1. 1 CHAPTER THREE CHEMICAL EQUATIONS & REACTION STOICHIOMETRY

2. 2 Chapter Three Goals 1. Chemical Equations 2. Calculations Based on Chemical Equations 3. The Limiting Reactant Concept 4. Percent Yields from Chemical Reactions 5. Sequential Reactions 6. Concentrations of Solutions 7. Dilution of solutions 8. Using Solutions in Chemical Reactions 9. Synthesis Question

3. 3 Chemical Equations Symbolic representation of a chemical reaction that shows: 1. reactants on left side of reaction 2. products on right side of equation 3. relative amounts of each using stoichiometric coefficients

4. 4 Chemical Equations Attempt to show on paper what is happening at the laboratory and molecular levels.

5. 5 Chemical Equations Look at the information an equation provides:

6. 6 Chemical Equations Look at the information an equation provides: reactants yields products

7. 7 Chemical Equations Look at the information an equation provides: reactants yields products 1 formula unit 3 molecules 2 atoms 3 molecules

8. 8 Chemical Equations Look at the information an equation provides: reactants yields products 1 formula unit 3 molecules 2 atoms 3 molecules 1 mole 3 moles 2 moles 3 moles

9. 9 Chemical Equations Look at the information an equation provides: reactants yields products 1 formula unit 3 molecules 2 atoms 3 molecules 1 mole 3 moles 2 moles 3 moles 159.7 g 84.0 g 111.7 g 132 g

10. 10 Chemical Equations Law of Conservation of Matter – There is no detectable change in quantity of matter in an ordinary chemical reaction. – Balanced chemical equations must always include the same number of each kind of atom on both sides of the equation. – This law was determined by Antoine Lavoisier. Propane,C 3 H 8, burns in oxygen to give carbon dioxide and water.

11. 11 Law of Conservation of Matter NH 3 burns in oxygen to form NO & water You do it!

12. 12 Law of Conservation of Matter NH 3 burns in oxygen to form NO & water

13. 13 Law of Conservation of Matter C 7 H 16 burns in oxygen to form carbon dioxide and water. You do it!

14. 14 Law of Conservation of Matter C 7 H 16 burns in oxygen to form carbon dioxide and water.

15. 15 Law of Conservation of Matter C 7 H 16 burns in oxygen to form carbon dioxide and water. Balancing equations is a skill acquired only with lots of practice – work many problems

16. 16 Calculations Based on Chemical Equations Can work in moles, formula units, etc. Frequently, we work in mass or weight (grams or kg or pounds or tons).

17. 17 Calculations Based on Chemical Equations Example 3-1: How many CO molecules are required to react with 25 formula units of Fe 2 O 3 ?

18. 18 Calculations Based on Chemical Equations Example 3-2: How many iron atoms can be produced by the reaction of 2.50 x 105 formula units of iron (III) oxide with excess carbon monoxide?

19. 19 Calculations Based on Chemical Equations Example 3-2: How many iron atoms can be produced by the reaction of 2.50 x 105 formula units of iron (III) oxide with excess carbon monoxide?

20. 20 Calculations Based on Chemical Equations Example 3-2: How many iron atoms can be produced by the reaction of 2.50 x 105 formula units of iron (III) oxide with excess carbon monoxide?

21. 21 Calculations Based on Chemical Equations Example 3-3: What mass of CO is required to react with 146 g of iron (III) oxide?

22. 22 Calculations Based on Chemical Equations Example 3-3: What mass of CO is required to react with 146 g of iron (III) oxide?

23. 23 Calculations Based on Chemical Equations Example 3-3: What mass of CO is required to react with 146 g of iron (III) oxide?

24. 24 Calculations Based on Chemical Equations Example 3-4: What mass of carbon dioxide can be produced by the reaction of 0.540 mole of iron (III) oxide with excess carbon monoxide?

25. 25 Calculations Based on Chemical Equations Example 3-4: What mass of carbon dioxide can be produced by the reaction of 0.540 mole of iron (III) oxide with excess carbon monoxide?

