1. Stoichiometry

2. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction and allows for the calculation of the amount of a different element/compound in the same reaction. Obviously the place to start is…

3. 1.Predict and write a balanced chemical equation.

4. 2.Start a RR track and convert the given information to moles (if not already as moles).

5. 1.Predict and write a balanced chemical equation. 2.Start a RR track and convert the given information to moles (if not already as moles). a.Use the molar mass (from the PT) to change grams → moles

6. 1.Predict and write a balanced chemical equation. 2.Start a RR track and convert the given information to moles (if not already as moles). a.Use the molar mass (from the PT) to change grams → moles b.Use 6.022 x 10 23 to change atoms/molecules/formula units → moles

7. 1.Predict and write a balanced chemical equation. 2.Start a RR track and convert the given information to moles (if not already as moles). a.Use the molar mass (from the PT) to change grams → moles b.Use 6.022 x 10 23 to change atoms/molecules/formula units → moles c.Use 22.41 L (for gases only) to change L → moles

8. 1.Predict and write a balanced chemical equation. 2.Start a RR track and convert the given information to moles (if not already as moles). a.Use the molar mass (from the PT) to change grams → moles b.Use 6.022 x 10 23 to change atoms/molecules/formula units → moles c.Use 22.4 L (for gases only) to change L → moles d.For water and dilute solutions, 1 mL = 1 g

9. 3.Convert the moles of the given to moles of the unknown using the mole ratio from the balanced chemical equation a.Mole # of unknown Mole # of known

10. 3.Convert the moles of the given to moles of the unknown using the mole ratio from the balanced chemical equation a.Mole # of unknown Mole # of known 4. Convert the moles of the unknown to the desired unit (see possible conversion factors in step 2)

11. Example: Potassium is reacted with phosphoric acid Step 1: What is step 1, always ?

12. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g)

13. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) a.If we started with 125 g of potassium, how much potassium phosphate can be produced, in grams?

14. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) a.If we started with 125 g of potassium, how much potassium phosphate can be produced, in grams? 125 g K Always start by copying the given information!

15. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) a.If we started with 125 g of potassium, how much potassium phosphate can be produced, in grams? 125 g K What MUST go here?

16. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) a.If we started with 125 g of potassium, how much potassium phosphate can be produced, in grams? 125 g K g Where do we always go to get grams information?

17. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) a.If we started with 125 g of potassium, how much potassium phosphate can be produced, in grams? 125 g K1 mol K 39.1 g Why “1 mole”?

18. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) a.If we started with 125 g of potassium, how much potassium phosphate can be produced, in grams? 125 g K1 mol K 39.1 g What MUST go here?

19. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) a.If we started with 125 g of potassium, how much potassium phosphate can be produced, in grams? 125 g K1 mol K 39.1 g mol K What compound/ element are we looking for?

20. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) a.If we started with 125 g of potassium, how much potassium phosphate can be produced, in grams? 125 g K1 mol K mol K 3 PO 4 39.1 g mol K What compound/ element are we looking for?

21. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) a.If we started with 125 g of potassium, how much potassium phosphate can be produced, in grams? 125 g K1 mol K mol K 3 PO 4 39.1 g mol K Where can we find a mole ratio for these?

22. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) a.If we started with 125 g of potassium, how much potassium phosphate can be produced, in grams? 125 g K1 mol K mol K 3 PO 4 39.1 g mol K

23. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) a.If we started with 125 g of potassium, how much potassium phosphate can be produced, in grams? 125 g K1 mol K2 mol K 3 PO 4 39.1 g6 mol K REDUCE?

24. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) a.If we started with 125 g of potassium, how much potassium phosphate can be produced, in grams? 125 g K1 mol K2 mol K 3 PO 4 39.1 g6 mol K What MUST go here?

25. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) a.If we started with 125 g of potassium, how much potassium phosphate can be produced, in grams? 125 g K1 mol K2 mol K 3 PO 4 39.1 g6 mol K mol K 3 PO 4 What specific information about potassium phosphate is wanted?

26. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) a.If we started with 125 g of potassium, how much potassium phosphate can be produced, in grams? 125 g K1 mol K2 mol K 3 PO 4 g 39.1 g6 mol K mol K 3 PO 4 Where do we always go to get grams information?

27. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) a.If we started with 125 g of potassium, how much potassium phosphate can be produced, in grams? 125 g K1 mol K2 mol K 3 PO 4 136 g 39.1 g6 mol K 1 mol K 3 PO 4 Why “1 mole”?

28. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) a.If we started with 125 g of potassium, how much potassium phosphate can be produced, in grams? 125 g K1 mol K2 mol K 3 PO 4 136 g 39.1 g6 mol K 1 mol K 3 PO 4 How should this be calculated?

29. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) a.If we started with 125 g of potassium, how much potassium phosphate can be produced, in grams? 125 g K1 mol K2 mol K 3 PO 4 136 g 39.1 g6 mol K 1 mol K 3 PO 4 125 × 2 × 136 ÷ 39.1 ÷ 6 = 144.928

30. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) a.If we started with 125 g of potassium, how much potassium phosphate can be produced, in grams? 125 g K1 mol K2 mol K 3 PO 4 136 g = 145 g K 3 PO 4 39.1 g6 mol K 1 mol K 3 PO 4 125 × 2 × 136 ÷ 39.1 ÷ 6 = 144.928 Why 3 Sig Digs?

31. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) b.If we started with 35.4 g of potassium, how many formula units of phosphoric acid are needed?

32. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) b.If we started with 35.4 g of potassium, how many formula units of phosphoric acid are needed? 35.4 g K

33. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) b.If we started with 35.4 g of potassium, how many formula units of phosphoric acid are needed? 35.4 g K g

34. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) b.If we started with 35.4 g of potassium, how many formula units of phosphoric acid are needed? 35.4 g K1 mol K 39.1 g

35. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) b.If we started with 35.4 g of potassium, how many formula units of phosphoric acid are needed? 35.4 g K1 mol K 39.1 g mol K

36. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) b.If we started with 35.4 g of potassium, how many formula units of phosphoric acid are needed? 35.4 g K1 mol K mol H 3 PO 4 39.1 g mol K

37. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) b.If we started with 35.4 g of potassium, how many formula units of phosphoric acid are needed? 35.4 g K1 mol K2 mol H 3 PO 4 39.1 g6 mol K

38. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) b.If we started with 35.4 g of potassium, how many formula units of phosphoric acid are needed? 35.4 g K1 mol K2 mol H 3 PO 4 39.1 g6 mol K mol H 3 PO 4

39. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) b.If we started with 35.4 g of potassium, how many formula units of phosphoric acid are needed? 35.4 g K1 mol K2 mol H 3 PO 4 6.022 × 10 23 formula units 39.1 g6 mol K1 mol H 3 PO 4

40. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) b.If we started with 35.4 g of potassium, how many formula units of phosphoric acid are needed? 35.4 g K1 mol K2 mol H 3 PO 4 6.022 × 10 23 formula units 39.1 g6 mol K1 mol H 3 PO 4 35.4 × 2 × 6.022 E 23 ÷ 39.1 ÷ 6 = 1.81738 E 23

41. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) b.If we started with 35.4 g of potassium, how many formula units of phosphoric acid are needed? 1.82 × 10 23 formula units H 3 PO 4 35.4 g K1 mol K2 mol H 3 PO 4 6.022 × 10 23 formula units = 39.1 g6 mol K1 mol H 3 PO 4

42. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) c.If we were able to collect 6.92 L of hydrogen gas, how many grams of phosphoric acid was used? 6.92 L H 2

43. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) c.If we were able to collect 6.92 L of hydrogen gas, how many grams of phosphoric acid was used? 6.92 L H 2 L H 2

44. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) c.If we were able to collect 6.92 L of hydrogen gas, how many grams of phosphoric acid was used? 6.92 L H 2 1 mol H 2 22.4 L H 2

45. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) c.If we were able to collect 6.92 L of hydrogen gas, how many grams of phosphoric acid was used? 6.92 L H 2 1 mol H 2 22.4 L H 2 mol H 2

46. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) c.If we were able to collect 6.92 L of hydrogen gas, how many grams of phosphoric acid was used? 6.92 L H 2 1 mol H 2 mol H 3 PO 4 22.4 L H 2 mol H 2

47. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) c.If we were able to collect 6.92 L of hydrogen gas, how many grams of phosphoric acid was used? 6.92 L H 2 1 mol H 2 2 mol H 3 PO 4 22.4 L H 2 3 mol H 2

48. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) c.If we were able to collect 6.92 L of hydrogen gas, how many grams of phosphoric acid was used? 6.92 L H 2 1 mol H 2 2 mol H 3 PO 4 22.4 L H 2 3 mol H 2 mol H 3 PO 4

49. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) c.If we were able to collect 6.92 L of hydrogen gas, how many grams of phosphoric acid was used? 6.92 L H 2 1 mol H 2 2 mol H 3 PO 4 g H 3 PO 4 22.4 L H 2 3 mol H 2 mol H 3 PO 4

50. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) c.If we were able to collect 6.92 L of hydrogen gas, how many grams of phosphoric acid was used? 6.92 L H 2 1 mol H 2 2 mol H 3 PO 4 98.0 g H 3 PO 4 22.4 L H 2 3 mol H 2 1 mol H 3 PO 4

51. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) c.If we were able to collect 6.92 L of hydrogen gas, how many grams of phosphoric acid was used? 6.92 L H 2 1 mol H 2 2 mol H 3 PO 4 98.0 g H 3 PO 4 = 22.4 L H 2 3 mol H 2 1 mol H 3 PO 4 6.92 × 2 × 98 ÷ 22.4 ÷ 3 = 20.1833

52. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) c.If we were able to collect 6.92 L of hydrogen gas, how many grams of phosphoric acid was used? 6.92 L H 2 1 mol H 2 2 mol H 3 PO 4 98.0 g H 3 PO 4 = 20.2 g H 3 PO 4 22.4 L H 2 3 mol H 2 1 mol H 3 PO 4 6.92 × 2 × 98 ÷ 22.4 ÷ 3 = 20.1833

53. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) d.If we were able to collect 1.455 L of hydrogen gas, how many moles of potassium were used? 1.455 L H 2

54. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) d.If we were able to collect 1.455 L of hydrogen gas, how many moles of potassium were used? 1.455 L H 2 1 mol H 2 22.41 L

55. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) d.If we were able to collect 1.455 L of hydrogen gas, how many moles of potassium were used? 1.455 L H 2 1 mol H 2 6 mols K 22.41 L3 mols H 2

56. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) d.If we were able to collect 1.455 L of hydrogen gas, how many moles of potassium were used? 1.455 L H 2 1 mol H 2 6 mols K 22.41 L3 mols H 2 1.455 × 6 ÷ 22.4 ÷ 3 = 0.129911

57. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) d.If we were able to collect 1.455 L of hydrogen gas, how many moles of potassium were used? 1.455 L H 2 1 mol H 2 6 mols K = 0.1299 mols K 22.41 L3 mols H 2 1.455 × 6 ÷ 22.41 ÷ 3 = 0.12983

58. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) e.Your turn: How many liters of hydrogen gas will be formed along with 3.20 × 10 24 formula units of potassium phosphate? 5 minutes remain…

59. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) e.Your turn: How many liters of hydrogen gas will be formed along with 3.20 × 10 24 formula units of potassium phosphate? 4 minutes remain… 3.20 × 10 24 formula units K 3 PO 4

60. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) e.Your turn: How many liters of hydrogen gas will be formed along with 3.20 × 10 24 formula units of potassium phosphate? 3 minutes remain… 3.20 × 10 24 formula units K 3 PO 4 1 mol K 3 PO 4 6.02 × 10 23 formula units

61. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) e.Your turn: How many liters of hydrogen gas will be formed along with 3.20 × 10 24 formula units of potassium phosphate? 2 minutes remain… 3.20 × 10 24 formula units K 3 PO 4 1 mol K 3 PO 4 3 mol H 2 6.02 × 10 23 formula units 2 mol K 3 PO 4

62. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) e.Your turn: How many liters of hydrogen gas will be formed along with 3.20 × 10 24 formula units of potassium phosphate? 1 minute remains… 3.20 × 10 24 formula units K 3 PO 4 1 mol K 3 PO 4 3 mol H 2 22.4 L = 6.02 × 10 23 formula units 2 mol K 3 PO 4 1 mol H 2

63. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) e.Your turn: How many liters of hydrogen gas will be formed along with 3.20 × 10 24 formula units of potassium phosphate? 3.20 × 10 24 formula units K 3 PO 4 1 mol K 3 PO 4 3 mol H 2 22.4 L = 6.02 × 10 23 formula units 2 mol K 3 PO 4 1 mol H 2 3.20 E 24 × 3 × 22.4 ÷ 6.02 E 23 ÷ 2 = 178.605

64. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) e.Your turn: How many liters of hydrogen gas will be formed along with 3.20 × 10 24 formula units of potassium phosphate? 179 L of H 2 3.20 × 10 24 formula units K 3 PO 4 1 mol K 3 PO 4 3 mol H 2 22.4 L = 6.02 × 10 23 formula units 2 mol K 3 PO 4 1 mol H 2

65. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) f.Your turn: How many atoms of potassium will be used along with 1.9 × 10 15 formula units of phosphoric acid? 5 minutes remain…

66. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) f.Your turn: How many atoms of potassium will be used along with 1.9 × 10 15 formula units of phosphoric acid? 4 minutes remain… 1.9 × 10 15 formula units H 3 PO 4

67. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) f.Your turn: How many atoms of potassium will be used along with 1.9 × 10 15 formula units of phosphoric acid? 3 minutes remain… 1.9 × 10 15 formula units H 3 PO 4 1 mol H 3 PO 4 6.02 × 10 23 formula units

68. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) f.Your turn: How many atoms of potassium will be used along with 1.9 × 10 15 formula units of phosphoric acid? 2 minutes remain… 1.9 × 10 15 formula units H 3 PO 4 1 mol H 3 PO 4 6 mol K 6.02 × 10 23 formula units 2 mol H 3 PO 4

69. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) f.Your turn: How many atoms of potassium will be used along with 1.9 × 10 15 formula units of phosphoric acid? 1 minute remains… 1.9 × 10 15 formula units H 3 PO 4 1 mol H 3 PO 4 6 mol K 6.02 × 10 23 atoms K = 6.02 × 10 23 formula units 2 mol H 3 PO 4 1 mol K

70. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) f.Your turn: How many atoms of potassium will be used along with 1.9 × 10 15 formula units of phosphoric acid? 1.9 E 15 × 6 × 6.02 E 23 ÷ 6.02 E 23 ÷ 2 = 5.7 E 15 1.9 × 10 15 formula units H 3 PO 4 1 mol H 3 PO 4 6 mol K 6.02 × 10 23 atoms K = 6.02 × 10 23 formula units 2 mol H 3 PO 4 1 mol K

71. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) f.Your turn: How many atoms of potassium will be used along with 1.9 × 10 15 formula units of phosphoric acid? 5.7 × 10 15 atoms K 1.9 × 10 15 formula units H 3 PO 4 1 mol H 3 PO 4 6 mol K 6.02 × 10 23 atoms K = 6.02 × 10 23 formula units 2 mol H 3 PO 4 1 mol K

72. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) g.Your turn: How many liters of hydrogen gas could be formed from 36.4 moles of potassium? 3 minutes remain…

73. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) g.Your turn: How many liters of hydrogen gas could be formed from 36.4 moles of potassium? 2 minutes remain… 36.4 mol K

74. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) g.Your turn: How many liters of hydrogen gas could be formed from 36.4 moles of potassium? 1 minute remains… 36.4 mol K3 mols H 2 6 mols K

75. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) g.Your turn: How many liters of hydrogen gas could be formed from 36.4 moles of potassium? 36.4 mol K3 mols H 2 22.4 L H 2 = 408 L H 2 6 mols K1 mol H 2 36.4 × 3 × 22.4 ÷ 6 = 407.68

76. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) h.Your turn: How many moles of potassium phosphate will be formed along with 7.811 grams of H 2 gas? 3 minutes remain…

77. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) h.Your turn: How many moles of potassium phosphate will be formed along with 7.811 grams of H 2 gas? 2 minutes remain… 7.811 g H 2

78. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) h.Your turn: How many moles of potassium phosphate will be formed along with 7.811 grams of H 2 gas? 1 minute remains… 7.811 g H 2 1 mol H 2 2.016 g H 2

79. Example: Potassium is reacted with phosphoric acid Step 1: 2H 3 PO 4 (aq) + 6K (s) → 2K 3 PO 4 (aq) + 3H 2 (g) h.Your turn: How many moles of potassium phosphate will be formed along with 7.811 grams of H 2 gas? 7.811 g H 2 1 mol H 2 2 mols K 3 PO 4 = 2.583 mols K 3 PO 4 2.016 g H 2 3 mols H 2 7.811 × 2 ÷ 2.016 ÷ 3 = 2.583

80. Limiting Reactant

82. Rarely in real life does a reaction start with exactly proportional amounts of the reactants, as the smallest differences in mass still add up to more than trillions of atoms, and often there are big differences in the amounts of the reactants available.

83. To determine which reactant will limit the reaction, the initial amount of all reactants must be known. Then take each react in turn and solve for the same product (it will not matter which product, so choose something convenient). The reactant the can produce the smallest amount of products is the limiting reactant, and all the others are the excess reactants.

84. Example: 10.0 grams of sodium is added to 50.0 grams of water. Which reactant is the limiting reactant and which one is the excess reactant?

85. Step 1: 2Na + 2HOH → 2NaOH + H 2

86. Step 2: Use both reactants to solve for the same product (let’s choose NaOH).

87. 10.0 g Na

88. Step 2: Use both reactants to solve for the same product (let’s choose NaOH). 10.0 g Na1 mol Na 23.0 g Na

89. Step 2: Use both reactants to solve for the same product (let’s choose NaOH). 10.0 g Na1 mol Na2 mol NaOH 23.0 g Na2 mol Na

90. Step 2: Use both reactants to solve for the same product (let’s choose NaOH). 10.0 g Na1 mol Na2 mol NaOH40.0 g NaOH = 23.0 g Na2 mol Na1 mol NaOH

91. Step 2: Use both reactants to solve for the same product (let’s choose NaOH). 10.0 g Na1 mol Na2 mol NaOH40.0 g NaOH = 17.4 g NaOH 23.0 g Na2 mol Na1 mol NaOH

92. Step 2: Use both reactants to solve for the same product (let’s choose NaOH). 10.0 g Na1 mol Na2 mol NaOH40.0 g NaOH = 17.4 g NaOH 23.0 g Na2 mol Na1 mol NaOH 50.0 g H 2 O

93. Step 2: Use both reactants to solve for the same product (let’s choose NaOH). 10.0 g Na1 mol Na2 mol NaOH40.0 g NaOH = 17.4 g NaOH 23.0 g Na2 mol Na1 mol NaOH 50.0 g H 2 O1 mol H 2 O 18.0 g H 2 O

94. Step 2: Use both reactants to solve for the same product (let’s choose NaOH). 10.0 g Na1 mol Na2 mol NaOH40.0 g NaOH = 17.4 g NaOH 23.0 g Na2 mol Na1 mol NaOH 50.0 g H 2 O1 mol H 2 O2 mol NaOH 18.0 g H 2 O2 mol H 2 O

