1. 9:47 PM Unit 1: Stoichiometry Chemistry 2202 1

2. 9:47 PM Stoichiometry Stoichiometry deals with quantities used in OR produced by a chemical reaction 2

3. 9:47 PM 3 Parts Mole Calculations (Chp. 2 & 3) Stoichiometry and Chemical Equations (Chp. 4) Solution Stoichiometry (Chp. 6) 3

4. 9:47 PM PART 1 - Mole Calculations Isotopes and Atomic Mass (pp. 43 - 46) Avogadro’s number (pp. 47 – 49) Mole Conversions (pp. 50 - 74) M, V stp, N A, n, m, v, N 4

5. 9:47 PM Questions p. 45 #’s 1 – 4 p. 46 #’s 1 – 6 p. 75 #’s 9 – 12 p. 51-53 #’s 5 – 15 p. 57 #’s 16 – 19 p. 59,60 #’s 20 – 27 p. 63,64 #’s 28 - 37 p. 54 #’s 5 – 8 p. 65 #’s 2, 4, 5 p. 75 #’s 13, 14 p. 76 #’s 15, 17–19, 21-23 p. 73 #’s 38 – 43 p. 74 #’s 1 – 4 p. 76 #’s 26, 27 5

6. 9:47 PM PART 1 - Mole Calculations Percent composition: - given mass (p. 79 - 82) - given the chemical formula (p. 83 - 86) Empirical Formulas (pp. 87 - 94) Molecular Formulas (pp. 95 - 98) Lab: Formula of a Hydrate 6

7. 9:47 PM Questions p. 82 #’s 1 – 4 p. 85 #’s 5 – 8 p. 89 #’s 9 – 12 p. 91 #’s 13 – 16 p. 97 #’s 17 - 20 p. 103 #’s 23 – 24 p. 86 #’s 1, 3 – 6 p. 94 #’s 1 - 7 p. 106 #’s 1 - 3, 6, 7 p. 107 – 109 #’s 5 – 23, 25 7

8. 9:47 PM Isotopes and Atomic Mass atomic number - the number of protons in an atom or ion mass number - the sum of the protons and neutrons in an atom isotopes - atoms which have the same number of protons and electrons but different numbers of neutrons 8

9. 9:47 PM Isotopes and Atomic Mass eg. 9

10. 9:47 PM Isotopes and Atomic Mass 10

11. 9:47 PM Isotopes and Atomic Mass not all isotopes are created equal 79 % 10 % 11 % 11

12. 9:47 PM Isotopes and Atomic Mass 12

13. 9:47 PM Isotopes and Atomic Mass atomic mass unit (AMU - p.43) - a unit used to describe the mass of individual atoms - the symbol for the AMU is u - 1 u is 1/12 of the mass of a carbon-12 atom 13

14. 9:47 PM Isotopes and Atomic Mass average atomic mass (AAM) - the AAM is the weighted average of all the isotopes of an element (p. 45) p. 14# 5 p. 45 #’s 1 – 3 p. 75 #’s 9 – 11 p. 46#’s 2 & 4 14

15. 9:47 PM Finding % Abundance eg. Br has two naturally occurring isotopes. Br-79 has a mass of 78.92 u and Br-81 has a mass of 80.92 u. If the AAM of Br is 79.90 u, determine the percentage abundance of each isotope. 15

16. 9:47 PM Let x = fraction of Br-79 Let y = fraction of Br-81 x + y = 1 78.92x + 80.92y = 79.90 Finding % Abundance y = 1 - x 16

17. 9:47 PM x + y = 1 78.92x + 80.92y = 79.90 17

18. 9:47 PM Avogadro’s Number (p. 47) p. 48 18

19. 9:47 PM Avogadro’s Number The MOLE is the number of atoms contained in exactly 12 g of carbon-12. 1 mole = 6.02214199 x 10 23 particles 1 mol = 6.02 x 10 23 particles 19

