</mtext>
</msub>
<mo>=</mo>
<mtext> 1</mtext>
<mtext>.51 </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<msub>
<mrow>
<mtext>mol CO</mtext>
</mrow>
<mtext>2</mtext>
</msub>
</mrow>
</menclose>
<mo>×</mo>
<mfrac>
<mrow>
<mtext>44</mtext>
<msub>
<mrow>
<mtext>.010 g CO</mtext>
</mrow>
<mtext>2</mtext>
</msub>
</mrow>
<mrow>
<mtext>1 </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<msub>
<mrow>
<mtext>mol CO</mtext>
</mrow>
<mtext>2</mtext>
</msub>
</mrow>
</menclose>
</mrow>
</mfrac>
<mo>=</mo>
<mtext> 66</mtext>
<msub>
<mrow>
<mtext>.5 g CO</mtext>
</mrow>
<mtext>2</mtext>
</msub>
</mrow>
</semantics>
</math>
</span>
</li>
</ol>
<p class="para editable block" id="averill_1.0-ch03_s04_s01_s01_p05">We can summarize these operations as follows:</p>
<span class="informalequation block">
<math xml:id="averill_1.0-ch03_m045" display="block">
<semantics>
<mrow>
<mtext>45</mtext>
<mtext>.3 g glucose</mtext>
<mo>×</mo>
<munder>
<mrow>
<mfrac>
<mrow>
<mtext>1 mol glucose</mtext>
</mrow>
<mrow>
<mtext>180</mtext>
<mtext>.2 g glucose</mtext>
</mrow>
</mfrac>
</mrow>
<mrow>
<mtext>step 1</mtext>
</mrow>
</munder>
<mo>×</mo>
<munder>
<mrow>
<mfrac>
<mrow>
<msub>
<mrow>
<mtext>6 mol CO</mtext>
</mrow>
<mtext>2</mtext>
</msub>
</mrow>
<mrow>
<mtext>1 mol glucose</mtext>
</mrow>
</mfrac>
</mrow>
<mrow>
<mtext>step 2</mtext>
</mrow>
</munder>
<mo>×</mo>
<munder>
<mrow>
<mfrac>
<mrow>
<mtext>44</mtext>
<msub>
<mrow>
<mtext>.010 g CO</mtext>
</mrow>
<mtext>2</mtext>
</msub>
</mrow>
<mrow>
<msub>
<mrow>
<mtext>1 mol CO</mtext>
</mrow>
<mtext>2</mtext>
</msub>
</mrow>
</mfrac>
</mrow>
<mrow>
<mtext>step 3</mtext>
</mrow>
</munder>
<mo>=</mo>
<mtext> 66</mtext>
<msub>
<mrow>
<mtext>.4 g CO</mtext>
</mrow>
<mtext>2</mtext>
</msub>
</mrow>
</semantics>
</math>
</span>
<p class="para editable block" id="averill_1.0-ch03_s04_s01_s01_p06">Discrepancies between the two values are attributed to rounding errors resulting from using stepwise calculations in steps 1–3. (For more information about rounding and significant digits, see Essential Skills 1 in <a class="xref" href="averill_1.0-ch01#averill_1.0-ch01">Chapter 1 "Introduction to Chemistry"</a>, <a class="xref" href="averill_1.0-ch01_s09#averill_1.0-ch01_s09">Section 1.9 "Essential Skills 1"</a>.) In <a class="xref" href="averill_1.0-ch10#averill_1.0-ch10">Chapter 10 "Gases"</a>, you will discover that this amount of gaseous carbon dioxide occupies an enormous volume—more than 33 L. We could use similar methods to calculate the amount of oxygen consumed or the amount of water produced.</p>
<p class="para editable block" id="averill_1.0-ch03_s04_s01_s01_p07">We just used the balanced chemical equation to calculate the mass of product that is formed from a certain amount of reactant. We can also use the balanced chemical equation to determine the masses of reactants that are necessary to form a certain amount of product or, as shown in Example 11, the mass of one reactant that is required to consume a given mass of another reactant.</p>
<div class="exercises block" id="averill_1.0-ch03_s04_s01_s01_n01">
<h3 class="title">Example 11</h3>
