Chapter 9. Table of Contents. Chapter 9. Lesson Starter. Chapter 9. Objective. Stoichiometry. Section 1 Introduction to Stoichiometry

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1 Stoichiometry Table of Contents Section 1 Introduction to Stoichiometry Section 3 Limiting Reactants and Percentage Yield Section 1 Introduction to Stoichiometry Lesson Starter Mg(s) + 2HCl(aq)? MgCl 2 (aq) + H 2 (g) If 2 mol of HCl react, how many moles of H 2 are obtained? How many moles of Mg will react with 2 mol of HCl? If 4 mol of HCl react, how many mol of each product are produced? How would you convert from moles of substances to masses? Section 1 Introduction to Stoichiometry Objective Define stoichiometry. Describe the importance of the mole ratio in stoichiometric calculations. Write a mole ratio relating two substances in a chemical equation.

2 Section 1 Introduction to Stoichiometry Stoichiometry Definition Composition stoichiometry deals with the mass relationships of elements in compounds. Reaction stoichiometry involves the mass relationships between reactants and products in a chemical reaction. Visual Concepts Stoichiometry Click below to watch the Visual Concept. Visual Concept Section 1 Introduction to Stoichiometry Reaction Stoichiometry Problems Problem Type 1: Given and unknown quantities are amounts in moles. Amount of given substance (mol) Problem Type 2: Given is an amount in moles and unknown is a mass Amount of given substance (mol) Amount of unknown substance (mol) Amount of unknown substance (mol) Mass of unknown substance (g)

3 Section 1 Introduction to Stoichiometry Reaction Stoichiometry Problems, continued Problem Type 3: Given is a mass and unknown is an amount in moles. Mass of given substance (g) Amount of given substance (mol) Amount of unknown substance (mol) Problem Type 4: Given is a mass and unknown is a mass. Mass of a given substance (g) Amount of given substance (mol) Amount of unknown substance (mol) Mass of unknown substance (g) Section 1 Introduction to Stoichiometry Mole Ratio A mole ratio is a conversion factor that relates the amounts in moles of any two substances involved in a chemical reaction Example: 2Al 2 O 3 (l)? 4Al(s) + 3O 2 (g),, Mole Ratios: 2 mol Al 2 O 3 2 mol Al 2 O 3 4 mol Al 4 mol Al 3 mol O 2 3 mol O 2 Section 1 Introduction to Stoichiometry Converting Between Amounts in Moles

4 Section 1 Introduction to Stoichiometry Stoichiometry Visual Concepts Molar Mass as a Conversion Factor Click below to watch the Visual Concept. Visual Concept Lesson Starter Acid-Base Neutralization Reaction Demonstration What is the equation for the reaction of HCl with NaOH? What is the mole ratio of HCl to NaOH?

5 Objective Calculate the amount in moles of a reactant or a product from the amount in moles of a different reactant or product. Calculate the mass of a reactant or a product from the amount in moles of a different reactant or product. Objectives, continued Calculate the amount in moles of a reactant or a product from the mass of a different reactant or product. Calculate the mass of a reactant or a product from the mass of a different reactant or product. Conversions of Quantities in Moles

6 Visual Concepts Conversion of Quantities in Moles Click below to watch the Visual Concept. Visual Concept Solving Mass-Mass Stoichiometry Problems Conversions of Quantities in Moles, continued Sample Problem A In a spacecraft, the carbon dioxide exhaled by astronauts can be removed by its reaction with lithium hydroxide, LiOH, according to the following chemical equation. CO 2 (g) + 2LiOH(s)? Li 2 CO 3 (s) + H 2 O(l) How many moles of lithium hydroxide are required to react with 20 mol CO 2, the average amount exhaled by a person each day?

7 Conversions of Quantities in Moles, continued Sample Problem A Solution CO 2 (g) + 2LiOH(s)? Li 2 CO 3 (s) + H 2 O(l) Given: amount of CO 2 = 20 mol Unknown: amount of LiOH (mol) Solution: mol ratio mol LiOH mol CO 2?? mol LiOH mol CO 2 2 mol LiOH 20 mol CO 2?? 40 mol LiOH 1 mol CO 2 Conversions of Amounts in Moles to Mass Solving Stoichiometry Problems with Moles or Grams

