1. CHAPTER 9 Stoichiometry More Conversions!

2. Stoichiometry Needs a balanced equation Use the balanced eqn to predict ending and / or starting amounts Coefficients are now mole ratios

3. Mole Ratios Example: 2Al 2 O 3 (l)  4Al(s) + 3O 2 (g) Mole Ratios: 2 mol Al 2 O 3 4 mol Al 2 mol Al 2 O 3 3 mol O 2 4 mol Al 3 mol O 2

4. Mole Ratios

5. Mole – Mole Problem A How many moles of lithium hydroxide are required to react with 20 mol CO 2, the average amount exhaled by a person each day? CO 2 (g) + 2LiOH(s)  Li 2 CO 3 (s) + H 2 O(l)

6. Mole – Mole Problem A 20 mol X mol 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? Step 1: Start with what you know from the problem.

7. Mole – Mole Problem A 20 mol X mol CO 2 (g) + 2LiOH(s)  Li 2 CO 3 (s) + H 2 O(l) 1 mol 2 mol How many moles of lithium hydroxide are required to react with 20 mol CO 2, the average amount exhaled by a person each day? Step 2: Determine what you know from the balanced equation.

8. Mole – Mole Problem A 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? Step 3: Set-up so units cancel & solve. 20 mol CO 2 = X mol LiOH 1 mol CO 2 2 mol LiOH

9. Mole – Mole Problem A 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? X = 40 mol LiOH

10. Mole – Mole Problem B 5.3 molX mol CO 2 (g) + 2LiOH(s)  Li 2 CO 3 (s) + H 2 O(l) How many moles of lithium carbonate are produced when 5.3 mol CO 2 are reacted? Step 1: Start with what you know from the problem.

11. Mole – Mole Problem B 5.3 molX mol CO 2 (g) + 2LiOH(s)  Li 2 CO 3 (s) + H 2 O(l)1 mol How many moles of lithium carbonate are produced when 5.3 mol CO 2 are reacted? Step 2: Determine what you know from the balanced eqn.

12. Mole – Mole Problem B 5.3 molX mol CO 2 (g) + 2LiOH(s)  Li 2 CO 3 (s) + H 2 O(l) 1 mol How many moles of lithium carbonate are produced when 5.3 mol CO 2 are reacted? Step 3: Set-up units to cancel out & solve. 5.3 mol CO 2 = X mol Li 2 CO 3 1 mol CO 2 1 mol Li 2 CO 3

13. Mole – Mole Problem B CO 2 (g) + 2LiOH(s)  Li 2 CO 3 (s) + H 2 O(l) How many moles of lithium carbonate are produced when 5.3 mol CO 2 are reacted? X = 5.3 mol Li 2 CO 3

14. Mole – Mass Problems mol – g

15. mol – Mass Problems

16. 1.Start with what you know. 2.Write a ratio of what you know from the problem above the equation. 3.Write a ratio of what you know from the balanced equation below. 4.Set the ratios together and Solve.

17. Balanced Equation: 6CO 2 (g) + 6H 2 O(l)  C 6 H 12 O 6 (s) + 6O 2 (g) What is the mass of glucose (C 6 H 12 O 6 ) produced from 3.00 mol of water (H 2 O)? mol – Mass Problem C

18. Balanced Equation: 3.00 mol X grams 6CO 2 (g) + 6H 2 O(l)  C 6 H 12 O 6 (s) + 6O 2 (g) What is the mass of glucose (C 6 H 12 O 6 ) produced from 3.00 mol of water (H 2 O)? mol – Mass Problem C Step 1: Start with what you know from the problem. 3.00 mol H 2 O = X g C 6 H 12 O 6

19. Balanced Equation: 3.00 mol X grams 6CO 2 (g) + 6H 2 O(l)  C 6 H 12 O 6 (s) + 6O 2 (g) 6 mol 180.18 g What is the mass of glucose (C 6 H 12 O 6 ) produced from 3.00 mol of water (H 2 O)? mol – Mass Problem C Step 2: Use the balanced equation to find the mol of known and the grams of unknown. 6 mol H 2 O = 180.18 g C 6 H 12 O 6

