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

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Presentation on theme: "Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Chapter 12 Stoichiometry 12.1 The Arithmetic of Equations 12.2 Chemical Calculations."— Presentation transcript:

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

2 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. CHEMISTRY & YOU What determines how much product you can make? If a carpenter had two tabletops and seven table legs, he would have difficulty building more than one functional four- legged table.

3 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Limiting and Excess Reagents To make tacos, you need enough meat, cheese, lettuce, tomatoes, sour cream, salsa, and seasonings. If you have only 2 taco shells, the quantity of taco shells will limit the number of tacos you can make. Thus, the taco shells are the limiting ingredient.

4 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Limiting and Excess Reagents In a chemical reaction, an insufficient quantity of any of the reactants will limit the amount of product that forms.

5 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Limiting and Excess Reagents A balanced chemical equation is a chemist’s recipe. What would happen if two molecules (moles) of N 2 reacted with three molecules (moles) of H 2 ? Experimental Conditions ReactantsProducts Before reaction 2 molecules N 2 3 molecules H 2 0 molecules NH 3 Chemical Equations N2(g)N2(g)+3H 2 (g)2NH 3 (g) “Microscopic recipe”1 molecule N 2 +3 molecules H 2 2 molecules NH 3 “Macroscopic recipe”1 mol N 2 +3 mol H 2 2 mol NH 3

6 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Limiting and Excess Reagents Before the reaction takes place, N 2 and H 2 are present in a 2:3 molecule (mole) ratio. As the reaction takes place, one molecule (mole) of N 2 reacts with 3 molecules (moles) of H 2 to produce two molecules (moles) of NH 3. Experimental Conditions ReactantsProducts Before reaction 2 molecules N 2 3 molecules H 2 0 molecules NH 3 After reaction 1 molecule N 2 0 molecules H 2 2 molecules NH 3

7 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Limiting and Excess Reagents Experimental Conditions ReactantsProducts Before reaction 2 molecules N 2 3 molecules H 2 0 molecules NH 3 After reaction 1 molecule N 2 0 molecules H 2 2 molecules NH 3 All the H 2 has now been used up, and the reaction stops. One molecule (mole) of unreacted N 2 is left in addition to the two molecules (moles) of NH 3 that have been produced by the reaction.

8 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Limiting and Excess Reagents Experimental Conditions ReactantsProducts Before reaction 2 molecules N 2 3 molecules H 2 0 molecules NH 3 After reaction 1 molecule N 2 0 molecules H 2 2 molecules NH 3 In this reaction, only the hydrogen is completely used up. H 2 is the limiting reagent, or the reactant that determines the amount of product that can be formed by a reaction.

9 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Limiting and Excess Reagents Experimental Conditions ReactantsProducts Before reaction 2 molecules N 2 3 molecules H 2 0 molecules NH 3 After reaction 1 molecule N 2 0 molecules H 2 2 molecules NH 3 The reactant that is not completely used up in a reaction is called the excess reagent. In this example, nitrogen is the excess reagent because some nitrogen remains unreacted.

10 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. CHEMISTRY & YOU What determines how much product you can make in a chemical reaction? A limited quantity of any of the reactants that are needed to make a product will limit the amount of product that forms.

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

12 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Analyze List the knowns and the unknown. 1 The number of moles of each reactant must first be found. The balanced equation is used to calculate the number of moles of one reactant needed to react with the given amount of the other reactant. KNOWNS mass of copper = 80.0 g Cu mass of sulfur = 25.0 g S molar mass of Cu = 63.5 g/mol molar mass of S = 32.1 g/mol 1 mol S/2 mol Cu UNKNOWN limiting reagent = ? Sample Problem 12.8

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

14 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Analyze List the knowns and the unknown. 1 The limiting reagent, which was determined in the previous sample problem, is used to calculate the maximum amount of Cu 2 S formed. KNOWNS limiting reagent = 1.26 mol Cu (from sample problem 12.8) 1 mol Cu 2 S = g Cu 2 S (molar mass) 1 mol Cu 2 S/2 mol Cu (mole ratio from balanced equation) UNKNOWN Yield = ? g Cu 2 S Sample Problem 12.9

15 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Finish the calculation by converting from moles to mass of product. Calculate Solve for the unknown. 2 Sample Problem g Cu 2 S 1 mol Cu 2 S 1.26 mol Cu  2 mol Cu 1 mol Cu 2 S  = 1.00  10 2 g Cu 2 S

16 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Copper is the limiting reagent in this reaction. The maximum number of grams of Cu 2 S produced should be more than the amount of copper that initially reacted because copper is combining with sulfur. However, the mass of Cu 2 S produced should be less than the total mass of the reactants (105.0 g) because sulfur was in excess. Evaluate Do the results make sense? 3 Sample Problem 12.9

17 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Only 2.26 g H 2 O are needed to react with 7.0 g Fe. Therefore, Fe is the limiting reagent g Fe    = 2.26 g H mol Fe g Fe 18.0 g H 2 O 1 mol H 2 O 2 mol Fe 2 mol H 2 O Rust forms when iron, oxygen, and water react. One chemical equation for the formation of rust is 2Fe + O 2 + 2H 2 O  2Fe(OH) 2 If 7.0 g of iron and 9.0 g of water are available to react, which is the limiting reagent?

18 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Percent Yield What does the percent yield of a reaction measure? A batting average is actually a percent yield.