20. 9:47 PM Avogadro’s Number In honor of Amedeo Avogadro, the number of particles in 1 mol has been called Avogadro’s number. N A = 6.02 x 10 23 particles/mol 20

21. 9:47 PM Avogadro’s Number Formulas: n = # of moles N = # of particles (atoms, ions, molecules, or formula units) N A = Avogadro’s # N = n x N A 21

22. 9:47 PM eg. How many moles are contained in the following? a) 2.56 x 10 28 Pb atoms b) 7.19 x 10 21 CO 2 molecules Avogadro’s Number 22

23. 9:47 PM Avogadro’s Number eg. Calculate the number of moles in 4.98 x 10 25 atoms of Al. eg.How many formula units of Na 2 SO 4 are in 5.69 mol of Na 2 SO 4 ? # of Na ions? # of Oxygen atoms? 23

24. 9:47 PM Avogadro’s Number eg. How many molecules of glucose are in 0.435 mol of C 6 H 12 O 6 ? How many carbon atoms? 24

25. 9:47 PM Avogadro’s Number eg. Calculate the number of moles in a sample of glucose that has 3.56 x 10 22 glucose molecules. 25

26. 9:47 PM Avogadro’s Number pp. 51 – 53: #’s 5 – 15 p. 54: #’s 4 - 8 26

27. 9:47 PM Molar Mass The mass of one mole of a substance is the molar mass of the substance eg. 1 mole of Pb → 207.19 g 1 mole of Ag → 107.87 g 27

28. 9:47 PM Molar Mass symbol for molar mass is M unit is g/mol eg. M Pb = 207.19 g/mol M Ag = 107.87 g/mol 28

29. 9:47 PM Molar Mass The molar mass of a compound is the sum of the molar masses of the elements in the compound eg. Calculate the molar mass of: a) H 2 Ob) C 6 H 12 O 6 c) Ca(OH) 2 29

30. 9:47 PM Molar Mass H 2 O has 2 hydrogens and 1 oxygen 2 x 1.01 = 2.02 1 X 16.00 = 16.00 18.02 g/mol 30

31. 9:47 PM Molar Mass C 6 H 12 O 6 6 x 12.01 = 72.06 12 x 1.01 = 12.12 6 x 16.00 = 96.00 180.18 g/mol 31

32. 9:47 PM Molar Mass Ca(OH) 2 1 x 40.08 = 40.08 2 x 16.00 = 32.00 2 x 1.01 = 2.02 74.10 g/mol Your calculator may not show the zeroes. There should be 2 digits after the decimal when adding molar masses 32

33. 9:47 PM Molar Mass p. 57: #’s 16 – 19 & Molar Masses Handout 1.151.92 g/mol 7. 58.44 g/mol 2.120.38 g/mol 8. 100.09 g/mol 3.105.99 g/mol 9. 44.02 g/mol 4.100.40 g/mol10. 248.22 g/mol 5.74.44 g/mol11. 115.04 g/mol 6.78.01 g/mol 33

34. 9:47 PM Molar Mass Calculations N = n x N A mass molar mass m = n x M Avogadro’s # 34

35. 9:47 PM Molar Mass Calculations N = n x N A m = n x M 35

36. 9:47 PM Molar Mass Calculations eg.How many moles are in 25.3 g of NO 2 ? m = 25.3 g M NO2 = 46.01 g/mol 36

37. 9:47 PM Molar Mass Calculations eg.What is the mass of 4.69 mol of water? n = 4.69 mol M water = 18.02 g/mol m = n x M = (4.69 mol)(18.02 g/mol) = 84.5 g 37

38. 9:47 PM Molar Mass Calculations Practice:p. 59 #’s 20 - 23 p. 60 #’s 24 - 27 38

39. particles (N) Moles (n) Mass (m) 5.98 x 10 26 Cu atoms 4.50 g H 2 O 6.15 mol O 3 Particle–Mole-Mass Conversions 9:47 PM39

40. 9:47 PM Molar Mass Calculations Practice:p. 63 #’s 28 p. 64 #’s 34 p. 76 # 15 moles (n) mass (m) particles (N) x M ÷ M ÷ N A x N A 40