<p class="para" id="averill_1.0-ch03_s04_s01_s01_p08">The combustion of hydrogen with oxygen to produce gaseous water is extremely vigorous, producing one of the hottest flames known. Because so much energy is released for a given mass of hydrogen or oxygen, this reaction was used to fuel the NASA (National Aeronautics and Space Administration) space shuttles, which have recently been retired from service. NASA engineers calculated the exact amount of each reactant needed for the flight to make sure that the shuttles did not carry excess fuel into orbit. Calculate how many tons of hydrogen a space shuttle needed to carry for each 1.00 tn of oxygen (1 tn = 2000 lb).</p>
<div class="informalfigure small">
<img src="section_07/6f35d591685db159716649783fa155e7.jpg">
<p class="para"><strong class="emphasis bold">The US space shuttle <em class="emphasis bolditalic">Discovery</em> during liftoff.</strong> The large cylinder in the middle contains the oxygen and hydrogen that fueled the shuttle’s main engine.</p>
</div>
<p class="para" id="averill_1.0-ch03_s04_s01_s01_p09"><strong class="emphasis bold">Given: </strong>reactants, products, and mass of one reactant</p>
<p class="para" id="averill_1.0-ch03_s04_s01_s01_p10"><strong class="emphasis bold">Asked for: </strong>mass of other reactant</p>
<p class="para" id="averill_1.0-ch03_s04_s01_s01_p11">
<strong class="emphasis bold">Strategy:</strong>
</p>
<p class="para" id="averill_1.0-ch03_s04_s01_s01_p12"><strong class="emphasis bold">A</strong> Write the balanced chemical equation for the reaction.</p>
<p class="para" id="averill_1.0-ch03_s04_s01_s01_p13"><strong class="emphasis bold">B</strong> Convert mass of oxygen to moles. From the mole ratio in the balanced chemical equation, determine the number of moles of hydrogen required. Then convert the moles of hydrogen to the equivalent mass in tons.</p>
<p class="para" id="averill_1.0-ch03_s04_s01_s01_p14">
<strong class="emphasis bold">Solution:</strong>
</p>
<p class="para" id="averill_1.0-ch03_s04_s01_s01_p15">We use the same general strategy for solving stoichiometric calculations as in the preceding example. Because the amount of oxygen is given in tons rather than grams, however, we also need to convert tons to units of mass in grams. Another conversion is needed at the end to report the final answer in tons.</p>
<p class="para" id="averill_1.0-ch03_s04_s01_s01_p16"><strong class="emphasis bold">A</strong> We first use the information given to write a balanced chemical equation. Because we know the identity of both the reactants and the product, we can write the reaction as follows:</p>
<span class="informalequation">
<span class="mathphrase">H<sub class="subscript">2</sub>(g) + O<sub class="subscript">2</sub>(g) → H<sub class="subscript">2</sub>O(g)</span>
</span>
<p class="para" id="averill_1.0-ch03_s04_s01_s01_p17">This equation is not balanced because there are two oxygen atoms on the left side and only one on the right. Assigning a coefficient of 2 to both H<sub class="subscript">2</sub>O and H<sub class="subscript">2</sub> gives the balanced chemical equation:</p>
<span class="informalequation">
<span class="mathphrase">2H<sub class="subscript">2</sub>(g) + O<sub class="subscript">2</sub>(g) → 2H<sub class="subscript">2</sub>O(g)</span>
</span>
<p class="para" id="averill_1.0-ch03_s04_s01_s01_p18">Thus 2 mol of H<sub class="subscript">2</sub> react with 1 mol of O<sub class="subscript">2</sub> to produce 2 mol of H<sub class="subscript">2</sub>O.</p>