20. Balanced Equation: 3.00 mol X grams 6CO 2 (g) + 6H 2 O(l)  C 6 H 12 O 6 (s) + 6O 2 (g) 6 mol 180.18 g What is the mass of glucose (C 6 H 12 O 6 ) produced from 3.00 mol of water (H 2 O)? mol – Mass Problem C Step 3: Set-up and solve. 3.00 mol H 2 O = X g C 6 H 12 O 6 6 mol H 2 O 180.18 g C 6 H 12 O 6

21. Balanced Equation: 6CO 2 (g) + 6H 2 O(l)  C 6 H 12 O 6 (s) + 6O 2 (g) What is the mass of glucose (C 6 H 12 O 6 ) produced from 3.00 mol of water (H 2 O)? mol – Mass Problem C 6 X = (3.00)(180.18) 6

22. Balanced Equation: 6CO 2 (g) + 6H 2 O(l)  C 6 H 12 O 6 (s) + 6O 2 (g) What is the mass of glucose (C 6 H 12 O 6 ) produced from 3.00 mol of water (H 2 O)? mol – Mass Problem C X = (3.00)(180.18) 6 X = 90.09 g C 6 H 12 O 6

23. Balanced Equation: 6CO 2 (g) + 6H 2 O(l)  C 6 H 12 O 6 (s) + 6O 2 (g) What is the mass of oxygen (O 2 ) produced from 2.50 mol of water (H 2 O)? mol – Mass Problem D

24. Balanced Equation: 2.50 mol X grams 6CO 2 (g) + 6H 2 O(l)  C 6 H 12 O 6 (s) + 6O 2 (g) What is the mass of oxygen (O 2 ) produced from 2.50 mol of water (H 2 O)? mol – Mass Problem D Step 1: Start with what you know from the problem. 2.50 mol H 2 O = X g O 2

25. Balanced Equation: 2.50 mol X grams 6CO 2 (g) + 6H 2 O(l)  C 6 H 12 O 6 (s) + 6O 2 (g) 6 mol (6 * 32 = 192) What is the mass of oxygen (O 2 ) produced from 2.50 mol of water (H 2 O)? mol – Mass Problem D Step 2: Determine what you know from the balanced equation. 6 mol H 2 O = 192.00 g O 2

26. Balanced Equation: 2.50 mol X grams 6CO 2 (g) + 6H 2 O(l)  C 6 H 12 O 6 (s) + 6O 2 (g) 6 mol (6 * 32 = 192) What is the mass of oxygen (O 2 ) produced from 2.50 mol of water (H 2 O)? mol – Mass Problem D Step 3: Set-up and solve. 2.50 mol H 2 O = X g O 2 6 mol H 2 O 192.00 g O 2

27. Balanced Equation: 6CO 2 (g) + 6H 2 O(l)  C 6 H 12 O 6 (s) + 6O 2 (g) What is the mass of oxygen (O 2 ) produced from 2.50 mol of water (H 2 O)? mol – Mass Problem D 6 X = (2.50)(192.00) 6

28. Balanced Equation: 6CO 2 (g) + 6H 2 O(l)  C 6 H 12 O 6 (s) + 6O 2 (g) What is the mass of oxygen (O 2 ) produced from 2.50 mol of water (H 2 O)? mol – Mass Problem D X = 80.0 g O 2

29. Mass – mol Problems g - mol

30. Mass – mol Problems

31. 1.Start with what you know. 2.Write a ratio of what you know from the problem above the equation. 3.Write a ratio of what you know from the balanced equation below. 4.Set the ratios together and Solve.

