1. CHEMISTRY 11 TEACHER’S NOTESLESSON

2. FOR THE TEACHER Lesson Objectives  Review, reinforce, and extend the concepts in Chapter 7: Stoichiometry in Chemical Reactions. This lesson relates to expectations D1.1, D1.2, D2.1, D2.3, D2.5, D2.6, and D3.4. Lesson Notes Slide 5 The key insight for this slide, and for the whole chapter, is that amounts of reactants and products in a given chemical reaction are always in the same ratio. As a follow up, discuss with students whether they think that masses of reactants and products are also in a constant ratio. Slide 6 The first of three slides for 7.2 develops the idea of constant amount ratios in terms of stoichiometric amounts. The reaction is chosen in part because of its architectural application (see Discussion tab).

3. FOR THE TEACHER Lesson Notes Slide 7 and 8 This calculation uses the chain of reasoning: mass of reactant → amount of reactant → amount of product → mass of product. After completing the calculation, you may wish to draw this process out explicitly through class discussion. Slide 9 The objective of this slide, including the Discussion tab, is to reinforce the connection between students' concepts of limiting and excess reagents on the one hand, and industrial and environmental applications on the other. Slide 10 and 11 This calculation links ideas about limiting reagents to the principle of constant ratios of amounts. Slide 12 To extend student learning, you may wish to discuss with students why a reverse reaction occurs in this example.

4. Quantities in a Chemical Reaction Complete the table of quantities in this possible chemical equation for the incomplete combustion of methane. # CH 4 molecules# O 2 molecules# CO molecules# H 2 O molecules 3 50100 2 × 6.02 × 10 23 = 2 mol 3 mol mol 20 mol Section 7.1

5. Amount of Relationships for Aluminum Oxidation Aluminum oxidizes extremely rapidly in air. In fact, this makes aluminum a good material for cladding buildings. 1. Balance the chemical equation for this reaction. 2. Based on the previous reaction, identify which of these pairs of amounts are stoichiometric amounts of reactants. 2 mol Al, 3 mol O 2 12.0 mol Al, 36.0 mol O 2 7.2 mol Al, 4.8 mol O 2 Section 7.2

6. Discussion: Amount of Relationships for Aluminum Oxidation Why might you think aluminum was useless for cladding buildings? Why would this be a wrong conclusion? Section 7.2

7. Mass Relationships for Aluminum Oxidation What mass of aluminum is required to produce 274 g of aluminum oxide? Complete the calculation. Given: = _____ ; = _____ /mol; = _____ /mol Required: _____ Solution: Step 1: Step 2 Section 7.2

8. Mass Relationships for Aluminum Oxidation Step 2: Convert mass of given substance into amount of given substance. = __________ × = __________ [carry four decimals] 1 mol Step 3: Convert amount of given substance into amount of required substance. = __________ × mol Al = __________ mol [carry four decimals] Step 4 Section 7.2

9. Mass Relationships for Aluminum Oxidation Step 4: Convert amount of the required substance into mass of required substance. = __________ mol Al × = __________ g Statement: The mass of aluminum required to produce 274 g of aluminum oxide is __________. mol Al Section 7.2

10. Excess and Limiting Reagents Complete the following statements using “excess” or “limiting.” 1.Sodium bicarbonate can be used as a reagent to clean up spills of toxic materials. The spilled material is the __________ reagent. The sodium bicarbonate is the __________ reagent. 2.When a Bunsen burner is first ignited, it burns with an orange flame because of incomplete combustion. The gas fuel is the __________ reagent. The oxygen is the __________ reagent. Section 7.3

11. Discussion: Excess and Limiting Reagents What are other situations in which you would want a particular reagent to be limiting, and why? Section 7.3

12. Calculating the Predicted Amount of Product Determine the amount of water produced when 3.6 mol of hydrochloric acid is combined with 1.4 mol of manganese (IV) oxide. Complete the calculation. Given: = __________ ; = __________ Required: _____ Solution: Step 1: Step 2 Section 7.4

13. Step 2: Use the amount of one reactant to find the stoichiometric amount of the other. = __________ × = __________ mol mol HCl Therefore, hydrochloric acid is the __________ reagent and manganese (IV) oxide is the __________ reagent. mol MnO 2 Step 3 Calculating the Predicted Amount of Product Section 7.4

14. Step 3: Use the amount of the limiting reagent to determine the amount of water produced. From Step 2, the limiting reagent is ________________. = __________ × = __________ mol mol HCl Statement: When 3.6 mol of hydrochloric acid is combined with 1.4 mol of manganese (IV) oxide, the amount of water produced is __________. mol H2OH2O Calculating the Predicted Amount of Product Section 7.4

15. Reaction Yield The actual yield of a chemical reaction is usually less than it theoretical yield. There are a number of factors that may account for this. Consider the reaction of nitrogen and hydrogen, which produces ammonia. Click the image below to play an animation. What term describes a reaction with the property illustrated? For which direction of the reaction would you want to maximize the yield, and why? Section 7.5

