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9 For Mg(OH)2: Step 2 Using the molar mass of Mg(OH)2, which is 24.31 g ! 2(16.00 g) ! 2(1.008 g) # 58.33 g, we determine the moles of Mg(OH)2 in 1.00 g of Mg(OH)2. mol Mg(OH)2 1.00 g Mg(OH)2 " 1 ________________ # 0.0171 mol Mg(OH)2 58.33 g Mg(OH)2 # 1.71 " 10!2 mol Mg(OH)2 Step 3 To determine the moles of HCl that react with this amount of 2 mol HCl Mg(OH)2, we use the mole ratio _______________ . 1 mol Mg(OH)2 2 mol HCl 1.71 " 10!2 mol Mg(OH)2 " _______________ 1 mol Mg(OH)2 # 3.42 " 10!2 mol HCl Step 4 1.00 g of Mg(OH)2 neutralizes 3.42 " 10!2 mol of HCl. We have already calculated that 1.00 g of NaHCO3 neutralizes only 1.19 " 10!2 mol of HCl. Therefore, Mg(OH)2 is a more effective antacid than NaHCO3 on a mass basis. Active Reading Question In the problem above, we added equal masses (1.0 g) of each antacid. Why would one antacid consume more stomach acid than the other? SECTION 9.2 REVIEW QUESTIONS 1 Why do we need to convert mass to moles in stoichiometry problems? 2 Solutions of sodium hydroxide cannot be kept for very long because they absorb carbon dioxide from the air, forming sodium carbonate. The unbalanced equation is NaOH(aq) ! CO2(g) n Na2CO3(aq) ! H2O(l) Calculate the number of grams of carbon dioxide that can be absorbed by complete reaction with a solution that contains 5.00 g of sodium hydroxide. 3 Show how the steps in solving a stoichiometry problem are similar to a method that could be used in a supermarket to determine which of two products would be the best buy. 4 You react 10.0 g of nitrogen gas with hydrogen gas according to the following reaction. N2(g) ! 3H2(g) n 2NH3(g) a. What mass of hydrogen is required to completely react with 10.0 g sample of nitrogen gas? b. What mass of ammonia is produced from 10.0 g of nitrogen gas and sufficient hydrogen gas? 5 You react hydrogen gas with chlorine gas according to the following reaction. H2(g) ! Cl2(g) n 2HCl(g) What mass of HCl(g) can be produced from 2.5 " 103 g of hydrogen gas with an excess of chlorine gas? 6 Determine what mass of carbon monoxide and what mass of hydrogen are required to form 6.0 kg of methanol by the reaction CO(g) ! 2H2(g) n CH3OH(l). RESEARCH LINKS 9.2 • Using Chemical Equations to Calculate Mass • 295 SECTION 9.3 Limiting Reactants and Percent Yield Key Terms • Limiting reactant (Limiting reagent) • Theoretical yield • Percent yield Objectives • To understand the concept of limiting reactants • To learn to recognize the limiting reactant in a reaction • To learn to use the limiting reactant to do stoichiometric calculations • To learn to calculate percent yield A. The Concept of Limiting Reactants Earlier in this chapter, we discussed making sandwiches. Recall that the sandwich-making process could be described as follows: 2 pieces bread ! 3 slices meat ! 1 slice cheese n sandwich In our earlier discussion, we always purchased the ingredients in the correct ratios so that we used all of the components, with nothing left over. Now assume that you came to work one day and found the following quantities of ingredients: 20 slices of bread 24 slices of meat 12 slices of cheese How many sandwiches can you make? What will be left over? To solve this problem, let’s see how many sandwiches we can make with each component. Bread: 1 sandwich 20 slices bread " ________________ # 10 sandwiches 2 slices of bread Meat: 1 sandwich 24 slices meat " _______________ # 8 sandwiches 3 slices of meat Cheese: 1 sandwich 12 slices cheese " ________________ # 12 sandwiches 1 slice of cheese How many sandwiches can you make? The answer is 8. When you run out of meat, you must stop making sandwiches. The meat is the limiting ingredient. What do you have left over? Making 8 sandwiches requires 