19 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Percent Yield When a balanced chemical equation is used to calculate the amount of product that will form during a reaction, the calculated value represents the theoretical yield. The theoretical yield is the maximum amount of product that could be formed from given amounts of reactants. The amount of product that actually forms when the reaction is carried out in the laboratory is called the actual yield.

20 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Percent Yield The percent yield is the ratio of the actual yield to the theoretical yield expressed as a percent. Because the actual yield of a chemical reaction is often less than the theoretical yield, the percent yield is often less than 100%. percent yield = actual yield theoretical yield  100%

21 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. The percent yield is a measure of the efficiency of a reaction carried out in the laboratory. Percent Yield The mass of the reactant is measured. The mass of one of the products, the actual yield, is measured. The percent yield is calculated. The reactant is heated.

22 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Percent Yield Many factors cause percent yields to be less than 100%. Reactions do not always go to completion; when a reaction is incomplete, less than the calculated amount of product is formed. Impure reactants and competing side reactions may cause unwanted products to form. Actual yield can be lower than the theoretical yield due to a loss of product during filtration or in transferring between containers. If reactants or products have not been carefully measured, a percent yield of 100% is unlikely.

23 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Calcium carbonate, which is found in seashells, is decomposed by heating. The balanced equation for this reaction is Sample Problem Calculating the Theoretical Yield of a Reaction CaCO 3 (s)  CaO(s) + CO 2 (g)  What is the theoretical yield of CaO if 24.8 g CaCO 3 is heated?

24 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Analyze List the knowns and the unknown. 1 Calculate the theoretical yield using the mass of the reactant. KNOWNS mass of CaCO 3 = 24.8 g CaCO 3 1 mol CaCO 3 = g CaCO 3 (molar mass) 1 mol CaO = 56.1 g CaO (molar mass) 1 mol CaO/1 mol CaCO 3 (mole ratio from balanced equation) UNKNOWN theoretical yield = ? g CaO Sample Problem 12.10

25 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Finish by converting from moles to mass of the product. Calculate Solve for the unknown. 2 Sample Problem g CaCO 3    g CaCO 3 1 mol CaCO 3 1 mol CaO 56.1 g CaO = 13.9 g CaO If there is an excess of a reactant, then there is more than enough of that reactant and it will not limit the yield of the reaction.

26 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. The mole ratio of CaO to CaCO 3 is 1:1. The ratio of their masses in the reaction should be the same as the ratio of their molar masses, which is slightly greater than 1:2. The result of the calculations shows that the mass of CaO is slightly greater than half the mass of CaCO 3. Evaluate Does the result make sense? 3 Sample Problem 12.10

27 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. What is the percent yield if 13.1 g CaO is actually produced when 24.8 g CaCO 3 is heated? Sample Problem Calculating the Percent Yield of a Reaction CaCO 3 (s)  CaO(s) + CO 2 (g)  Calculate the theoretical yield first. Then you can calculate the percent yield.

28 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Analyze List the knowns and the unknown. 1 Use the equation for percent yield. The theoretical yield for this problem was calculated in Sample Problem UNKNOWN percent yield = ? % Sample Problem KNOWNS actual yield = 13.1 g CaO theoretical yield = 13.9 g CaO (from sample problem 12.10) percent yield = actual yield theoretical yield  100%

29 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Substitute the values for actual yield and theoretical yield into the equation for percent yield. Calculate Solve for the unknown. 2 Sample Problem percent yield =  100% = 94.2% 13.1 g CaO 13.9 g CaO

30 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. In this example, the actual yield is slightly less than the theoretical yield. Therefore, the percent yield should be slightly less than 100%. Evaluate Does the result make sense? 3 Sample Problem 12.11

31 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Carbon tetrachloride, CCl 4, is a solvent that was once used in large amounts in dry cleaning. One reaction that produces carbon tetrachloride is CS 2 + 3Cl 2  CCl 4 + S 2 Cl 2 What is the percent yield of CCl 4 if 617 kg is produced from the reaction of 312 kg of CS 2 ?

32 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. CS 2 + 3Cl 2  CCl 4 + S 2 Cl 2 What is the percent yield of CCl 4 if 617 kg is produced from the reaction of 312 kg of CS 2 ? 3.12  10 5 g CS 2    g CS 2 1 mol CS 2 1 mol CCl 4 1 mol CS g CCl 4 1 mol CCl 4 = 6.30  10 5 g CCl 4 = 630 kg CCl 4 Percent yield =  100% = 97.9% 617 kg CCl kg CCl 4

33 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Key Concepts and Key Equation In a chemical reaction, an insufficient quantity of any of the reactants will limit the amount of product that forms. The percent yield is a measure of the efficiency of a reaction performed in the laboratory. percent yield = actual yield theoretical yield  100%

34 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Glossary Terms limiting reagent: any reactant that is used up first in a chemical reaction; it determines the amount of product that can be formed in the reaction excess reagent: a 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 in a chemical reaction

35 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. theoretical yield: the amount of product that could form during a reaction calculated from a balanced chemical equation; it represents the maximum amount of product that could be formed from a given amount of reactant actual yield: the amount of product that forms when a reaction is carried out in the laboratory percent yield: the ratio of the actual yield to the theoretical yield for a chemical reaction expressed as a percentage; a measure of the efficiency of a reaction Glossary Terms

36 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. The Mole and Quantifying Matter BIG IDEA The percent yield of a reaction can be calculated from the actual yield and theoretical yield of the reaction.


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