<ol class="orderedlist" id="averill_1.0-ch03_s04_s01_s01_l03">
<li>
<p class="para"><strong class="emphasis bold">B</strong> To convert tons of oxygen to units of mass in grams, we multiply by the appropriate conversion factors:</p>
<span class="informalequation">
<math xml:id="averill_1.0-ch03_m046" display="block">
<semantics>
<mrow>
<msub>
<mrow>
<mtext>mass of O</mtext>
</mrow>
<mtext>2</mtext>
</msub>
<mo>=</mo>
<mtext> 1</mtext>
<mtext>.00 </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<mtext>tn</mtext>
</mrow>
</menclose>
<mo>×</mo>
<mfrac>
<mrow>
<mtext>2000 </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<mtext>lb</mtext>
</mrow>
</menclose>
</mrow>
<mrow>
<menclose notation="updiagonalstrike">
<mrow>
<mtext>tn</mtext>
</mrow>
</menclose>
</mrow>
</mfrac>
<mo>×</mo>
<mfrac>
<mrow>
<mn>453.6</mn>
<mtext> g</mtext>
</mrow>
<mrow>
<menclose notation="updiagonalstrike">
<mrow>
<mtext>lb</mtext>
</mrow>
</menclose>
</mrow>
</mfrac>
<mo>=</mo>
<mn>9.07</mn>
<mo>×</mo>
<msup>
<mrow>
<mn>10</mn>
</mrow>
<mn>5</mn>
</msup>
<msub>
<mrow>
<mtext> g O</mtext>
</mrow>
<mtext>2</mtext>
</msub>
</mrow>
</semantics>
</math>
</span>
<p class="para" id="averill_1.0-ch03_s04_s01_s01_p19">Using the molar mass of O<sub class="subscript">2</sub> (32.00 g/mol, to four significant figures), we can calculate the number of moles of O<sub class="subscript">2</sub> contained in this mass of O<sub class="subscript">2</sub>:</p>
<span class="informalequation">
<math xml:id="averill_1.0-ch03_m047" display="block">
<semantics>
<mrow>
<msub>
<mrow>
<mtext>mol O</mtext>
</mrow>
<mtext>2</mtext>
</msub>
<mo>=</mo>
<mn>9.07</mn>
<mo>×</mo>
<msup>
<mrow>
<mn>10</mn>
</mrow>
<mn>5</mn>
</msup>
<mtext> </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<msub>
<mrow>
<mtext>g O</mtext>
</mrow>
<mtext>2</mtext>
</msub>
</mrow>
</menclose>
<mo>×</mo>
<mfrac>
<mrow>
<msub>
<mrow>
<mtext>1 mol O</mtext>
</mrow>
<mtext>2</mtext>
</msub>
</mrow>
<mrow>
<mtext>32</mtext>
<mtext>.00 </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<msub>
<mrow>
<mtext>g O</mtext>
</mrow>
<mtext>2</mtext>
</msub>
</mrow>
</menclose>
</mrow>
</mfrac>
<mo>=</mo>
<mtext> 2</mtext>
<mtext>.83</mtext>
<mo>×</mo>
<msup>
<mrow>
<mtext>10</mtext>
</mrow>
<mtext>4</mtext>
</msup>
<msub>
<mrow>
<mtext> mol O</mtext>
</mrow>
<mtext>2</mtext>
</msub>
</mrow>
</semantics>
</math>
</span>
</li>
<li>
<p class="para">Now use the coefficients in the balanced chemical equation to obtain the number of moles of H<sub class="subscript">2</sub> needed to react with this number of moles of O<sub class="subscript">2</sub>:</p>
<span class="informalequation">
<math xml:id="averill_1.0-ch03_m048" display="block">
<semantics>
<mtable columnalign="left">
<mtr>
<mtd>
<msub>
<mtext>mol H</mtext>
<mtext>2</mtext>
</msub>
<mo>=</mo>
<msub>
<mtext>mol O</mtext>
<mtext>2</mtext>
</msub>
<mo>×</mo>
<mfrac>
<mrow>
<msub>
<mrow>
<mtext>2 mol H</mtext>
</mrow>
<mtext>2</mtext>
</msub>
</mrow>
<mrow>
<msub>
<mrow>
<mtext>1 mol O</mtext>
</mrow>
<mtext>2</mtext>
</msub>
</mrow>
</mfrac>
</mtd>
</mtr>
<mtr>
<mtd>
<mtext> </mtext>
<mo>=</mo>
<mtext> 2</mtext>
<mtext>.83</mtext>
<mo>×</mo>
<msup>
<mtext>10</mtext>