16 pieces of bread. You started with 20 pieces, so you have 4 pieces of bread left. You also used 8 pieces of cheese for the 8 sandwiches, so you have 12 $ 8 # 4 pieces of cheese left. Active Reading Question There are more slices of meat than of bread or cheese. So why does the meat limit the number of sandwiches you can make? In this example, the ingredient present in the largest number (the meat) was actually the component that limited the number of sandwiches you could make. This situation arose because each sandwich required 3 slices of meat—more than the quantity required of any other ingredient. 296 • Chapter 9 • Chemical Quantities 9 A Closer Look When molecules react with each other to form products, considerations very similar to those involved in making sandwiches arise. We can illustrate these ideas with the reaction of N2(g) and H2(g) to form NH3(g): N2(g) ! 3H2(g) n 2NH3(g) Consider the following container of N2(g) and H2(g): = H2 = N2 What will this container look like if the reaction between N2 and H2 proceeds to completion? To answer this question, you need to remember that each N2 requires 3 H2 molecules to form 2 NH3. To make things clear, we will circle groups of reactants: = H2 = N2 = NH3 Before the reaction After the reaction In this case, the mixture of N2 and H2 contained just the number of molecules needed to form NH3 with nothing left over. That is, the ratio of the number of H2 molecules to N2 molecules was 15 H2 3H ______ " _____2 5 N2 1 N2 This ratio exactly matches the numbers in the balanced equation 3H2(g) ! N2(g) n 2NH3(g). This type of mixture is called a stoichiometric mixture—one that contains the relative amounts of reactants that matches the numbers in the balanced equation. In this case all reactants will be consumed to form products. 9.3 • Limiting Reactants and Percent Yield • 297 Now consider another container of N2(g) and H2(g): = H2 = N2 What will the container look like if the reaction between N2(g) and H2(g) proceeds to completion? Remember that each N2 requires 3 H2. Circling groups of reactants we have = H2 = N2 = NH3 Before the reaction After the reaction In this case, the hydrogen (H2) is limiting. That is, the H2 molecules are used up before all of the N2 molecules are consumed. In this situation the amount of hydrogen limits the amount of product (ammonia) that can form—hydrogen is the limiting reactant. Some N2 molecules are left over in this case because the reaction runs out of H2 molecules first. To determine how much product can be formed from a given mixture of reactants, we have to look for the reactant that is limiting—the one that runs out first and thus limits the amount of product that can form. In some cases, the mixture of reactants might be stoichiometric—that is, all reactants run out at the same time. In general, however, you cannot assume that a given mixture of reactants is a stoichiometric mixture, so you must determine whether one of the reactants is limiting. Limiting reactant (limiting reagent) The reactant that is completely used up when a reaction is run to completion The reactant that runs out first and thus limits the amounts of products that can form is called the limiting reactant (limiting reagent). To this point, we have considered examples where the numbers of reactant molecules could be counted. In “real life” you can’t count the molecules directly—you can’t see them and, even if you could, there would be far too many to count. Instead, you must count by weighing. We must therefore explore how to find the limiting reactant, given the masses of the reactants. 