26. 26 Calculations Based on Chemical Equations Example 3-4: What mass of carbon dioxide can be produced by the reaction of 0.540 mole of iron (III) oxide with excess carbon monoxide?

27. 27 Calculations Based on Chemical Equations Example 3-5: What mass of iron (III) oxide reacted with excess carbon monoxide if the carbon dioxide produced by the reaction had a mass of 8.65 grams? You do it!

28. 28 Calculations Based on Chemical Equations Example 3-5: What mass of iron (III) oxide reacted with excess carbon monoxide if the carbon dioxide produced by the reaction had a mass of 8.65 grams?

29. 29 Calculations Based on Chemical Equations Example 3-6: How many pounds of carbon monoxide would react with 125 pounds of iron (III) oxide? You do it!

30. 30 Calculations Based on Chemical Equations YOU MUST BE PROFICIENT WITH THESE TYPES OF PROBLEMS!!!

31. 31 Limiting Reactant Concept Kitchen example of limiting reactant concept. 1 packet of muffin mix + 2 eggs + 1 cup of milk  12 muffins How many muffins can we make with the following amounts of mix, eggs, and milk?

32. 32 Limiting Reactant Concept Mix PacketsEggsMilk 11 dozen1 gallon limiting reactant is the muffin mix 21 dozen1 gallon 31 dozen1 gallon 41 dozen1 gallon 51 dozen1 gallon 61 dozen1 gallon 71 dozen1 gallon limiting reactant is the dozen eggs

33. 33 Limiting Reactant Concept Example 3-7: Suppose a box contains 87 bolts, 110 washers, and 99 nuts. How many sets, each consisting of one bolt, two washers, and one nut, can you construct from the contents of one box?

34. 34 Limiting Reactant Concept Look at a chemical limiting reactant situation. Zn + 2 HCl  ZnCl 2 + H 2

35. 35 Limiting Reactant Concept Example 3-8: What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g of carbon disulfide with 110. g of oxygen?

36. 36 Limiting Reactant Concept Example 3-8: What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g of carbon disulfide with 110. g of oxygen?

37. 37 Limiting Reactant Concept Example 3-8: What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g of carbon disulfide with 110. g of oxygen?

38. 38 Limiting Reactant Concept Example 3-8: What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g of carbon disulfide with 110. g of oxygen?

39. 39 Limiting Reactant Concept What do we do next? You do it!

40. 40 Limiting Reactant Concept Which is limiting reactant? Limiting reactant is O 2. What is maximum mass of sulfur dioxide? Maximum mass is 147 g.

41. 41 Percent Yields from Reactions Theoretical yield is calculated by assuming that the reaction goes to completion. – Determined from the limiting reactant calculation. Actual yield is the amount of a specified pure product made in a given reaction. – In the laboratory, this is the amount of product that is formed in your beaker, after it is purified and dried. Percent yield indicates how much of the product is obtained from a reaction.

42. 42 Percent Yields from Reactions Example 3-9: A 10.0 g sample of ethanol, C 2 H 5 OH, was boiled with excess acetic acid, CH 3 COOH, to produce 14.8 g of ethyl acetate, CH 3 COOC 2 H 5. What is the percent yield?

43. 43 Percent Yields from Reactions

44. 44 Percent Yields from Reactions

45. 45 Percent Yields from Reactions

46. 46 Percent Yields from Reactions

47. 47 Sequential Reactions Example 3-10: Starting with 10.0 g of benzene (C 6 H 6 ), calculate the theoretical yield of nitrobenzene (C 6 H 5 NO 2 ) and of aniline (C 6 H 5 NH 2 ).

48. 48 Sequential Reactions

49. 49 Sequential Reactions Next calculate the mass of aniline produced. You do it!

50. 50 Sequential Reactions

51. 51 Sequential Reactions

52. 52 Sequential Reactions If 6.7 g of aniline is prepared from 10.0 g of benzene, what is the percentage yield? You do it!

