Chapter 12 Stoichiometry 12.3 Limiting Reagent and Percent Yield 12.1 The Arithmetic of Equations 12.2 Chemical Calculations 12.3 Limiting Reagent and Percent Yield Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) Do Now: Methane burns in air by the following reaction: CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) What volume of water vapor is produced at STP by burning 501 g of methane? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) Do Now: Methane burns in air by the following reaction: CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) What volume of water vapor is produced at STP by burning 501 g of methane? = 1.40  103 L H2O Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

If I have: 100 Graham Crackers 75 Marshmallows 250 Chocolate Pieces 2 GC + 1 M + 4 Cp  1 Sm If I have: 100 Graham Crackers 75 Marshmallows 250 Chocolate Pieces How many S’mores can you make? What is the limiting ingredient? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Glossary Terms theoretical yield: maximum amount of product that could be formed from given reactants limiting reagent: reactant that is used up first; determines amount of product that can be made excess reagent: reagent present in a quantity that is more than sufficient to react with a limiting reagent; any reactant that remains after the limiting reagent is used up Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

How many moles of Cu2S could we make? What is the limiting reagent? 2Cu + S  Cu2S If we have: 100 moles Cu 75 moles S How many moles of Cu2S could we make? What is the limiting reagent? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

2Cu + S  Cu2S Determining the Limiting Reagent in a Reaction Sample Problem 12.8 Determining the Limiting Reagent in a Reaction Copper reacts with sulfur to form copper(I) sulfide according to the following balanced equation: 2Cu + S  Cu2S What is the limiting reagent when 80.0 g Cu reacts with 25.0 g S? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Determining Limiting Reagent Sample Problem 12.8 Determining Limiting Reagent Determine how much product you can make from each of the reactants. The reactant that produces the smallest amount of product is the limiting reagent. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Determining Limiting Reagent Sample Problem 12.8 Determining Limiting Reagent Copper 80.0 g Cu  63.5 g Cu 1 mol Cu 1 mol Cu2S  = 0.63 mol Cu2S 2 mol Cu Sulfur  = 0.78 mol Cu2S 1 mol S Copper is the limiting reagent. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Sample Problem 12.9 Using Limiting Reagent to Find the Quantity of a Product What is the maximum number of grams of Cu2S that can be formed when 80.0 g Cu reacts with 25.0 g S? 2Cu(s) + S(s)  Cu2S(s) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Convert from moles to mass of product. Sample Problem 12.9 Convert from moles to mass of product. 159.1 g Cu2S 1 mol Cu2S 0.63 mol Cu2S  = 100 g Cu2S Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Rust forms when iron, oxygen, and water react Rust forms when iron, oxygen, and water react. One chemical equation for the formation of rust is 2Fe + O2 + 2H2O  2Fe(OH)2 If 7.0 g of iron and 9.0 g of water are available to react, which is the limiting reagent? (Assume O2 is available in excess.) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Rust forms when iron, oxygen, and water react Rust forms when iron, oxygen, and water react. One chemical equation for the formation of rust is 2Fe + O2 + 2H2O  2Fe(OH)2 If 7.0 g of iron and 9.0 g of water are available to react, which is the limiting reagent? 7.00 g Fe   = 0.13 mol Fe(OH)2 1 mol Fe 55.85 g Fe 2 mol Fe 2 mol Fe(OH)2 9.00 g H2O   = 0.50 mol Fe(OH)2 1 mol H2O 18 g H2O 2 mol H2O 2 mol Fe(OH)2 Fe is the limiting reagent. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

CaCO3(s)  CaO(s) + CO2(g) Sample Problem 12.10 Calculating the Theoretical Yield of a Reaction Calcium carbonate, which is found in seashells, is decomposed by heating. The balanced equation for this reaction is CaCO3(s)  CaO(s) + CO2(g) D What is the theoretical yield of CaO if 24.8 g CaCO3 is heated? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Calculate Solve for the unknown. Sample Problem 12.10 Calculate Solve for the unknown. 2 24.8 g CaCO3    100.1 g CaCO3 1 mol CaCO3 1 mol CaO 56.1 g CaO = 13.9 g CaO Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Percent Yield theoretical yield : maximum amount of product that could be formed from given amounts of reactants. actual yield : amount of product that actually forms when the reaction is carried out in the laboratory. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

CaCO3(s)  CaO(s) + CO2(g) Sample Problem 12.11 Calculating the Percent Yield of a Reaction What is the percent yield if 13.1 g CaO is actually produced when 24.8 g CaCO3 is heated? Calculate the theoretical yield first. Then you can calculate the percent yield. CaCO3(s)  CaO(s) + CO2(g) D Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

13.1 g CaO percent yield =  100% = 94.2% 13.9 g CaO Sample Problem 12.11 percent yield =  100% = 94.2% 13.1 g CaO 13.9 g CaO Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Carbon tetrachloride, CCl4, is a solvent that was once used in large amounts in dry cleaning. One reaction that produces carbon tetrachloride is CS2 + 3Cl2  CCl4 + S2Cl2 What is the percent yield of CCl4 if 617 g is produced from the reaction of 312 g of CS2? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

What is the percent yield of CCl4 if 617 g is produced from the reaction of 312 g of CS2? CS2 + 3Cl2  CCl4 + S2Cl2 312 g CS2    76.142 g CS2 1 mol CS2 1 mol CCl4 153.81 g CCl4 Theoretical yield = 630 g CCl4 Percent yield =  100% = 97.9% 617 g CCl4 630 g CCl4 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

If 50. 0 g of silicon dioxide is heated with an excess of carbon, 27 If 50.0 g of silicon dioxide is heated with an excess of carbon, 27.9 g of silicon carbide is produced. SiO2 + 3C  SiC + 2CO What is the percent yield of this reaction? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

If 50. 0 g of silicon dioxide is heated with an excess of carbon, 27 If 50.0 g of silicon dioxide is heated with an excess of carbon, 27.9 g of silicon carbide is produced. SiO2 + 3C  SiC + 2CO What is the percent yield of this reaction? 83.5% Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

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