298 • Chapter 9 • Chemical Quantities 9 Active Reading Question What is meant by the term “limiting reactant”? B. Calculations Involving a Limiting Reactant Manufacturers of cars, bicycles, and appliances order parts in the same proportion as they are used in their products. For example, auto manufacturers order four times as many wheels as engines and bicycle manufacturers order twice as many pedals as seats. Likewise, when chemicals are mixed together so that they can undergo a reaction, they are often mixed in stoichiometric quantities—that is, in exactly the correct amounts so that all reactants “run out” (are used up) at the same time. Let’s consider the production of hydrogen for use in the manufacture of ammonia. Ammonia, a very important fertilizer itself and a starting material for other fertilizers, is made by combining nitrogen from the air with hydrogen. The hydrogen for this process is produced by the reaction of methane with water according to the balanced equation CH4(g) ! H2O(g) n 3H2(g) ! CO(g) Let’s consider the question, What mass of water is required to react exactly with 249 g of methane? Where do we want to go? How much water will just use up all of the 249 g of methane, leaving no water or methane remaining? What do we know? • CH4(g) ! H2O(g) n 3 H2(g) ! CO(g) • Ammonia being dissolved in irrigation water to provide fertilizer for a field of corn 249 g CH4 How do we get there? Drawing a map of the problem is helpful. CH4(g) ! H2O(g) 249 g CH4 Grams of H 2O Use molar mass of CH4 Use molar mass of H2O Moles of CH4 Use mole ratio from balanced equation n 3H2(g) ! CO(g) Moles of H 2O 9.3 • Limiting Reactants and Percent Yield • 299 Step 1 The equation is already balanced. Step 2 We first convert the mass of CH4 to moles, using the molar mass of CH4 (16.04 g/mol). 1 mol CH4 249 g CH4 ! ____________ " 15.5 mol CH4 16.04 g CH4 Step 3 Because in the balanced equation 1 mol of CH4 reacts with 1 mol of H2O, we have 1 mol H2O 15.5 mol CH4 ! __________ " 15.5 mol H2O 1 mol CH4 Step 4 Therefore, 15.5 mol of H2O will react exactly with the given mass of CH4. Converting 15.5 mol of H2O to grams of H2O (molar mass " 18.02 g/mol) gives 18.02 g H2O 15.5 mol H2O ! ____________ " 279 g H2O 1 mol H2O i nformation The reactant that is consumed first limits the amounts of products that can form. This result means that if 249 g of methane is mixed with 279 g of water, both reactants will “run out” at the same time. The reactants have been mixed in stoichiometric quantities. If, on the other hand, 249 g of methane is mixed with 300 g of water, the methane will be consumed before the water runs out. The water will be in excess. In this case, the quantity of products formed will be determined by the quantity of methane present. After the methane is consumed, no more products can be formed, even though some water still remains. In this situation, the amount of methane limits the amount of products that can be formed, and it is the limiting reactant. In any stoichiometry problem, where reactants are not mixed in stoichiometric quantities, it is essential to determine which reactant is limiting to calculate correctly the amounts of products that will be formed. This concept is illustrated in Figure 9.1. Note from this figure that because there are fewer water molecules than CH4 molecules, the water is consumed first. After the water molecules are gone, no more products can form. So in this case water is the limiting reactant. Figure 9.1 A mixture of 5CH4 and 3H2O molecules undergoes the reaction CH4(g) # H2O(g) n 3H2(g) # CO(g). Note that the H2O molecules are used up first, leaving two CH4 molecules unreacted. 300 • Chapter 9 • Chemical Quantities 9 E XA MP L E 9 . 7 Stoichiometric Calculations: Identifying the Limiting Reactant Suppose that 25.0 kg (2.50 ! 104 g) of nitrogen gas and 5.00 kg (5.00 ! 