41. 9:47 PM Molar Mass Calculations eg.How many molecules are in 26.9 g of water? m = 26.9 g M water = 18.02 g/mol N A = 6.022 x 10 23 molecules/mol Find N 41

42. 9:47 PM Molar Mass Calculations = 1.493 mol H 2 O N = n x N A = 1.493 X 6.022 x 10 23 = 8.99 x 10 23 molecules 42

43. 9:47 PM Molar Mass Calculations eg.How many molecules are in 4.78 g of glucose? m = 4.78 g M water = 180.18 g/mol N A = 6.022 x 10 23 molecules/mol Find N 9:47 PM43

44. 9:47 PM Molar Mass Calculations = 1.493 mol H 2 O N = n x N A = 1.493 X 6.022 x 10 23 = 8.99 x 10 23 molecules 9:47 PM44

45. 9:47 PM Molar Mass Calculations eg. A sample of Sn contains 4.69 x 10 28 atoms. Calculate its mass. N = 4.69 x 10 28 N A = 6.022 x 10 23 molecules/mol M Sn = 118.69 g/mol Find m 45

46. 9:47 PM Molar Mass Calculations = 77,881 mol m = n x M = 77881 mol x 118.69 g/mol = 9.24 x 10 6 g 46

47. 9:47 PM Molar Mass Calculations Practice:p. 63 #’s 28 - 33 p. 64 #’s 34 – 37 p. 76 # 15 47

48. 9:47 PM Molar Mass Calculations Practice: p. 54 #’s 5 - 8 p. 65 #’s 2, 4, 5 p. 75 #’s 13, 14, p. 76 #’s 15, 17 – 19, 21 -23 48

49. 9:47 PM Molar Volume The volume of a gas increases when temperature increases but decreases when pressure increases. The volume of gases is measured under conditions of Standard Temperature and Pressure (STP) 49

50. 9:47 PM Molar Volume Standard Pressure – 101.3 kPa Standard Temperature – 0 °C Avogadro hypothesized that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. 50

51. 9:47 PM Molar Volume Experimental evidence shows the volume of one mole of ANY GAS at STP is 22.4 L/mol OR V STP = 22.4 L/mol 51

52. 9:47 PM N = n x N A m = n x M v = n x V STP given volume in Litres 52 Molar Volume

53. 9:47 PM53 Sample Questions 1.Calculate the number of moles in 36.4 L of CO 2 gas at STP. v = 36.4 L V STP = 22.4 L/mol Find n

54. 9:47 PM54 Sample Questions 2.Calculate the volume of 0.479 mol of helium gas at STP. n = 0.479 mol V STP = 22.4 L/mol Find v v = n x V STP = 0.479 mol x 22.4 L/mol = 10.7 L

55. 9:47 PM Mole Calculations moles (n) mass (m) particles (N) x M ÷ M ÷ N A x N A volume (v) x V STP ÷ V STP 9:47 PM55

56. 9:47 PM56 Sample Questions 3.Calculate the volume of 86.4 g of CO 2 gas at STP. m = 86.4 g M = 44.01 g/mol V STP = 22.4 L/mol Find v 2 steps v = n x V STP

57. 9:47 PM57 Sample Questions 4.Calculate the mass of 168.9 L of propane gas, C 3 H 8, at STP. v = 168.9 L V STP = 22.4 L/mol M = 44.11 g/mol Find m 2 steps m = n x M

58. 9:47 PM Molar Volume p. 73 #’s 38 – 43 p. 74 #’s 1 – 4 p. 76 #’s 26, 27 58

59. 9:47 PM59

60. 9:47 PM Percent Composition (p. 79) The mass percent is the mass of an element in a compound expressed as a percent of the total mass of the compound. 60

61. 9:47 PM Percent Composition eg. 8.50 g of a compound was analyzed and found to contain 6.00 g of hydrogen and 2.50 g of carbon. Calculate the mass percent for each element. p. 82 #’s 1 - 4 61