32. Balanced Equation: 4NH 3 (g) + 5O 2 (g)  4NO(g) + 6H 2 O(g) How many moles of NO are formed from 824 g of NH 3 ? Mass – mol Problem E

33. Balanced Equation: 824 g X mol 4NH 3 (g) + 5O 2 (g)  4NO(g) + 6H 2 O(g) How many moles of NO are formed from 824 g of NH 3 ? Mass – mol Problem E Step 1: Start with what you know from the problem. 824 g NH 3 = X mol

34. Balanced Equation: 824 g X mol 4NH 3 (g) + 5O 2 (g)  4NO(g) + 6H 2 O(g) (4*17.01 = 68.04) 4 mol How many moles of NO are formed from 824 g of NH 3 ? Mass – mol Problem E Step 2: Determine what you know from the balanced equation. 68.04 g NH 3 = 4 mol NO

35. Balanced Equation: 824 g X mol 4NH 3 (g) + 5O 2 (g)  4NO(g) + 6H 2 O(g) (4*17.01 = 68.04) 4 mol How many moles of NO are formed from 824 g of NH 3 ? Mass – mol Problem E Step 3: Set-up your ratio. 824 g NH 3 = X mol NO 68.04 g NH 3 4 mol NO

36. Balanced Equation: 4NH 3 (g) + 5O 2 (g)  4NO(g) + 6H 2 O(g) How many moles of NO are formed from 824 g of NH 3 ? Mass – mol Problem E 824 g NH 3 = X mol NO 68.04 g NH 3 4 mol NO 68.04 X = (4)(824) 68.04 68.04

37. Balanced Equation: 4NH 3 (g) + 5O 2 (g)  4NO(g) + 6H 2 O(g) How many moles of NO are formed from 824 g of NH 3 ? Mass – mol Problem E X = 48.4 g NO

38. Balanced Equation: 4NH 3 (g) + 5O 2 (g)  4NO(g) + 6H 2 O(g) How many moles of H 2 O are formed from 412 g of NH 3 ? Mass – mol Problem F

39. Balanced Equation: 412 g X mol 4NH 3 (g) + 5O 2 (g)  4NO(g) + 6H 2 O(g) How many moles of H 2 O are formed from 412 g of NH 3 ? Mass – mol Problem F Step 1: Start with what you know from the problem. 412 g NH 3 = X mol H 2 O

40. Balanced Equation: 412 g X mol 4NH 3 (g) + 5O 2 (g)  4NO(g) + 6H 2 O(g) (4*17.03 = 68.12) 6 mol How many moles of H 2 O are formed from 412 g of NH 3 ? Mass – mol Problem F Step 2: Determine what is known from the balanced equation. 68.12 g NH 3 = 6 mol H 2 O

41. Balanced Equation: 412 g X mol 4NH 3 (g) + 5O 2 (g)  4NO(g) + 6H 2 O(g) (4*17.03 = 68.12) 6 mol How many moles of H 2 O are formed from 412 g of NH 3 ? Mass – mol Problem F Step 3: Set the ratios together and Solve. 412 g NH 3 = X mol H 2 O 68.12 g NH 3 6 mol H 2 O

42. Balanced Equation: 4NH 3 (g) + 5O 2 (g)  4NO(g) + 6H 2 O(g) How many moles of H 2 O are formed from 412 g of NH 3 ? Mass – mol Problem F 68.12 X = (412)(6) 68.12 68.12

43. Balanced Equation: 4NH 3 (g) + 5O 2 (g)  4NO(g) + 6H 2 O(g) How many moles of H 2 O are formed from 412 g of NH 3 ? Mass – mol Problem F X = 36.3 mol H 2 O

44. Mass – Mass Problems g - g

45. Gram – Gram Problems

46. 1.Start with what you know. 2.Write a ratio of what you know from the problem above the equation. 3.Write a ratio of what you know from the balanced equation below. 4.Set the ratios together and Solve.

47. Gram – Gram Problems

48. Gram – Gram Problem G Sn(s) + 2HF(g)  SnF 2 (s) + H 2 (g) How many grams of SnF 2 are produced from the reaction of 30.00 g HF?