16. Discussion: Reaction Yield What are some other factors that affect reaction yield? Section 7.5

17. ANSWERS TEACHER’S NOTESLESSON

18. Quantities in a Chemical Reaction Complete the table of quantities in this possible chemical equation for the incomplete combustion of methane. # CH4 molecules# O2 molecules# CO molecules# H2O molecules 2324 507550100 2 × 6.02 × 10 23 = 2 mol 3 mol2 mol4 mol 10 mol15 mol10 mol20 mol Section 7.1

19. Aluminum oxidizes extremely rapidly in air. In fact, this makes aluminum a good material for cladding buildings. 1. Balance the chemical equation for this reaction. 2. Based on the previous reaction, identify which of these pairs of amounts are stoichiometric amounts of reactants. 2 mol Al, 3 mol O 2 12.0 mol Al, 36.0 mol O 2 7.2 mol Al, 4.8 mol O 2 42 Amount of Relationships for Aluminum Oxidation Section 7.2

20. What mass of aluminum is required to produce 274 g of aluminum oxide? Complete the calculation. Given: = _____ ; = _____ /mol; = _____ /mol Required: _____ Solution: Step 1: Step 2 274 g26.98 g 101.96 g m Al 432 26.98 g 274 g 101.96 g/mol Mass Relationships for Aluminum Oxidation Section 7.2

21. Step 2: Convert mass of given substance into amount of given substance. = __________ × = __________ [carry four decimals] Step 3: Convert amount of given substance into amount of required substance. = __________ × mol Al = __________ mol [carry four decimals] Step 4 274 g 1 mol 101.96 g 2.6873 mol 2.6873 4 2 5.3746 Mass Relationships for Aluminum Oxidation Section 7.2

22. Step 4: Convert amount of the required substance into mass of required substance. = __________ mol Al × = __________ g Statement: The mass of aluminum required to produce 274 g of aluminum oxide is __________. 5.3746 145 145 g Mass Relationships for Aluminum Oxidation 26.98 g mol Al Section 7.2

23. Complete the following statements using “excess” or “limiting.” 1.Sodium bicarbonate can be used as a reagent to clean up spills of toxic materials. The spilled material is the __________ reagent. The sodium bicarbonate is the __________ reagent. 2.When a Bunsen burner is first ignited, it burns with an orange flame, because of incomplete combustion. The gas fuel is the __________ reagent. The oxygen is the __________ reagent. limiting excess limiting Excess and Limiting Reagents Section 7.3

24. Determine the amount of water produced when 3.6 mol of hydrochloric acid is combined with 1.4 mol of maganese (IV) oxide. Complete the calculation. Given: = __________ ; = __________ Required: _____ Solution: Step 1: Step 2 3.6 mol HCl 1.4 mol MnO 2 m H2OH2O 3.6 mol HCl 1.4 mol MnO 2 42 Calculating the Predicted Amount of Product m H2OH2O Section 7.4

25. = __________ × = __________ mol mol HCl Therefore, hydrochloric acid is the __________ reagent and manganese (IV) oxide is the __________ reagent. 3.6 mol HCl mol MnO 2 0.90 1 4 limiting excess Step 3 Step 2: Use the amount of one reactant to find the stoichiometric amount of the other. Calculating the Predicted Amount of Product Section 7.4

26. Step 3: Use the amount of the limiting reagent to determine the amount of water produced. From Step 2, the limiting reagent is ________________. = __________ × = __________ mol mol HCl Statement: When 3.6 mol of hydrochloric acid is combined with 1.4 mol of manganese (IV) oxide, the amount of water produced is __________. 3.6 mol HCl mol H2OH2O 1.8 2 4 1.8 mol hydrochloric acid Calculating the Predicted Amount of Product Section 7.4

27. Sample answer to Discussion: You might think that, as aluminum oxidizes very rapidly, it would corrode quickly. However, what actually happens is that a layer of aluminum oxide forms a tough skin which protects the rest of the aluminum from corroding. Answer for Discussion Question: Slide 6 Section 7.2

28. Sample answers to Discussion: In an industrial process, the most expensive reagent should be the limiting one. Treating fuel as the limiting reagent in complete combustion reduces both costs and emissions. When using a chemical reaction to recover expensive chemicals after an experiment, the most expensive reagent should be the limiting one, because you want to recover all of the expensive reagent; if the less expensive reagent remains in the solution, you are not losing as much money. Answer for Discussion Question: Slide 11 Section 7.3

29. Answer for question about reaction type and part: This is a reversible reaction. You would want to maximize the forward reaction yield, to produce as much ammonia as possible. Answer for Discussion Question: Slide 15 Section 7.5

30. Sample answer to Discussion: Other factors that affect reaction yield include spillages, splashes, and losses from isolation and/or purification during the experimental procedure; impurities, for example due to oxidation of reactive metals; competing side reactions that may form other products besides the desired one(s). Answer for Discussion Question: Slide 16 Section 7.5