53. 53 Concentration of Solutions Solution is a mixture of two or more substances dissolved in another. – Solute is the substance present in the smaller amount. – Solvent is the substance present in the larger amount. – In aqueous solutions, the solvent is water. The concentration of a solution defines the amount of solute dissolved in the solvent. – The amount of sugar in sweet tea can be defined by its concentration. One common unit of concentration is:

54. 54 Concentration of Solutions Example 3-12: Calculate the mass of 8.00% w/w NaOH solution that contains 32.0 g of NaOH.

55. 55 Concentration of Solutions Example 3-11: What mass of NaOH is required to prepare 250.0 g of solution that is 8.00% w/w NaOH?

56. 56 Concentration of Solutions Example 3-13: Calculate the mass of NaOH in 300.0 mL of an 8.00% w/w NaOH solution. Density is 1.09 g/mL. You do it!

57. 57 Concentrations of Solutions Example 3-14: What volume of 12.0% KOH contains 40.0 g of KOH? The density of the solution is 1.11 g/mL. You do it!

58. 58 Concentrations of Solutions Example 3-14: What volume of 12.0% KOH contains 40.0 g of KOH? The density of the solution is 1.11 g/mL.

59. 59 Concentrations of Solutions Second common unit of concentration:

60. 60 Concentrations of Solutions Example 3-15: Calculate the molarity of a solution that contains 12.5 g of sulfuric acid in 1.75 L of solution. You do it!

61. 61 Concentrations of Solutions Example 3-15: Calculate the molarity of a solution that contains 12.5 g of sulfuric acid in 1.75 L of solution.

62. 62 Concentrations of Solutions Example 3-15: Calculate the molarity of a solution that contains 12.5 g of sulfuric acid in 1.75 L of solution.

63. 63 Concentrations of Solutions Example 3-16: Determine the mass of calcium nitrate required to prepare 3.50 L of 0.800 M Ca(NO 3 ) 2. You do it!

64. 64 Concentrations of Solutions Example 3-16: Determine the mass of calcium nitrate required to prepare 3.50 L of 0.800 M Ca(NO 3 ) 2.

65. 65 Concentrations of Solutions One of the reasons that molarity is commonly used is because: M x L = moles solute and M x mL = mmol solute

66. 66 Concentrations of Solutions Example 3-17: The specific gravity of concentrated HCl is 1.185 and it is 36.31% w/w HCl. What is its molarity?

67. 67 Concentrations of Solutions Example 3-17: The specific gravity of concentrated HCl is 1.185 and it is 36.31% w/w HCl. What is its molarity?

68. 68 Concentrations of Solutions Example 3-17: The specific gravity of concentrated HCl is 1.185 and it is 36.31% w/w HCl. What is its molarity?

69. 69 Dilution of Solutions To dilute a solution, add solvent to a concentrated solution. – One method to make tea “less sweet.” – How fountain drinks are made from syrup. The number of moles of solute in the two solutions remains constant. The relationship M 1 V 1 = M 2 V 2 is appropriate for dilutions, but not for chemical reactions.

70. 70 Dilution of Solutions Common method to dilute a solution involves the use of volumetric flask, pipet, and suction bulb.

71. 71 Dilution of Solutions Example 3-18: If 10.0 mL of 12.0 M HCl is added to enough water to give 100. mL of solution, what is the concentration of the solution?

72. 72 Dilution of Solutions Example 3-19: What volume of 18.0 M sulfuric acid is required to make 2.50 L of a 2.40 M sulfuric acid solution? You do it!

73. 73 Dilution of Solutions Example 3-19: What volume of 18.0 M sulfuric acid is required to make 2.50 L of a 2.40 M sulfuric acid solution?

74. 74 Using Solutions in Chemical Reactions Combine the concepts of molarity and stoichiometry to determine the amounts of reactants and products involved in reactions in solution.

75. 75 Using Solutions in Chemical Reactions Example 3-20: What volume of 0.500 M BaCl 2 is required to completely react with 4.32 g of Na 2 SO 4 ?