<mtext>4</mtext>
</msup>
<mtext> </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<msub>
<mrow>
<mtext>mol O</mtext>
</mrow>
<mtext>2</mtext>
</msub>
</mrow>
</menclose>
<mo>×</mo>
<mfrac>
<mrow>
<msub>
<mrow>
<mtext>2 mol H</mtext>
</mrow>
<mtext>2</mtext>
</msub>
</mrow>
<mrow>
<mtext>1 </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<msub>
<mrow>
<mtext>mol O</mtext>
</mrow>
<mtext>2</mtext>
</msub>
</mrow>
</menclose>
</mrow>
</mfrac>
<mo>=</mo>
<mn>5.66</mn>
<mo>×</mo>
<msup>
<mn>10</mn>
<mn>4</mn>
</msup>
<msub>
<mtext> mol H</mtext>
<mtext>2</mtext>
</msub>
</mtd>
</mtr>
</mtable>
</semantics>
</math>
</span>
</li>
<li>
<p class="para">The molar mass of H<sub class="subscript">2</sub> (2.016 g/mol) allows us to calculate the corresponding mass of H<sub class="subscript">2</sub>:</p>
<span class="informalequation">
<math xml:id="averill_1.0-ch03_m049" display="block">
<semantics>
<mrow>
<msub>
<mrow>
<mtext>mass of H</mtext>
</mrow>
<mtext>2</mtext>
</msub>
<mo>=</mo>
<mtext> 5</mtext>
<mtext>.66</mtext>
<mo>×</mo>
<msup>
<mrow>
<mtext>10</mtext>
</mrow>
<mtext>4</mtext>
</msup>
<mtext> </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<msub>
<mrow>
<mtext>mol H</mtext>
</mrow>
<mtext>2</mtext>
</msub>
</mrow>
</menclose>
<mo>×</mo>
<mfrac>
<mrow>
<mtext>2</mtext>
<msub>
<mrow>
<mtext>.016 g H</mtext>
</mrow>
<mtext>2</mtext>
</msub>
</mrow>
<mrow>
<menclose notation="updiagonalstrike">
<mrow>
<msub>
<mrow>
<mtext>mol H</mtext>
</mrow>
<mtext>2</mtext>
</msub>
</mrow>
</menclose>
</mrow>
</mfrac>
<mo>=</mo>
<mn>1.14</mn>
<mo>×</mo>
<msup>
<mrow>
<mn>10</mn>
</mrow>
<mn>5</mn>
</msup>
<msub>
<mrow>
<mtext> g H</mtext>
</mrow>
<mtext>2</mtext>
</msub>
</mrow>
</semantics>
</math>
</span>
<p class="para" id="averill_1.0-ch03_s04_s01_s01_p20">Finally, convert the mass of H<sub class="subscript">2</sub> to the desired units (tons) by using the appropriate conversion factors:</p>
<span class="informalequation">
<math xml:id="averill_1.0-ch03_m050" display="block">
<semantics>
<mrow>
<msub>
<mrow>
<mtext>tons H</mtext>
</mrow>
<mtext>2</mtext>
</msub>
<mo>=</mo>
<mtext> 1</mtext>
<mtext>.14</mtext>
<mo>×</mo>
<msup>
<mrow>
<mtext>10</mtext>
</mrow>
<mtext>5</mtext>
</msup>
<mtext> </mtext>
<menclose notation="updiagonalstrike">
<mtext>g</mtext>
</menclose>
<msub>
<mrow>
<mtext> H</mtext>
</mrow>
<mtext>2</mtext>
</msub>
<mo>×</mo>
<mfrac>
<mrow>
<mtext>1</mtext>
<menclose notation="updiagonalstrike">
<mrow>
<mtext>lb</mtext>
</mrow>
</menclose>
</mrow>
<mrow>
<mtext>453</mtext>
<mtext>.6 </mtext>
<menclose notation="updiagonalstrike">
<mtext>g</mtext>
</menclose>
</mrow>
</mfrac>
<mo>×</mo>
<mfrac>
<mrow>
<mtext>1 tn</mtext>
</mrow>
<mrow>
<mtext>2000 </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<mtext>lb</mtext>
</mrow>
</menclose>
</mrow>
</mfrac>
<mo>=</mo>
<mn>0.126</mn>
<msub>
<mrow>
<mtext> tn H</mtext>
</mrow>
<mtext>2</mtext>
</msub>
</mrow>
</semantics>
</math>
</span>
<p class="para" id="averill_1.0-ch03_s04_s01_s01_p21">The space shuttle had to be designed to carry 0.126 tn of H<sub class="subscript">2</sub> for each 1.00 tn of O<sub class="subscript">2</sub>. Even though 2 mol of H<sub class="subscript">2</sub> are needed to react with each mole of O<sub class="subscript">2</sub>, the molar mass of H<sub class="subscript">2</sub> is so much smaller than that of O<sub class="subscript">2</sub> that only a relatively small mass of H<sub class="subscript">2</sub> is needed compared to the mass of O<sub class="subscript">2</sub>.</p>