8 Conversions of Amounts in Moles to Mass, continued Sample Problem B In photosynthesis, plants use energy from the sun to produce glucose, C 6 H 12 O 6, and oxygen from the reaction of carbon dioxide and water. What mass, in grams, of glucose is produced when 3.00 mol of water react with carbon dioxide? Conversions of Amounts in Moles to Mass, continued Sample Problem B Solution Given: amount of H 2 O = 3.00 mol Unknown: mass of C 6 H 12 O 6 produced (g) Solution: Balanced Equation: 6CO 2 (g) + 6H 2 O(l)? C 6 H 12 O 6 (s) + 6O 2 (g) mol ratio molar mass factor mol H 2 O? mol C 6 H 12 O 6 mol H 2 O? g C 6 H 12 O 6 mol C 6 H 12 O 6? g C 6 H 12 O mol H 2 O? 1 mol C 6 H 12 O 6 6 mol H 2 O? g C 6 H 12 O 6 1 mol C 6 H 12 O 6 = 90.1 g C 6 H 12 O 6 Conversions of Mass to Amounts in Moles

9 Visual Concepts Mass and Number of Moles of an Unknown Click below to watch the Visual Concept. Visual Concept Conversions of Mass to Amounts in Moles, continued Sample Problem D The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH 3 (g) + O 2 (g)? NO(g) + H 2 O(g) (unbalanced) The reaction is run using 824 g NH 3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H 2 O are formed? Conversions of Mass to Amounts in Moles, continued Sample Problem D Solution Given: mass of NH 3 = 824 g Unknown: a. amount of NO produced (mol) b. amount of H 2 O produced (mol) Solution: Balanced Equation: 4NH 3 (g) + 5O 2 (g)? 4NO(g) + 6H 2 O(g) molar mass factor mol ratio a. g NH 3? mol NH 3 mol NO?? mol NO g NH 3 mol NH 3 b. g NH 3? mol NH 3 g NH 3? mol H 2 O mol NH 3? mol H 2 O

10 Conversions of Mass to Amounts in Moles, continued Sample Problem D Solution, continued a. 824 g NH 3? molar mass factor 1 mol NH g NH 3? mol ratio 4 mol NO 4 mol NH 3? 48.4 mol NO b. 824 g NH 3? 1 mol NH 3? 6 mol H O 2? 72.5 mol H g NH 3 4 mol NH 2 O 3 Mass-Mass to Visual Concepts Mass-Mass Click below to watch the Visual Concept. Visual Concept

11 Solving Mass-Mass Problems Mass-Mass to, continued Sample Problem E Tin(II) fluoride, SnF 2, is used in some toothpastes. It is made by the reaction of tin with hydrogen fluoride according to the following equation. Sn(s) + 2HF(g)? SnF 2 (s) + H 2 (g) How many grams of SnF 2 are produced from the reaction of g HF with Sn? Mass-Mass to, continued Sample Problem E Solution Given: amount of HF = g Unknown: mass of SnF 2 produced (g) Solution: g HF? g HF? molar mass factor mol ratio molar mass factor mol HF g HF? mol SnF 2 mol HF 1 mol HF g HF? 1 mol SnF 2 2 mol HF = g SnF 2? g SnF 2 mol SnF 2? g SnF 2? g SnF 2 1 mol SnF 2

12 Solving Various Types of Stoichiometry Problems Solving Various Types of Stoichiometry Problems Solving Volume-Volume Problems

13 Solving Particle Problems Section 3 Limiting Reactants and Percentage Yield Objectives Describe a method for determining which of two reactants is a limiting reactant. Calculate the amount in moles or mass in grams of a product, given the amounts in moles or masses in grams of two reactants, one of which is in excess. Distinguish between theoretical yield, actual yield, and percentage yield. Calculate percentage yield, given the actual yield and quantity of a reactant. Section 3 Limiting Reactants and Percentage Yield Limiting Reactants The limiting reactant is the reactant that limits the amount of the other reactant that can combine and the amount of product that can form in a chemical reaction. The excess reactant is the substance that is not used up completely in a reaction.

14 Visual Concepts Limiting Reactants and Excess Reactants Click below to watch the Visual Concept. Visual Concept Section 3 Limiting Reactants and Percentage Yield Limited Reactants, continued Sample Problem F Silicon dioxide (quartz) is usually quite unreactive but reacts readily with hydrogen fluoride according to the following equation. SiO 2 (s) + 4HF(g)? SiF 4 (g) + 2H 2 O(l) If 6.0 mol HF is added to 4.5 mol SiO 2, which is the limiting reactant? Section 3 Limiting Reactants and Percentage Yield Limited Reactants, continued Sample Problem F Solution SiO 2 (s) + 4HF(g)? SiF 4 (g) + 2H 2 O(l) Given: amount of HF = 6.0 mol amount of SiO 2 = 4.5 mol Unknown: limiting reactant Solution: mole ratio mol HF? mol SiF 4 mol HF mol SiO 2? mol SiF 4 mol SiO 2? mol SiF 4 produced? mol SiF 4 produced

