Chapter 9. Preview. Lesson Starter Objective Stoichiometry Definition Reaction Stoichiometry Problems Mole Ratio Stoichiometry Calculations

Transcription

1 Preview Lesson Starter Objective Stoichiometry Definition Reaction Stoichiometry Problems Mole Ratio Stoichiometry

2 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?

3 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.

4 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.

5 Section 1 Introduction to Stoichiometry Stoichiometry Click below to watch the Visual Concept. Visual Concept

6 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)

7 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)

8 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

9 Section 1 Introduction to Stoichiometry Converting Between Amounts in Moles

10 Section 1 Introduction to Stoichiometry Stoichiometry

11 Section 1 Introduction to Stoichiometry Molar Mass as a Conversion Factor Click below to watch the Visual Concept. Visual Concept

12 Preview Lesson Starter Objective Conversions of Quantities in Moles Conversions of Amounts in Moles to Mass Mass-Mass to Solving Various Types of Stoichiometry Problems

13 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?

14 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.

15 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.

16 Conversions of Quantities in Moles

17 Conversion of Quantities in Moles Click below to watch the Visual Concept. Visual Concept

18 Solving Mass-Mass Stoichiometry Problems

19 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?

20 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 mol LiOH = 1 mol CO 2 2 mol LiOH 40 mol LiOH

21 Conversions of Amounts in Moles to Mass

22 Solving Stoichiometry Problems with Moles or Grams

23 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?

24 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

25 Conversions of Mass to Amounts in Moles

26 Mass and Number of Moles of an Unknown Click below to watch the Visual Concept. Visual Concept

27 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?

28 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) Solution: b. amount of H 2 O produced (mol) 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

29 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 3 = mol NO b. 824 g NH 3 1 mol NH 6 mol H O 3 2 = g NH 4 mol NH mol H O 2

30 Mass-Mass to

31 Mass-Mass Click below to watch the Visual Concept. Visual Concept

32 Solving Mass-Mass Problems

33 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?

34 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 mol SnF g SnF g HF mol HF mol SnF 2 2 = 1 mol HF 1 mol SnF g SnF g HF 2 mol HF 1 mol SnF = g SnF g SnF 2

35 Solving Various Types of Stoichiometry Problems

36 Solving Various Types of Stoichiometry Problems

37 Solving Volume-Volume Problems

38 Solving Particle Problems

39 Section 3 Limiting Reactants and Percentage Yield Preview Objectives Limiting Reactants Percentage Yield

40 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.

41 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.

42 Section 3 Limiting Reactants and Percentage Yield Limiting Reactants and Excess Reactants Click below to watch the Visual Concept. Visual Concept

43 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?

44 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

45 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.

46 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

47 Visual Concepts Comparing Actual and Theoretical Yield Click below to watch the Visual Concept. Visual Concept

48 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?

49 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

50 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 = 100 = 53.0 g 73.2%

51 End of Chapter 9 Show