</li>
</ol>
<p class="simpara">Exercise</p>
<p class="para" id="averill_1.0-ch03_s04_s01_s01_p22">Alchemists produced elemental mercury by roasting the mercury-containing ore cinnabar (HgS) in air:</p>
<span class="informalequation">
<span class="mathphrase">HgS(s) + O<sub class="subscript">2</sub>(g) → Hg(l) + SO<sub class="subscript">2</sub>(g)</span>
</span>
<p class="para" id="averill_1.0-ch03_s04_s01_s01_p23">The volatility and toxicity of mercury make this a hazardous procedure, which likely shortened the life span of many alchemists. Given 100 g of cinnabar, how much elemental mercury can be produced from this reaction?</p>
<p class="para" id="averill_1.0-ch03_s04_s01_s01_p24"><strong class="emphasis bold">Answer: </strong>86.2 g</p>
</div>
</div>
</div>
<div class="section" id="averill_1.0-ch03_s04_s02">
<h2 class="title editable block">Limiting Reactants</h2>
<p class="para editable block" id="averill_1.0-ch03_s04_s02_p01">In all the examples discussed thus far, the reactants were assumed to be present in stoichiometric quantities. Consequently, none of the reactants was left over at the end of the reaction. This is often desirable, as in the case of a space shuttle, where excess oxygen or hydrogen was not only extra freight to be hauled into orbit but also an explosion hazard. More often, however, reactants are present in mole ratios that are not the same as the ratio of the coefficients in the balanced chemical equation. As a result, one or more of them will not be used up completely but will be left over when the reaction is completed. In this situation, the amount of product that can be obtained is limited by the amount of only one of the reactants. The reactant that restricts the amount of product obtained is called the <span class="margin_term"><a class="glossterm">limiting reactant</a><span class="glossdef">The reactant that restricts the amount of product obtained in a chemical reaction.</span></span>. The reactant that remains after a reaction has gone to completion is <em class="emphasis">in excess</em>.</p>
<p class="para editable block" id="averill_1.0-ch03_s04_s02_p02">To be certain you understand these concepts, let’s first consider a nonchemical example. Assume you have invited some friends for dinner and want to bake brownies for dessert. You find two boxes of brownie mix in your pantry and see that each package requires two eggs. The balanced equation for brownie preparation is thus</p>
<div class="equation block" id="averill_1.0-ch03_s04_s02_eq01">
<p class="title editable"><span class="title-prefix">Equation 3.21</span> </p>
<span class="mathphrase">1 box mix + 2 eggs → 1 batch brownies</span>
</div>