103 g) of hydrogen gas are mixed and reacted to form ammonia. Calculate the mass of ammonia produced when this reaction is run to completion. Solution Where do we want to go? How much ammonia will be produced when the reaction is complete? What do we know? • N2(g) " H2(g) n NH3(g) • • 2.5 ! 104 g N2 5.00 ! 103 g H2 How do we get there? Step 1 The balanced equation for this reaction is N2(g) " 3H2(g) n 2NH3(g) This problem is different from the others we have done so far in that we are mixing specified amounts of two reactants. To know how much product forms, we must determine which reactant is consumed first. That is, we must determine which is the limiting reactant in this experiment. To do so, we must add a step to our normal procedure. We can map this process as follows: N2(g) " 3H2(g) 2.50 ! 104 g N2 5.00 ! 103 g H2 Use molar mass of N2 Use molar mass of H2 Moles of N2 Use mole ratios to determine limiting reactant n 2NH3(g) Moles of H2 Moles of limiting reactant We will use the moles of the limiting reactant to calculate the moles and then the grams of the product. 9.3 • Limiting Reactants and Percent Yield • 301 Step 2 We first calculate the moles of the two reactants present: 1 mol N2 2.50 ! 104 g N2 ! __________ " 8.92 ! 102 mol N2 28.02 g N2 1 mol H2 5.00 ! 103 g H2 ! __________ " 2.48 ! 103 mol H2 2.016 g H2 Step 3 Now we must determine which reactant is limiting (will be consumed first). We have 8.92 ! 102 mol of N2. Let’s determine how many moles of H2 are required to react with this much N2. Because 1 mol of N2 reacts with 3 mol of H2, the number of moles of H2 we need to react completely with 8.92 ! 102 mol of N2 is determined as follows: 8.92 ! 102 mol N2 3 mol H2 1 mol N2 Moles of H2 required 3 mol H 8.92 ! 102 mol N2 ! _________2 " 2.68 ! 103 mol H2 1 mol N2 Is N2 or H2 the limiting reactant? The answer comes from the comparison Moles of H2 available 2.48 ! 103 less than Moles of H2 required 2.68 ! 103 We see that 8.92 ! 102 mol of N2 requires 2.68 ! 103 mol of H2 to react completely. However, only 2.48 ! 103 mol of H2 is present. This means that the hydrogen will be consumed before the nitrogen runs out, so hydrogen is the limiting reactant in this particular situation. Note that in our effort to determine the limiting reactant, we could have started instead with the given amount of hydrogen and calculated the moles of nitrogen required. 2.48 ! 103 mol H2 1 mol N2 3 mol H2 Moles of N2 required 1 mol N 2.48 ! 103 mol H2 ! _________2 " 8.27 ! 102 mol N2 3 mol H2 Thus, 2.48 ! 103 mol of H2 requires 8.27 ! 102 mol of N2. Because 8.92 ! 102 mol of N2 is actually present, the nitrogen is in excess. Moles of N2 available 8.92 ! 102 302 • Chapter 9 • Chemical Quantities greater than Moles of N2 required 8.27 ! 102 If nitrogen is in excess, hydrogen will “run out” first; again we find that hydrogen limits the amount of ammonia formed. H2 is limiting reactant. i N2 is in excess. nformation Always check to see which, if any, reactant is limiting when you are given the amounts of two or more reactants. Step 4 Because the moles of H2 present are limiting, we must use this quantity to determine the moles of NH3 that can form. 2 mol NH 2.48 ! 103 mol H2 ! ___________3 " 1.65 ! 103 mol NH3 3 mol H2 Step 5 Next, we convert moles of NH3 to mass of NH3. 17.03 g NH 1.65 ! 103 mol NH3 ! ____________3 " 2.81 ! 104 g NH3 1 mol NH3 " 28.1 kg NH3 Therefore, 25.0 kg of N2 and 5.00 kg of H2 can form 28.1 kg of NH3. The strategy used in Example 9.7 is summarized in Figure 9.2. Grams of H2 Molar mass of H2 Moles of H2 Figure 9.2 A map of the procedure used in Example 9.7 H2 limiting Grams of N2 Molar mass of N2 Moles of H2 2 mol NH3 3 mol H2 Moles of NH3 Molar mass of NH3 Grams of NH3 Moles of N2 L e t ’s R e v i e w Steps for Solving Stoichiometry Problems Involving Limiting Reactants Step 1 Write and balance the equation for the reaction. Step 2 Convert known masses of reactants to moles. Step 3 Using the numbers of moles of reactants and the appropriate mole ratios, determine which reactant is limiting. Step 4 Using the amount of the limiting reactant and the appropriate mole ratios, compute the number of moles of the desired product. Step 5 Convert from moles of product to grams of product, using the molar mass (if this is required by the problem). 