62. 9:47 PM Percent Composition mass percent may be found using the formula & the molar mass of a compound. eg. Find the percentage composition for CH 4 62

63. 9:47 PM Percent Composition M = 12.01 g/mol + 4(1.01 g/mol) = 12.01 g/mol + 4.04 g/mol = 16.05 g/mol % H = (4.04/16.05) X 100% = 25.2 % % C = (12.01/16.05) X 100 % = 74.8 % p. 85 #’s 5 - 8 63

64. p. 85 #’s 5 - 8 p. 86 #’s 1, 3 – 6 p. 107 #’s 6 – 10 9:47 PM64

65. 9:47 PM Empirical Formulas An empirical formula gives the simplest ratio of elements in a compound. A molecular formula shows the actual number of atoms in one molecule of the compound. Ionic compounds are always written as empirical formulas 65

66. 9:47 PM Empirical Formulas CompoundMolecular Formula Empirical Formula butaneC 4 H 10 glucoseC 6 H 12 O 6 waterH2OH2O benzeneC6H6C6H6 C2H5C2H5 CH 2 O H2OH2O CH 66

67. 9:47 PM Empirical Formulas (p.87) 67

68. 9:47 PM Determining Empirical Formulas The empirical formula of a compound may be determined by using the % composition of a given compound. 68

69. 9:47 PM Empirical Formulas Method: assume you have 100.0 g of the compound (ie. change % to g) calculate the moles (n) for each element divide each n by the smallest n to get the ratio for the empirical formula 69

70. 9:47 PM Empirical Formulas eg. A compound was analyzed and found to contain 87.4% N and 12.6 % H by mass. Determine the empirical formula of the compound. p. 89 #’s 9 - 12 70

71. 9:47 PM Empirical Formulas  When finding the EF, the mole ratio may not be a whole number ratio. eg. A compound contains 84.73% N and 15.27 % H by mass. Determine the empirical formula of the compound. 71

72. 9:47 PM p. 90 72

73. 9:47 PM Empirical Formulas Determine the empirical formula of each compound. a) 89.91% C and 10.08 % H. b) 53.4% O and 46.6% P. p. 91 #’s 13 – 16 p. 94 #’s 2-4, 6, 7 Answers on p. 109 73

74. 9:47 PM MgO Lab 74

75. 9:47 PM Molecular Formulas The molecular formula of a compound is a multiple of the empirical formula. See p. 95 75

76. 9:47 PM Molecular Formulas To find the molecular formula we need the empirical formula and the molar mass of the compound eg. The empirical formula of hydrazine is NH 2. The molar mass of hydrazine is 32.06 g/mol. What is the molecular formula for hydrazine? 76

77. 9:47 PM Molecular Formulas eg. A hydrocarbon contains 85.6% carbon and 14.4% hydrogen by mass. The molar mass of the compound is 56.12 g/mol. Determine the molecular formula for the hydrocarbon? 77

78. 9:47 PM Molecular Formulas p. 97 #’s 17 – 20 p. 107, 108 #’s 11 - 14 78

79. 9:47 PM CHC analyzer (p. 99 – 101) 1.Describe the operation of a carbon hydrogen combustion analyzer. 2.22.0 g of carbon dioxide and 10.8 g of water is collected in a CHC analysis. Determine the empirical formula of the hydrocarbon. p. 101 #’s 21, 22 79

80. 9:47 PM Formula of a hydrate Heating a hy drate removes the water of hydration resulting in an anhydrous ionic compound. By comparing the mass of the hydrate and the anhydrous compound we can find % water in a hydrate AND the degree of hydration. 80

81. 9:47 PM Formula of a hydrate eg. Use the data below to determine the value of x in LiCl xH2O. mass of crucible = 26.35 g crucible + hydrate = 42.15 g crucible + anhydrous compound= 34.94 g 81

82. 9:47 PM Formula of a hydrate To determine the formula of a hydrate: - calculate the moles of water - calculate the moles of anhydrous compound - determine the simplest ratio 82