<p class="para editable block" id="averill_1.0-ch03_s04_s02_p03">If you have a dozen eggs, which ingredient will determine the number of batches of brownies that you can prepare? Because each box of brownie mix requires two eggs and you have two boxes, you need four eggs. Twelve eggs is eight more eggs than you need. Although the ratio of eggs to boxes in <a class="xref" href="#averill_1.0-ch03_s04_s02_eq01" xrefstyle="select:label">Equation 3.21</a> is 2:1, the ratio in your possession is 6:1. Hence the eggs are the ingredient (reactant) present in excess, and the brownie mix is the limiting reactant (<a class="xref" href="#averill_1.0-ch03_s04_s02_f01">Figure 3.12 "The Concept of a Limiting Reactant in the Preparation of Brownies"</a>). Even if you had a refrigerator full of eggs, you could make only two batches of brownies.</p>
<div class="figure large editable block" id="averill_1.0-ch03_s04_s02_f01">
<p class="title"><span class="title-prefix">Figure 3.12</span> The Concept of a Limiting Reactant in the Preparation of Brownies</p>
<img src="section_07/c790a451e58d96bd1e04dcf1ecc7057d.jpg">
</div>
<p class="para editable block" id="averill_1.0-ch03_s04_s02_p04">Let’s now turn to a chemical example of a limiting reactant: the production of pure titanium. This metal is fairly light (45% lighter than steel and only 60% heavier than aluminum) and has great mechanical strength (as strong as steel and twice as strong as aluminum). Because it is also highly resistant to corrosion and can withstand extreme temperatures, titanium has many applications in the aerospace industry. Titanium is also used in medical implants and portable computer housings because it is light and resistant to corrosion. Although titanium is the ninth most common element in Earth’s crust, it is relatively difficult to extract from its ores. In the first step of the extraction process, titanium-containing oxide minerals react with solid carbon and chlorine gas to form titanium tetrachloride (TiCl<sub class="subscript">4</sub>) and carbon dioxide. Titanium tetrachloride is then converted to metallic titanium by reaction with magnesium metal at high temperature:</p>
<div class="equation block" id="averill_1.0-ch03_s04_s02_eq02">
<p class="title editable"><span class="title-prefix">Equation 3.22</span> </p>
<span class="mathphrase">TiCl<sub class="subscript">4</sub>(g) + 2Mg(l) → Ti(s) + 2MgCl<sub class="subscript">2</sub>(l)</span>
</div>
<p class="para editable block" id="averill_1.0-ch03_s04_s02_p05">Because titanium ores, carbon, and chlorine are all rather inexpensive, the high price of titanium (about $100 per kilogram) is largely due to the high cost of magnesium metal. Under these circumstances, magnesium metal is the limiting reactant in the production of metallic titanium.</p>
<div class="informalfigure small block">
<img src="section_07/aee687834e11bbe430825273ab3698a3.jpg">
<p class="para"><strong class="emphasis bold">Medical use of titanium.</strong> Here is an example of its successful use in joint replacement implants.</p>
</div>
<p class="para editable block" id="averill_1.0-ch03_s04_s02_p06">Suppose you have 1.00 kg of titanium tetrachloride and 200 g of magnesium metal. How much titanium metal can you produce according to <a class="xref" href="#averill_1.0-ch03_s04_s02_eq02" xrefstyle="select:label">Equation 3.22</a>? Solving this type of problem requires that you carry out the following steps:</p>
<ol class="orderedlist editable block" id="averill_1.0-ch03_s04_s02_l01">
<li>Determine the number of moles of each reactant.</li>
