1. Basic Stoichiometry Pisgah High School M. Jones Revision history: 5/16/03, 02/04/12, 04/27/12 Part One

2. The word stoichiometry comes from the Greek words stoicheion which means “element” and metron which means “measure”.

3. Stoichiometry deals with the amounts of reactants and products in a chemical reaction.

4. Stoichiometry deals with moles.

5. Recall that … 1 mole = 22.4 L of any gas at STP 1 mole = the molar mass molecules 1 mole = 6.022 x 10 23 atoms or

6. The word mole is one that represents a very large number. … “mole” means 6.022 x 10 23 Much like “dozen” means 12,

7. The key to doing stoichiometry is the balanced chemical equation. 2 H 2 + O 2  2 H 2 O 22

8. The coefficients give the relative number of atoms or molecules of each reactant or product … as well as the number of moles. 2 H 2 + O 2  2 H 2 O

9. 2 molecules of hydrogen 1 molecule of oxygen 2 molecules of water Two molecules of hydrogen combine with one molecule of oxygen to make two molecules of water.

10. 2 H 2 + O 2  2 H 2 O The balanced chemical equation also gives the smallest integer number of moles. 2 molecules of hydrogen 1 molecule of oxygen 2 molecules of water

11. 2 H 2 + O 2  2 H 2 O The balanced chemical equation also gives the smallest integer number of moles. 2 moles of hydrogen 1 mole of oxygen 2 moles of water

12. 2 H 2 + O 2  2 H 2 O 2 moles of hydrogen 1 mole of oxygen 2 moles of water Two moles of hydrogen combine with one mole of oxygen to make two moles of water.

13. How can we show that this is really true? Consider the combustion of hydrogen in oxygen …

14. 2 H 2 + O 2  2 H 2 O 2 moles of hydrogen 1 mole of oxygen 2 moles of water What do each of the reactants and product weigh?

15. 2 H 2 + O 2  2 H 2 O 2 moles of hydrogen 1 mole of oxygen 2 moles of water 2 x 2.0 g1 x 32.0 g2 x 18.0 g 4.0 g32.0 g36.0 g +=

16. 2 H 2 + O 2  2 H 2 O 2 moles of hydrogen 1 mole of oxygen 2 moles of water 4.0 g32.0 g36.0 g += The Law of Conservation of Matter

17. 2 H 2 + O 2  2 H 2 O 2 moles of hydrogen 1 mole of oxygen 2 moles of water 4.0 g32.0 g36.0 g += Matter can neither be created nor destroyed, only changed in form.

18. The oxidation of iron

19. Consider the oxidation of iron: Fe(s) + O 2 (g)  Fe 2 O 3 (s) 432 4 moles 3 moles2 moles The coefficients give the ratio of moles.

20. Consider the oxidation of iron: Fe(s) + O 2 (g)  Fe 2 O 3 (s) 432 If these react … then we have the following: 8 moles 6 moles4 moles 2 moles 1.5 moles1 mole 0.50 mol 0.375 mol0.25 mol 4 moles 3 moles2 moles

21. Consider the oxidation of iron: Fe(s) + O 2 (g)  Fe 2 O 3 (s) 432 0.375 mole O 2 The 0.375 moles was not as easy to predict.

22. Consider the oxidation of iron: Fe(s) + O 2 (g)  Fe 2 O 3 (s) 432 Use a conversion factor to determine the number of moles of an unknown. 0.375 mole O 2 The 0.375 moles was not as easy to predict.

23. Consider the oxidation of iron: Fe(s) + O 2 (g)  Fe 2 O 3 (s) 432 The conversion factor comes from the coefficients in the balanced equation. 0.375 mole O 2

24. Back to the oxidation of iron: Fe(s) + O 2 (g)  Fe 2 O 3 (s) 432 4 moles 3 moles2 moles Calculate the masses of these moles.

25. Back to the oxidation of iron: Fe(s) + O 2 (g)  Fe 2 O 3 (s) 432 4 moles 3 moles2 moles 4 x 55.8 g3 x 32.0 g 2 x 159.6 g 319.2 g96.0 g223.2 g += Mass is conserved.