83. m water = 42.15 – 34.94 = 7.21 g H 2 O m LiCl = 34.94 – 26.35 = 8.59 g LiCl 9:47 PM83

84. 9:47 PM eg.Na 2 CO 3. xH 2 O crucible= 15.96 g crucible + hydrate = 22.19 g crucible + anhydrous compound = 19.67 g 84

85. 9:47 PM eg. CoCl 2.xH 2 O crucible= 151.96 g crucible + hydrate = 164.35 g crucible + anhydrous compound = 158.23 g p. 103; # 24Lab: pp. 104-105 85

86. 9:47 PM Formula of a hydrate mass of empty beaker mass of beaker & hydrate mass of beaker & anhydrous compound 86

87. 9:47 PM Review – Chp. 3 p. 86 #’s 1, 3 – 6 p. 94 #’s 1 – 7 p. 103 # 23 p. 106 # 7 pp. 107–109 #’s 5 – 23, 25 87

88. 9:47 PM Test p. 45 #’s 1 – 4 p. 46 #’s 1 – 6 p. 75 #’s 9 – 12 p. 51-53 #’s 5 – 15 p. 57 #’s 16 – 19 p. 59,60 #’s 20 – 27 p. 63,64 #’s 28 - 37 p. 54 #’s 5 – 8 p. 65 #’s 2, 4, 5 p. 75 #’s 13, 14 p. 76 #’s 15, 17–19, 21-23 p. 73 #’s 38 – 43 p. 74 #’s 1 – 4 p. 76 #’s 26, 27 88

89. 9:47 PM Test p. 82 #’s 1 – 4 p. 85 #’s 5 – 8 p. 89 #’s 9 – 12 p. 91 #’s 13 – 16 p. 97 #’s 17 - 20 p. 103 #’s 23 – 24 p. 86 #’s 1, 3 – 6 p. 94 #’s 1 - 7 p. 106 #’s 1 - 3, 6, 7 p. 107 – 109 #’s 5 – 23, 25 89

90. 9:47 PM Stoichiometry (Chp.4)  Stoichiometry is the determination of quantities needed for, or produced by, chemical reactions.  Ratios from balanced chemical equations are used to predict quantities. 90

91. 9:47 PM Stoichiometry – p. 111 Clubhouse sandwich recipe 91

92. 9:47 PM Clubhouse sandwich recipe Slices of Toast Slices of Turkey Strips of Bacon # of Sandwiches 12 27 66 100 Fill in the missing quantities: 92

93. 9:47 PM Mole Ratios A mole ratio is a mathematical expression that shows the relative amounts of two species involved in a chemical change. 93

94. 9:47 PM Mole Ratios A mole ratio Comes from a balanced chemical equation Shows the relative amounts of the reactants/products in moles Is the coefficient for the required species in the numerator and the coefficient for the given species in the denominator. 94

95. 9:47 PM Mole Ratios 95

96. 9:47 PM N 2(g) + 3 H 2 (g) → 2 NH 3 (g) 20 66 140 81 9:47 PM96

97. 9:47 PM C 3 H 8 + 5 O 2 → 3 CO 2 + 4 H 2 O 1. How many moles of CO 2 are produced when 31.5 mol of O 2 react? 97

98. 9:47 PM C 3 H 8 + 5 O 2 → 3 CO 2 + 4 H 2 O 2. How many moles of H 2 O are produced when 1.35 mol of O 2 react? 9:47 PM98

99. 9:47 PM C 3 H 8 + 5 O 2 → 3 CO 2 + 4 H 2 O 3. How many moles of O 2 are needed to produce 31.5 mol of CO 2 ? 9:47 PM99

100. 9:47 PM C 3 H 8 + 5 O 2 → 3 CO 2 + 4 H 2 O 4. How many moles of C 3 H 8 are needed to react with 0.369 mol of O 2 ? 9:47 PM100