15 Section 3 Limiting Reactants and Percentage Yield Limited Reactants, continued Sample Problem F Solution, continued SiO 2 (s) + 4HF(g)? SiF 4 (g) + 2H 2 O(l) 6.0 mol HF? 1 mol SiF 4 4 mol HF? 1.5 mol SiF 4 produced 4.5 mol SiO 2? 1 mol SiF 4 1 mol SiO 2? 4.5 mol SiF 4 produced HF is the limiting reactant. Section 3 Limiting Reactants and Percentage Yield Percentage Yield The theoretical yield is the maximum amount of product that can be produced from a given amount of reactant. The actual yield of a product is the measured amount of that product obtained from a reaction. The percentage yield is the ratio of the actual yield to the theoretical yield, multiplied by 100. percentage yield? actual yield theorectical yield? 100 Visual Concepts Comparing Actual and Theoretical Yield Click below to watch the Visual Concept. Visual Concept

16 Section 3 Limiting Reactants and Percentage Yield Percentage Yield, continued Sample Problem H Chlorobenzene, C 6 H 5 Cl, is used in the production of many important chemicals, such as aspirin, dyes, and disinfectants. One industrial method of preparing chlorobenzene is to react benzene, C 6 H 6, with chlorine, as represented by the following equation. C 6 H 6 (l) + Cl 2 (g)? C 6 H 5 Cl(l) + HCl(g) When 36.8 g C 6 H 6 react with an excess of Cl 2, the actual yield of C 6 H 5 Cl is 38.8 g. What is the percentage yield of C 6 H 5 Cl? Section 3 Limiting Reactants and Percentage Yield Percentage Yield, continued Sample Problem H Solution C 6 H 6 (l) + Cl 2 (g)? C 6 H 5 Cl(l) + HCl(g) Given: mass of C 6 H 6 = 36.8 g mass of Cl 2 = excess actual yield of C 6 H 5 Cl = 38.8 g Unknown: percentage yield of C 6 H 5 Cl Solution: Theoretical yield molar mass factor mol ratio molar mass? mol C H Cl 6 5? mol C 6 H 6 g C 6 H 6? mol C 6 H 6 g C 6 H 6 g C 6 H 5 Cl? g C mol C 6 H 5 Cl 6 H 5 Cl Section 3 Limiting Reactants and Percentage Yield Percentage Yield, continued Sample Problem H Solution, continued C 6 H 6 (l) + Cl 2 (g)? C 6 H 5 Cl(l) + HCl(g) Theoretical yield 1 mol C 36.8 g C 6 H 6? 6 H 6? 1 mol C H Cl 6 5? g C H Cl g C 6 H 6 1 mol C 6 H 6 1 mol C 6 H 5 Cl? 53.0 g C 6 H 5 Cl Percentage yield percentage yield C 6 H 5 Cl? percentage yield? 38.8 g 53.0 g actual yield theorectical yield? 100? 100? 73.2%

17 End of Show 1. In stoichiometry, chemists are mainly concerned with A. the types of bonds found in compounds. B. mass relationships in chemical reactions. C. energy changes occurring in chemical reactions. D. the speed with which chemical reactions occur. 1. In stoichiometry, chemists are mainly concerned with A. the types of bonds found in compounds. B. mass relationships in chemical reactions. C. energy changes occurring in chemical reactions. D. the speed with which chemical reactions occur.

18 2. Assume ideal stoichiometry in the reaction CH 4 + 2O 2? CO 2 + 2H 2 O. If you know the mass of CH 4, you can calculate A. only the mass of CO 2 produced. B. only the mass of O 2 reacting. C. only the mass of CO 2 + H 2 O produced. D. the mass of O 2 reacting and CO 2 + H 2 O produced. 2. Assume ideal stoichiometry in the reaction CH 4 + 2O 2? CO 2 + 2H 2 O. If you know the mass of CH 4, you can calculate A. only the mass of CO 2 produced. B. only the mass of O 2 reacting. C. only the mass of CO 2 + H 2 O produced. D. the mass of O 2 reacting and CO 2 + H 2 O produced. 3. Which mole ratio for the equation 6Li + N 2? 2Li 3 N is incorrect? A. 6 mol Li C. 2 mol Li 3 N 2 mol N 2 1 mol N 2 B. 1 mol N 2 2 mol Li D. 3 N 6 mol Li 6 mol Li