9.3 • Limiting Reactants and Percent Yield • 303 EXA MP LE 9 .8 Stoichiometric Calculations: Reactions Involving the Masses of Two Reactants Nitrogen gas can be prepared by passing gaseous ammonia over solid copper(II) oxide at high temperatures. The other products of the reaction are solid copper and water vapor. How many grams of N2 are formed when 18.1 g of NH3 is reacted with 90.4 g of CuO? Solution Where do we want to go? How much nitrogen will be produced when the reaction is complete? What do we know? Copper(II) oxide reacting with ammonia in a heated tube • NH3(g) ! CuO(s) n N2(g) ! Cu(s) ! H2O(g) • 18.1 g NH3 • 90.4 g CuO How do we get there? Step 1 Balance the equation: 2NH3(g) ! 3CuO(s) n N2(g) ! 3Cu(s) ! 3H2O(g) Step 2 From the masses of reactants available, we must compute the moles of NH3 (molar mass " 17.03 g) and of CuO (molar mass " 79.55 g). 1 mol NH3 18.1 g NH3 # ____________ 17.03 g NH3 " 1.06 mol NH3 1 mol CuO " 1.14 mol CuO 90.4 g CuO # ____________ 79.55 g CuO Step 3 To determine which reactant is limiting, we use the mole ratio between CuO and NH3 to determine how CuO is required. mol CuO " 1.59 mol CuO 1.06 mol NH3 # 3___________ 2 mol NH3 Then we compare how much CuO we have with how much of it we need. Moles of CuO available 1.14 less than Moles of CuO needed to react with all the NH3 1.59 Therefore, 1.59 mol of CuO is required to react with 1.06 mol of NH3, but only 1.14 mol of CuO is actually present. So the amount of CuO is limiting; CuO will run out before NH3 does. Note that CuO is limiting even though the original mass of CuO was much greater than the original mass of NH3. 304 • Chapter 9 • Chemical Quantities 9 Step 4 CuO is the limiting reactant, so we must use the amount of CuO in calculating the amount of N2 formed. Using the mole ratio between CuO and N2 from the balanced equation, we have 1 mol N2 1.14 mol CuO ! ___________ " 0.380 mol N2 3 mol CuO Step 5 Using the molar mass of N2 (28.02), we can now calculate the mass of N2 produced. 28.02 g N 0.380 mol N2 ! __________2 " 10.6 g N2 1 mol N2 Practice Problem • Exercise 9.8 Lithium nitride, an ionic compound containing the Li! and N3" ions, is prepared by the reaction of lithium metal and nitrogen gas. Calculate the mass of lithium nitride formed from 56.0 g of nitrogen gas and 56.0 g of lithium in the unbalanced reaction: Li(s) # N2(g) n Li3N(s) Active Reading Question In the example, how can CuO be limiting if we started with a greater mass of CuO than NH3? C. Percent Yield Previously, we calculated the amount of product formed when specified amounts of reactants were mixed together. In doing these calculations, we used the fact that the amount of product is controlled by the limiting reactant. Products stop forming when one reactant runs out. The amount of product calculated in this way is called the theoretical yield of that product. It is the amount of product predicted from the amounts of reactants used. For instance, in Example 9.8, 10.6 g of nitrogen represents the theoretical yield. This is the maximum amount of nitrogen that can be produced from the quantities of reactants used. Actually, however, the amount of product predicted (the theoretical yield) is seldom obtained. One reason for this is the presence of side reactions (other reactions that consume one or more of the reactants or products). The actual yield of product, which is the amount of product actually obtained, is often compared to the theoretical yield. This comparison, usually expressed as a percent, is called the percent yield. Actual yield ________________ ! 