49. Gram – Gram Problem G 30.00 g X g Sn(s) + 2HF(g)  SnF 2 (s) + H 2 (g) How many grams of SnF 2 are produced from the reaction of 30.00 g HF? Step 1: Start with what you know from the problem. 30.00 g HF = X g SnF 2

50. Gram – Gram Problem G 30.00 g X g Sn(s) + 2HF(g)  SnF 2 (s) + H 2 (g) (2*20.01=40.02) 156.69 How many grams of SnF 2 are produced from the reaction of 30.00 g HF? Step 2: Determine what you know from the balanced equation. 40.02 g HF = 156.69 g SnF 2

51. Gram – Gram Problem G 30.00 g X g Sn(s) + 2HF(g)  SnF 2 (s) + H 2 (g) (2*20.01=40.02) 156.69 How many grams of SnF 2 are produced from the reaction of 30.00 g HF? Step 3: Put the ratios together and Solve. 30.00 g HF = X g SnF 2 40.02 g HF 156.69 g SnF 2

52. Gram – Gram Problem G Sn(s) + 2HF(g)  SnF 2 (s) + H 2 (g) How many grams of SnF 2 are produced from the reaction of 30.00 g HF? 40.02 X = (30.00)(156.69) 40.02 40.02

53. Gram – Gram Problem G Sn(s) + 2HF(g)  SnF 2 (s) + H 2 (g) How many grams of SnF 2 are produced from the reaction of 30.00 g HF? X = 117.5 g SnF 2

54. Gram – Gram Problem H Sn(s) + 2HF(g)  SnF 2 (s) + H 2 (g) How many grams of HF are produced from the reaction of 150.5 g H 2 ?

55. Gram – Gram Problem H X g 150.5 g Sn(s) + 2HF(g)  SnF 2 (s) + H 2 (g) How many grams of HF are produced from the reaction of 150.5 g H 2 ? Step 1: Start with what you know from the problem. 150.5 g H 2 = X g HF

56. Gram – Gram Problem H X g 150.5 g Sn(s) + 2HF(g)  SnF 2 (s) + H 2 (g) (2*20.01=40.02) (2.02) How many grams of HF are produced from the reaction of 150.5 g H 2 ? Step 2: Determine what you know from the balanced equation. 2.02 g H 2 = 40.02 g HF

57. Gram – Gram Problem H X g 150.5 g Sn(s) + 2HF(g)  SnF 2 (s) + H 2 (g) (2*20.01=40.02) (2.02) How many grams of HF are produced from the reaction of 150.5 g H 2 ? Step 3: Set-up the ratios and Solve. 150.5 g H 2 = X g HF 2.02 g H 2 40.02 g HF

58. Gram – Gram Problem H Sn(s) + 2HF(g)  SnF 2 (s) + H 2 (g) How many grams of HF are produced from the reaction of 150.5 g H 2 ? 2.02 X = (150.5)(40.02) 2.02

59. Gram – Gram Problem H Sn(s) + 2HF(g)  SnF 2 (s) + H 2 (g) How many grams of HF are produced from the reaction of 150.5 g H 2 ? X = 2982 g HF

60. To Review … When solving: Known Amount & Unit Given in Problem Known Amount & Unit from Equation = Unknown Unit Given in Problem Unknown Amount & Unit from Equation

61. Comparing Actual and Theoretical Yield Visual Concepts Chapter 9

62. Percent Yield Visual Concepts Chapter 9

63. 1.Using the mole ratio of the reactants and products in a chemical reaction, what will you most likely be able to determine? A.rate of the reaction B.energy absorbed or released by the reaction C.chemical names of the reactants and products D.mass of a product produced from a known mass of reactants

64. 1.Using the mole ratio of the reactants and products in a chemical reaction, what will you most likely be able to determine? A.rate of the reaction B.energy absorbed or released by the reaction C.chemical names of the reactants and products D.mass of a product produced from a known mass of reactants

65. 3.What is the mole ratio of CO 2 to C 6 H 12 O 6 in the combustion reaction: C 6 H 12 O 6 + 6O 2 6CO 2 + 6H 2 O? A.1:1 B.1:2 C.1:6 D.6:1

66. 3.What is the mole ratio of CO 2 to C 6 H 12 O 6 in the combustion reaction: C 6 H 12 O 6 + 6O 2 6CO 2 + 6H 2 O? A.1:1 B.1:2 C.1:6 D.6:1