26. The decomposition of ammonium carbonate

27. Now consider the decomposition of solid ammonium carbonate. (NH 4 ) 2 CO 3  2 NH 3 + CO 2 + H 2 O Suppose 96.0 grams of ammonium carbonate decomposes. How many grams of each of the gases will be produced?

28. Now consider the decomposition of solid ammonium carbonate. (NH 4 ) 2 CO 3  2 NH 3 + CO 2 + H 2 O 96.0 grams? g The coefficients tell moles, not grams. Convert 96.0 g (NH 4 ) 2 CO 3 to moles.

29. Now consider the decomposition of solid ammonium carbonate. (NH 4 ) 2 CO 3  2 NH 3 + CO 2 + H 2 O 96.0 grams? g Find the molar mass of ammonium carbonate. 2x14 + 8 +12 + 3x16 = 96.0 g/mol

30. Now consider the decomposition of solid ammonium carbonate. (NH 4 ) 2 CO 3  2 NH 3 + CO 2 + H 2 O 96.0 grams? g Isn’t that convenient! We have one mole of ammonium carbonate.

31. Now consider the decomposition of solid ammonium carbonate. (NH 4 ) 2 CO 3  2 NH 3 + CO 2 + H 2 O 96.0 grams? g 1 mol 2 mol 2 moles of ammonia are produced, along with 1 mole of carbon dioxide and 1 mole of water vapor.

32. Now consider the decomposition of solid ammonium carbonate. (NH 4 ) 2 CO 3  2 NH 3 + CO 2 + H 2 O 96.0 grams? g How many grams of each product are formed? 1 mol 2 mol 2 x 17.0g 44.0 g18.0 g

33. Now consider the decomposition of solid ammonium carbonate. (NH 4 ) 2 CO 3  2 NH 3 + CO 2 + H 2 O 96.0 grams? g 34.0 g + 44.0 g + 18.0 g = 1 mol 2 mol 2 x 17.0g 44.0 g18.0 g 96.0 g

34. The reaction between dinitrogen pentoxide and water

35. Consider the reaction between dinitrogen pentoxide and water. What kind of reaction is it? Is it a … Double replacement reaction? Combustion reaction? Decomposition reaction? Single replacement reaction? So it must be a synthesis reaction

36. Consider the reaction between dinitrogen pentoxide and water. Which kind of synthesis reaction is it? 1.Hydrogen + nonmetal = binary acid 2.Metal + nonmetal = salt 3.Metal + water = base 4.Nonmetal + water = ternary acid N 2 O 5 + HOH  2 HNO 3 HNO 3 is a ternary acid; HNO 3 is nitric acid

37. N 2 O 5 + HOH  2 HNO 3 Suppose we needed to make 100.0 grams of nitric acid. How many grams of dinitrogen pentoxide would we need to react with excess water?

38. N 2 O 5 + HOH  2 HNO 3 100.0 g??? g = 1.59 mole HNO 3 1.59 mole0.794 mole = 85.7 g N 2 O 5 1.59 mol/2

39. N 2 O 5 + HOH  2 HNO 3 100.0 g85.7 g 1.59 mole0.794 mole 1.59 mol/2 But. Suppose that we find out that there are only 60.0 grams of dinitrogen pentoxide. How many grams of nitric acid could we actually make?

40. N 2 O 5 + HOH  2 HNO 3 ??? g60.0 g = 70.0 g HNO 3 1.11 mole0.556 mole = 0.556 mol N 2 O 5 0.556 mol x 2

41. Description of stoichiometry problems

42. Stoichiometry problems will usually take one of the following forms: 1.Mole-mole problem where you might be given moles and asked to find moles of another substance. 2.Mole-mass problem where you might be given moles and asked find the mass of another substance.

43. 3.Mass-mass problem where you might be given a mass and asked to find the mass of another substance. 4.Mass-volume problem where you might be given a mass and asked to find the volume of a gas. 5.Volume-volume problem where you might be given a volume and asked to find another volume.

44. Volume-volume stoichiometry problems are easiest when you use Gay-Lussac’s Law. “The ratio between the volumes of the reactant gases and the products can be expressed in simple whole numbers.”