19 3. Which mole ratio for the equation 6Li + N 2? 2Li 3 N is incorrect? A. 6 mol Li C. 2 mol Li 3 N 2 mol N 2 1 mol N 2 B. 1 mol N 2 2 mol Li D. 3 N 6 mol Li 6 mol Li 4. For the reaction below, how many moles of N 2 are required to produce 18 mol NH 3? N 2 + 3H 2? 2NH 3 A. 4.5 B. 9.0 C. 18 D For the reaction below, how many moles of N 2 are required to produce 18 mol NH 3? N 2 + 3H 2? 2NH 3 A. 4.5 B. 9.0 C. 18 D. 36

20 5. What mass of NaCl can be produced by the reaction of 0.75 mol Cl 2? 2Na + Cl 2? 2NaCl A g B. 1.5 g C. 44 g D. 88 g 5. What mass of NaCl can be produced by the reaction of 0.75 mol Cl 2? 2Na + Cl 2? 2NaCl A g B. 1.5 g C. 44 g D. 88 g 6. What mass of CO 2 can be produced from 25.0 g CaCO 3 given the decomposition reaction CaCO 3? CaO + CO 2 A g B g C g D g

21 6. What mass of CO 2 can be produced from 25.0 g CaCO 3 given the decomposition reaction CaCO 3? CaO + CO 2 A g B g C g D g 7. If a chemical reaction involving substances A and B stops when B is completely used up, then B is referred to as the A. excess reactant. B. primary reactant. C. limiting reactant. D. primary product. 7. If a chemical reaction involving substances A and B stops when B is completely used up, then B is referred to as the A. excess reactant. B. primary reactant. C. limiting reactant. D. primary product.

22 8. If a chemist calculates the maximum amount of product that could be obtained in a chemical reaction, he or she is calculating the A. percentage yield. B. mole ratio. C. theoretical yield. D. actual yield. 8. If a chemist calculates the maximum amount of product that could be obtained in a chemical reaction, he or she is calculating the A. percentage yield. B. mole ratio. C. theoretical yield. D. actual yield. 9. What is the maximum number of moles of AlCl 3 that can be produced from 5.0 mol Al and 6.0 mol Cl 2? 2Al + 3Cl 2? 2AlCl 3 A. 2.0 mol AlCl 3 B. 4.0 mol AlCl 3 C. 5.0 mol AlCl 3 D. 6.0 mol AlCl 3

23 9. What is the maximum number of moles of AlCl 3 that can be produced from 5.0 mol Al and 6.0 mol Cl 2? 2Al + 3Cl 2? 2AlCl 3 A. 2.0 mol AlCl 3 B. 4.0 mol AlCl 3 C. 5.0 mol AlCl 3 D. 6.0 mol AlCl 3 Short Answer 10.Why is a balanced equation necessary to solve a mass-mass stoichiometry problem? Short Answer 10.Why is a balanced equation necessary to solve a mass-mass stoichiometry problem? Answer: The coefficients of the balanced equation are needed for the mole-mole ratio that is necessary to solve stoichiometric problems involving two different substances.

24 Short Answer 11.What data are necessary to calculate the percentage yield of a reaction? Short Answer 11.What data are necessary to calculate the percentage yield of a reaction? Answer: the theoretical yield and the actual yield of the product Extended Response 12.A student makes a compound in the laboratory and reports an actual yield of 120%. Is this result possible? Assuming that all masses were measured correctly, give an explanation.

25 Extended Response 12.A student makes a compound in the laboratory and reports an actual yield of 120%. Is this result possible? Assuming that all masses were measured correctly, give an explanation. Answer: The product was impure, so the mass contained the product and other substances. For example, if NaCl is made from HCl and NaOH but is not dried, the mass of NaCl can include water, which would result in a yield greater than 100%. Extended Response 13. Benzene, C 6 H 6, is reacted with bromine, Br 2, to produce bromobenzene, C 6 H 5 Br, and hydrogen bromide, HBr, as shown below. When 40.0 g of benzene are reacted with 95.0 g of bromine, 65.0 g of bromobenzene is produced. C 6 H 6 + Br 2? C 6 H 5 Br + HBr a. Which compound is the limiting reactant? b. What is the theoretical yield of bromobenzene? c. What is the reactant in excess, and how much remains after the reaction is completed? d. What is the percentage yield? Extended Response 13. Benzene, C 6 H 6, is reacted with bromine, Br 2, to produce bromobenzene, C 6 H 5 Br, and hydrogen bromide, HBr, as shown below. When 40.0 g of benzene are reacted with 95.0 g of bromine, 65.0 g of bromobenzene is produced. C 6 H 6 + Br 2? C 6 H 5 Br + HBr a. Which compound is the limiting reactant? benzene b. What is the theoretical yield of bromobenzene? 80.4 g c. What is the reactant in excess, and how much remains after the reaction is completed? bromine, 13.2 g d. What is the percentage yield? 80.8%