100% " percent yield Theoretical yield For example, if the reaction considered in Example 9.8 actually gave 6.63 g of nitrogen instead of the predicted 10.6 g, the percent yield of nitrogen would be 6.63 g N2 _________ ! 100% " 62.5% 10.6 g N2 Theoretical yield The maximum amount of a given product that can be formed when the limiting reactant is completely consumed Percent yield The actual yield of a product as the percentage of the theoretical yield i nformation Percent yield is important as an indicator of the efficiency of a particular reaction. 9.3 • Limiting Reactants and Percent Yield • 305 Active Reading Question Which do we calculate by using stoichiometry, the actual or the theoretical yield? EXA MP L E 9. 9 Stoichiometric Calculations: Determining Percent Yield Methanol can be produced by the reaction between carbon monoxide and hydrogen. Let’s consider this process again. Suppose 68.5 kg (6.85 ! 104 g) of CO(g) is reacted with 8.60 kg (8.60 ! 103 g) of H2(g). MATH • Calculate the theoretical yield of methanol. Part ! 100% Percent # ______ Whole • If 3.57 ! 104 g of CH3OH is actually produced, what is the percent yield of methanol? Solution Where do we want to go? • theoretical yield of CH3OH • percent yield of CH3OH What do we know? • CO(g) " H2(g) n CH3OH(l) • 8.60 ! 103 g H2 • 6.85 ! 104 g CO • 3.57 ! 104 g CH3OH produced How do we get there? Step 1 The balanced equation is 2H2(g) " CO(g) n CH3OH(l). Step 2 Next, we calculate the moles of reactants. 1 mol CO # 2.45 ! 103 mol CO 6.85 ! 104 g CO ! ___________ 28.01 g CO 1 mol H2 8.60 ! 103 g H2 ! __________ # 4.27 ! 103 mol H2 2.016 g H2 Step 3 Now we determine which reactant is limiting. Using the mole ratio between CO and H2 from the balanced equation, we have 2 mol H2 # 4.90 ! 103 mol H2 2.45 ! 103 mol CO ! __________ 1 mol CO Moles of H2 present 4.27 ! 103 306 • Chapter 9 • Chemical Quantities less than Moles of H2 needed to react with all the CO 4.90 ! 103 We see that 2.45 ! 103 mol of CO requires 4.90 ! 103 mol of H2. Because only 4.27 ! 103 mol of H2 is actually present, H2 is limiting. Step 4 We must therefore use the amount of H2 and the mole ratio between H2 and CH3OH to determine the maximum amount of methanol that can be produced in the reaction. 1 mol CH3OH 4.27 ! 103 mol H2 ! ______________ " 2.14 ! 103 mol CH3 2 mol H2 This represents the theoretical yield in moles. Step 5 Using the molar mass of CH3OH (32.04 g), we can calculate the theoretical yield in grams. 32.04 g CH3OH 2.14 ! 103 mol CH3OH ! _______________ " 6.86 ! 104 g CH3OH 1 mol CH3OH So, from the amounts of reactants given, the maximum amount of CH3OH that can be formed is 6.85 ! 104 g. This is the theoretical yield. The percent yield is Actual yield (grams) ________________________ ! 100% Theoretical yield (grams) 3.57 ! 104 g CH3OH " ____________________ ! 100% 6.86 ! 104 g CH3OH " 52.0% Practice Problem • Exercise 9.9 Titanium(IV) oxide is a white compound used as a coloring pigment. In fact, the page you are now reading is white because of the presence of this compound in the paper. Solid titanium(IV) oxide can be prepared by reacting gaseous titanium(IV) chloride with oxygen gas. A second product of this reaction is chlorine gas. TiCl4(g) # O2(g) n TiO2(s) # Cl2(g) a. Suppose that 6.71 ! 103 g of titanium(IV) chloride is reacted with 2.45 ! 103 g of oxygen. Calculate the maximum mass of titanium(IV) oxide that can form. b. If the percent yield of TiO2 is 75%, what mass is actually formed? 9.3 • Limiting Reactants and Percent Yield • 307