45. Simply put, Gay-Lussac’s Law says this: The volumes of the gases are in the same ratio as the coefficients in the balanced equation. 2 H 2 + O 2  2 H 2 O 2 moles1 mole2 moles 2 L1 L2 L

46. Applications of Gay-Lussac’s Law

47. C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O(g ) 1 mol 5 mols3 mols4 mols Suppose 3.5 L of propane gas at STP is burned in oxygen, how many L at STP of oxygen will be required? 17.5 L O 2 1 L 5 L3 L4 L

48. C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O(g ) 1 mol 5 mols3 mols4 mols Suppose 3.5 L of propane gas at STP is burned in oxygen, how many L at STP of oxygen will be required? 17.5 L O 2 How many liters of carbon dioxide gas and water vapor at STP would be produced? 10.5 L CO 2 and 14.0 L H 2 O 1 L 5 L3 L4 L

49. CH 4 (g) + 4 Cl 2 (g)  CCl 4 (g) + 4 HCl(g) When methane gas is allowed to react with an excess of chlorine gas, tetrachloromethane and hydrogen chloride gas will be produced. How many L of methane will react with 0.800 L of chlorine gas at STP? 0.200 L Cl 2

50. Stoichiometry problems involving gases

51. Find the mass of HOCl that can be produced when 2.80 L of chlorine gas at STP reacts with excess hydrogen peroxide. Cl 2 (g) + H 2 O 2 (l)  2 HOCl (l) 2.80 L??? g Convert 2.80 L of Cl 2 gas at STP to moles

52. Cl 2 (g) + H 2 O 2 (l)  2 HOCl (l) 2.80 L??? g 0.125 mol Cl 2 0.125 mol.125 x 2 0.250 mol 13.1g HOCl

53. The reaction between copper and nitric acid

54. Will copper dissolve in acids? Cu + 2HCl  CuCl 2 + 2H 2 (g) No Reaction Most metals react with HCl to produce a metal chloride solution and H 2 gas. Not copper Consider hydrochloric acid

55. Will copper dissolve in acids? Cu + 2HCl  CuCl 2 + 2H 2 (g) No Reaction Consider hydrochloric acid Copper is below hydrogen in the activity series. Copper metal will only replace elements that are below it in the activity series.

56. What about other acids? The same is true for all acids except nitric acid Cu + HBr  NR Cu + HI  NR Cu + HF  NR Cu + H 2 SO 4  NR Cu + HC 2 H 3 O 2  NR

57. Nitric acid is the only acid that will dissolve copper. 3Cu + 8HNO 3  3Cu(NO 3 ) 2 + 2NO + 4H 2 O A beaker contains a penny and some nitric acid is added.

58. Nitric acid is the only acid that will dissolve copper. The penny begins to disappear and the solution turns blue-green and a brown gas is given off. 3Cu + 8HNO 3  3Cu(NO 3 ) 2 + 2NO + 4H 2 O

59. Nitric acid is the only acid that will dissolve copper. The gas produced in the reaction is NO, which is colorless. Reddish brown NO 2 forms when NO reacts with the oxygen in the air. 3Cu + 8HNO 3  3Cu(NO 3 ) 2 + 2NO + 4H 2 O

60. Nitric acid is the only acid that will dissolve copper. The penny is gone and the solution turns a dark blue. The brown NO 2 gas escapes from the open beaker. 3Cu + 8HNO 3  3Cu(NO 3 ) 2 + 2NO + 4H 2 O

61. Nitric acid is the only acid that will dissolve copper. Calculate the volume of NO gas at STP when 20.0 grams of copper dissolves. 3Cu + 8HNO 3  3Cu(NO 3 ) 2 + 2NO + 4H 2 O

62. Nitric acid is the only acid that will dissolve copper. 20.0 g??? L 0.315 mol Cu 0.315 mol 0.210 mol x 0.667 3Cu + 8HNO 3  3Cu(NO 3 ) 2 + 2NO + 4H 2 O

63. Nitric acid is the only acid that will dissolve copper. 20.0 g 4.70 L NO 0.315 mol ??? L 0.210 mol 3Cu + 8HNO 3  3Cu(NO 3 ) 2 + 2NO + 4H 2 O x 0.667

64. Part Two of Basic Stoichiometry will include the gas laws.