76. 76 Using Solutions in Chemical Reactions Example 3-20: What volume of 0.500 M BaCl 2 is required to completely react with 4.32 g of Na 2 SO 4 ?

77. 77 Using Solutions in Chemical Reactions Example 3-20: What volume of 0.500 M BaCl 2 is required to completely react with 4.32 g of Na 2 SO 4 ?

78. 78 Using Solutions in Chemical Reactions Example 3-21: (a)What volume of 0.200 M NaOH will react with 50.0 mL 0f 0.200 M aluminum nitrate, Al(NO 3 ) 3 ?

79. 79 Using Solutions in Chemical Reactions Example 3-20: (a)What volume of 0.200 M NaOH will react with 50.0 mL 0f 0.200 M aluminum nitrate?

80. 80 Using Solutions in Chemical Reactions (b)What mass of Al(OH) 3 precipitates in (a)? You do it!

81. 81 Using Solutions in Chemical Reactions (b) What mass of Al(OH) 3 precipitates in (a)?

82. 82 Using Solutions in Chemical Reactions Titrations are a method of determining the concentration of an unknown solutions from the known concentration of a solution and solution reaction stoichiometry. – Requires special lab glassware Buret, pipet, and flasks – Must have an an indicator also

83. 83 Using Solutions in Chemical Reactions Example 3-22: What is the molarity of a KOH solution if 38.7 mL of the KOH solution is required to react with 43.2 mL of 0.223 M HCl?

84. 84 Using Solutions in Chemical Reactions Example 3-22: What is the molarity of a KOH solution if 38.7 mL of the KOH solution is required to react with 43.2 mL of 0.223 M HCl?

85. 85 Using Solutions in Chemical Reactions Example 3-22: What is the molarity of a KOH solution if 38.7 mL of the KOH solution is required to react with 43.2 mL of 0.223 M HCl?

86. 86 Using Solutions in Chemical Reactions Example 3-22: What is the molarity of a KOH solution if 38.7 mL of the KOH solution is required to react with 43.2 mL of 0.223 M HCl?

87. 87 Using Solutions in Chemical Reactions Example 3-23: What is the molarity of a barium hydroxide solution if 44.1 mL of 0.103 M HCl is required to react with 38.3 mL of the Ba(OH) 2 solution?

88. 88 Using Solutions in Chemical Reactions Example 3-23: What is the molarity of a barium hydroxide solution if 44.1 mL of 0.103 M HCl is required to react with 38.3 mL of the Ba(OH) 2 solution?

89. 89 Using Solutions in Chemical Reactions Example 3-23: What is the molarity of a barium hydroxide solution if 44.1 mL of 0.103 M HCl is required to react with 38.3 mL of the Ba(OH) 2 solution?

90. 90 Using Solutions in Chemical Reactions Example 3-23: What is the molarity of a barium hydroxide solution if 44.1 mL of 0.103 M HCl is required to react with 38.3 mL of the Ba(OH) 2 solution?

91. 91 Synthesis Question Nylon is made by the reaction of hexamethylene diamine with adipic acid.

92. 92 Synthesis Question in a 1 to 1 mole ratio. The structure of nylon is: where the value of n is typically 450,000. On a daily basis, a DuPont factory makes 1.5 million pounds of nylon. How many pounds of hexamethylene diamine and adipic acid must they have available in the plant each day?

93. 93 Synthesis Question

94. 94 Synthesis Question

95. 95 Synthesis Question

96. 96 Synthesis Question

97. 97 Synthesis Question

98. 98 Synthesis Question

99. 99 Group Activity Manganese dioxide, potassium hydroxide and oxygen react in the following fashion: A mixture of 272.9 g of MnO 2, 26.6 L of 0.250 M KOH, and 41.92 g of O 2 is allowed to react as shown above. After the reaction is finished, 234.6 g of KMnO 4 is separated from the reaction mixture. What is the per cent yield of this reaction?

100. 100 End of Chapter 3