95. Step 2: Use both reactants to solve for the same product (let’s choose NaOH). 10.0 g Na1 mol Na2 mol NaOH40.0 g NaOH = 17.4 g NaOH 23.0 g Na2 mol Na1 mol NaOH 50.0 g H 2 O1 mol H 2 O2 mol NaOH40.0 g NaOH = 18.0 g H 2 O2 mol H 2 O1 mol NaOH

96. Step 2: Use both reactants to solve for the same product (let’s choose NaOH). 10.0 g Na1 mol Na2 mol NaOH40.0 g NaOH = 17.4 g NaOH 23.0 g Na2 mol Na1 mol NaOH 50.0 g H 2 O1 mol H 2 O2 mol NaOH40.0 g NaOH = 111 g NaOH 18.0 g H 2 O2 mol H 2 O1 mol NaOH

97. Step 2: Use both reactants to solve for the same product (let’s choose NaOH). Na is limiting and thus H 2 O is excess. 10.0 g Na1 mol Na2 mol NaOH40.0 g NaOH = 17.4 g NaOH 23.0 g Na2 mol Na1 mol NaOH 50.0 g H 2 O1 mol H 2 O2 mol NaOH40.0 g NaOH = 111 g NaOH 18.0 g H 2 O2 mol H 2 O1 mol NaOH

98. How much excess is there? Step 1: Starting with the limiting...

99. How much excess is there? Step 1: Starting with the limiting... 10.0 g Na

100. How much excess is there? Step 1: Starting with the limiting... 10.0 g Na1 mol Na 23.0 g Na

101. How much excess is there? Step 1: Starting with the limiting... 10.0 g Na1 mol Na2 mol H 2 O 23.0 g Na2 mol Na

102. How much excess is there? Step 1: Starting with the limiting... 10.0 g Na1 mol Na2 mol H 2 O18.0 g H 2 O = 23.0 g Na2 mol Na1 mol H 2 O

103. How much excess is there? Step 1: Starting with the limiting... 10.0 g Na1 mol Na2 mol H 2 O18.0 g H 2 O = 7.83 g H 2 O used 23.0 g Na2 mol Na1 mol H 2 O

104. How much excess is there? Step 1: Starting with the limiting... Step 2: 50.0 g H 2 O – 7.83 g H 2 O = 42.2 g H 2 O 10.0 g Na1 mol Na2 mol H 2 O18.0 g H 2 O = 7.83 g H 2 O used 23.0 g Na2 mol Na1 mol H 2 O

105. How many grams of base are produce?

106. 17.4 g NaOH

107. How many grams of base are produce? 17.4 g NaOH How many milliliters of gas is produced?

108. How many grams of base are produce? 17.4 g NaOH How many milliliters of gas is produced? 10.0 g Na

109. How many grams of base are produce? 17.4 g NaOH How many milliliters of gas is produced? 10.0 g Na1 mol Na 23.0 g Na

110. How many grams of base are produce? 17.4 g NaOH How many milliliters of gas is produced? 10.0 g Na1 mol Na1 mol H 2 23.0 g Na2 mol Na

111. How many grams of base are produce? 17.4 g NaOH How many milliliters of gas is produced? 10.0 g Na1 mol Na1 mol H 2 22. 4 L 23.0 g Na2 mol Na1 mol H 2 gas

112. How many grams of base are produce? 17.4 g NaOH How many milliliters of gas is produced? 10.0 g Na1 mol Na1 mol H 2 22. 4 L1000 mL = 23.0 g Na2 mol Na1 mol H 2 gas1 L

113. How many grams of base are produce? 17.4 g NaOH How many milliliters of gas is produced? 10.0 g Na1 mol Na1 mol H 2 22. 4 L1000 mL = 4870 mL 23.0 g Na2 mol Na1 mol H 2 gas1 L