<li>Compare the mole ratio of the reactants with the ratio in the balanced chemical equation to determine which reactant is limiting.</li>
<li>Calculate the number of moles of product that can be obtained from the limiting reactant.</li>
<li>Convert the number of moles of product to mass of product.</li>
</ol>
<p class="para editable block" id="averill_1.0-ch03_s04_s02_p07"> </p>
<ol class="orderedlist block" id="averill_1.0-ch03_s04_s02_l02">
<li>
<p class="para">To determine the number of moles of reactants present, you must calculate or look up their molar masses: 189.679 g/mol for titanium tetrachloride and 24.305 g/mol for magnesium. The number of moles of each is calculated as follows:</p>
<span class="informalequation">
<math xml:id="averill_1.0-ch03_m051" display="block">
<semantics>
<mtable columnalign="left">
<mtr columnalign="left">
<mtd columnalign="left">
<msub>
<mtext>moles TiCl</mtext>
<mtext>4</mtext>
</msub>
</mtd>
<mtd columnalign="left">
<mo>=</mo>
<mfrac>
<mrow>
<msub>
<mrow>
<mtext>mass TiCl</mtext>
</mrow>
<mtext>4</mtext>
</msub>
</mrow>
<mrow>
<msub>
<mrow>
<mtext>molar mass TiCl</mtext>
</mrow>
<mtext>4</mtext>
</msub>
</mrow>
</mfrac>
</mtd>
</mtr>
<mtr columnalign="left">
<mtd columnalign="left">
<mrow></mrow>
</mtd>
<mtd columnalign="left">
<mo>=</mo>
<mtext> 1000 </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<msub>
<mrow>
<mtext>g TiCl</mtext>
</mrow>
<mtext>4</mtext>
</msub>
</mrow>
</menclose>
<mo>×</mo>
<mfrac>
<mrow>
<msub>
<mrow>
<mtext>1 mol TiCl</mtext>
</mrow>
<mtext>4</mtext>
</msub>
</mrow>
<mrow>
<mtext>189</mtext>
<mtext>.679 </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<msub>
<mrow>
<mtext>g TiCl</mtext>
</mrow>
<mtext>4</mtext>
</msub>
</mrow>
</menclose>
</mrow>
</mfrac>
<mo>=</mo>
<mtext> 5</mtext>
<msub>
<mtext>.272 mol TiCl</mtext>
<mtext>4</mtext>
</msub>
</mtd>
</mtr>
<mtr columnalign="left">
<mtd columnalign="left">
<mtext>moles Mg </mtext>
</mtd>
<mtd columnalign="left">
<mo>=</mo>
<mfrac>
<mrow>
<mtext>mass Mg</mtext>
</mrow>
<mrow>
<mtext>molar mass Mg</mtext>
</mrow>
</mfrac>
</mtd>
</mtr>
<mtr columnalign="left">
<mtd columnalign="left">
<mrow></mrow>
</mtd>
<mtd columnalign="left">
<mo>=</mo>
<mtext> 200 </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<mtext>g Mg</mtext>
</mrow>
</menclose>
<mo>×</mo>
<mfrac>
<mrow>
<mtext>1 mol Mg</mtext>
</mrow>
<mrow>
<mtext>24</mtext>
<mtext>.305 </mtext>
<menclose notation="updiagonalstrike">
<mrow>
<mtext>g Mg</mtext>
</mrow>
</menclose>
</mrow>
</mfrac>
<mo>=</mo>
<mtext> 8</mtext>
<mtext>.23 mol Mg</mtext>
</mtd>
</mtr>
</mtable>
</semantics>
</math>
</span>
</li>
<li>
<p class="para">You have more moles of magnesium than of titanium tetrachloride, but the ratio is only</p>
<span class="informalequation">
<math xml:id="averill_1.0-ch03_m052" display="block">
<semantics>
<mrow>
<mfrac>
<mrow>
<mtext>mol Mg</mtext>
</mrow>
<mrow>
<msub>
<mrow>
<mtext>mol TiCl</mtext>
</mrow>
<mtext>4</mtext>
</msub>
</mrow>
</mfrac>
<mo>=</mo>
<mfrac>
<mrow>
<mtext>8</mtext>
<mtext>.23 mol</mtext>
</mrow>
<mrow>
<mtext>5</mtext>
<mtext>.272 mol</mtext>
</mrow>
</mfrac>
<mo>=</mo>
<mtext> 1</mtext>
<mtext>.56</mtext>
</mrow>
</semantics>
</math>
</span>
<p class="para" id="averill_1.0-ch03_s04_s02_p08">Because the ratio of the coefficients in the balanced chemical equation is</p>