Chapter 9. Table of Contents. Stoichiometry. Section 1 Introduction to Stoichiometry. Section 2 Ideal Stoichiometric Calculations

Transcription

1 Stoichiometry Table of Contents Section 1 Introduction to Stoichiometry Section 2 Ideal Stoichiometric Calculations Section 3 Limiting Reactants and Percentage Yield

2 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.

3 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.

4 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)

5 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) 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) Amount of unknown substance (mol) Mass of unknown substance (g)

6 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

7 Section 1 Introduction to Stoichiometry Converting Between Amounts in Moles

8 Section 1 Introduction to Stoichiometry Stoichiometry Calculations

9 Section 2 Ideal Stoichiometric Calculations 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.

10 Section 2 Ideal Stoichiometric Calculations 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.

11 Section 2 Ideal Stoichiometric Calculations Conversions of Quantities in Moles

12 Section 2 Ideal Stoichiometric Calculations Solving Mass-Mass Stoichiometry Problems

13 Section 2 Ideal Stoichiometric Calculations 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?

14 Section 2 Ideal Stoichiometric Calculations 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 CO 20 mol CO 2 2 mol ratio mol LiOH mol CO 2 2 mol LiOH 1 mol CO 2 mol LiOH 40 mol LiOH

15 Section 2 Ideal Stoichiometric Calculations Conversions of Amounts in Moles to Mass

16 Section 2 Ideal Stoichiometric Calculations Solving Stoichiometry Problems with Moles or Grams

17 Section 2 Ideal Stoichiometric Calculations 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?

18 Section 2 Ideal Stoichiometric Calculations 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 H O mol ratio molar mass factor mol C6H12O 6 g C6H12O 6 mol H O mol C H O g C H O mol H O 2 1 mol C H O g C H O 6 mol H O 1 mol C H O = 90.1 g C 6 H 12 O 6

19 Section 2 Ideal Stoichiometric Calculations Conversions of Mass to Amounts in Moles

20 Section 2 Ideal Stoichiometric Calculations 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?

21 Section 2 Ideal Stoichiometric Calculations 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) a. g NH 3 molar mass factor mol NH mol NO mol NH 3 g NH mol ratio 3 3 mol NO b. g NH mol NH mol H O mol H O g NH3 mol NH3

22 Section 2 Ideal Stoichiometric Calculations Conversions of Mass to Amounts in Moles, continued Sample Problem D Solution, continued molar mass factor mol ratio a. 824 g NH 3 1 mol NH 4 mol NO g NH 4 mol NH mol NO b. 824 g NH 3 1 mol NH 6 mol H O g NH 4 mol NH mol H O 2

23 Section 2 Ideal Stoichiometric Calculations Mass-Mass to Calculations

24 Section 2 Ideal Stoichiometric Calculations Solving Mass-Mass Problems

25 Section 2 Ideal Stoichiometric Calculations Mass-Mass to Calculations, 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?

26 Section 2 Ideal Stoichiometric Calculations Mass-Mass to Calculations, 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 mol SnF g SnF g HF mol HF mol SnF mol HF 1 mol SnF g SnF g HF 2 mol HF 1 mol SnF = g SnF g SnF 2

27 Section 2 Ideal Stoichiometric Calculations Solving Various Types of Stoichiometry Problems

28 Section 2 Ideal Stoichiometric Calculations Solving Various Types of Stoichiometry Problems

29 Section 2 Ideal Stoichiometric Calculations Solving Volume-Volume Problems

30 Section 2 Ideal Stoichiometric Calculations Solving Particle Problems

31 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.

32 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.

33 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?

34 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 SiF4 mol HF mol SiF4 produced mol HF mol SiO mol SiF mol SiO2 mol SiF produced

35 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) 1 mol SiF 4.5 mol SiO 4.5 mol SiF produced mol SiO2 1 mol SiF mol HF 1.5 mol SiF4 produced 4 mol HF HF is the limiting reactant.

36 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. actual yield percentage yield 100 theorectical yield

37 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 Cl2, the actual yield of C 6 H 5 Cl is 38.8 g. What is the percentage yield of C 6 H 5 Cl?

38 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 g C H molar mass factor mol ratio molar mass mol C H mol C H Cl g C H Cl g C H Cl g C6H6 mol C6H6 mol C6H5Cl

39 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 C6H6 1 mol C6H5Cl g C6H5Cl 36.8 g C6H g C H 1 mol C H 1 mol C H Cl Percentage yield g C H Cl actual yield percentage yield C6H5Cl 100 theorectical yield g percentage yield g 73.2%

40 End of Chapter 9 Show

41 Standardized Test Preparation Multiple Choice 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.

42 Standardized Test Preparation Multiple Choice 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.

43 Standardized Test Preparation Multiple Choice 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.

44 Standardized Test Preparation Multiple Choice 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.

45 Standardized Test Preparation Multiple Choice 3. Which mole ratio for the equation 6Li + N 2 2Li 3 N is incorrect? A. 6 mol Li 2 mol Li C. 3 N 2 mol N 2 1 mol N 2 1 mol N B. 2 D. 6 mol Li 2 mol Li 3 N 6 mol Li

46 Standardized Test Preparation Multiple Choice 3. Which mole ratio for the equation 6Li + N 2 2Li 3 N is incorrect? A. 6 mol Li 2 mol Li C. 3 N 2 mol N 2 1 mol N 2 1 mol N B. 2 D. 6 mol Li 2 mol Li 3 N 6 mol Li

47 Standardized Test Preparation Multiple Choice 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. 36

48 Standardized Test Preparation Multiple Choice 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. 36

49 Standardized Test Preparation Multiple Choice 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

50 Standardized Test Preparation Multiple Choice 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

51 Standardized Test Preparation Multiple Choice 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

52 Standardized Test Preparation Multiple Choice 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

53 Standardized Test Preparation Multiple Choice 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.

54 Standardized Test Preparation Multiple Choice 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.

55 Standardized Test Preparation Multiple Choice 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.

56 Standardized Test Preparation Multiple Choice 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.

57 Standardized Test Preparation Multiple Choice 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

58 Standardized Test Preparation Multiple Choice 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

59 Standardized Test Preparation Short Answer 10.Why is a balanced equation necessary to solve a mass-mass stoichiometry problem?

60 Standardized Test Preparation 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.

61 Standardized Test Preparation 11.What data are necessary to calculate the percentage yield of a reaction?

62 Standardized Test Preparation 11.What data are necessary to calculate the percentage yield of a reaction? Answer: the theoretical yield and the actual yield of the product

63 Standardized Test Preparation 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.

64 Standardized Test Preparation 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%.

65 Standardized Test Preparation 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?

66 Standardized Test Preparation 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%