101. 9:47 PM C 5 H 12 + O 2 → CO 2 + H 2 O 5. How many moles of CO 2 are produced when 6.35 mol of O 2 react? 9:47 PM101

102. 9:47 PM Al (s) + Br 2(l) → AlBr 3(s) 6. How many moles of Br 2 are needed to produce 0.315 mol of AlBr 3 ? p. 115 #’s 4 – 6 p. 117 #’s 9 & 10 9:47 PM102

103. 9:47 PM Mole to Mole Stoichiometry 1. How many moles of nitrogen gas are needed to produce 6.75 mol of NH 3 in a reaction with hydrogen gas? 2. How many moles of silver would be produced if 10.0 mol of silver nitrate reacts with copper metal? 3. How many moles of water are produced when 20.6 mol of CH 4 burns? 103

104. 9:47 PM Mole to Mole Stoichiometry 1. How many moles of copper would be produced if 20.5 mol of copper (II) oxide decomposes? 9:47 PM104

105. 9:47 PM Mole to Mole Stoichiometry 1. How many moles of copper would be produced if 20.5 mol of copper (II) oxide decomposes? 9:47 PM105

106. 9:47 PM Mass to Mole Stoichiometry 1. How many moles of water are produced when 20.6 g of CH 4 burns? 9:47 PM106

107. 9:47 PM Mass to Mole Stoichiometry 9:47 PM107

108. 9:47 PM Mole to Mass Stoichiometry 1. What mass of CaCl 2 is produced when 4.38 mol of Ca(NO 3 ) 2 reacts with NaCl? 108

109. 9:47 PM 109

110. 9:47 PM Mole to Mass Stoichiometry 1. What mass of CaCl 2 is produced when 4.38 mol of Ca(NO 3 ) 2 reacts with NaCl? 1. How many moles of copper would be produced if 20.5 g of copper (II) oxide decomposes? 2. How many moles of water are produced when 5.45 gl of C 3 H 8 burns? 9:47 PM110

111. 9:47 PM Mass to Mass Stoichiometry eg. Calculate the mass of HCl needed to react with 3.56 g of Fe to produce FeCl 2. 9:47 PM111

112. 9:47 PM Ca(NO3)2 + 2 NaCl → CaCl2 + 2 NaNO3 112

113. 9:47 PM 2 CuO → 2 Cu + O2 113

114. 9:47 PM C3H8 + 5 O2 → 3 CO2 + 4 H2O 114

115. 9:47 PM Stoichiometry (Chp.4) Four step stoichiometry: 1. Write a balanced chemical equation 2. Calculate moles given from mass 3. Mole ratio – find moles required 4. Calculate required quantity (mass) 115

116. 9:47 PM Stoichiometry (Chp.4) eg. What mass of CO 2 gas is produced when 45.9 g of CH 4 burns ? Step #1 CH 4 + 2 O 2 → CO 2 + 2 H 2 O 45.9 g ? g 116

117. 9:47 PM117

118. 9:47 PM118

119. 9:47 PM eg. How many moles of HCl needed to react with 3.56 g of Fe to produce FeCl 2. 119

120. 9:47 PM Mole Calculations (p. 121 #13) 3.56 g ? g = 0.06374 mol Fe Fe + 2 HCl → FeCl 2 + H 2 Step #2 Step #3 = 0.12748 mol HCl 120

121. 9:47 PM Mole Calculations p. 122 #15 Given 32.0 g of sulfur (M = 256.56 g/mol) Find mass of ZnS #2n = 0.1247 mol S 8 #3n = 0.9976 mol ZnS (M = 97.45 g/mol) #4m = 97.2 g ZnS 121

122. 9:47 PM Mole Calculations p. 123 #18 Given 33.5 g of H 3 PO 4 (M = 98.00 g/mol) Find mass of MgO #2n = 0.3418 mol H 3 PO 4 #3n = 0.5128 mol MgO (M = 40.31 g/mol) #4m = 20.7 g MgO 122

123. 9:47 PM Mole Calculations p. 123 #17 Given 25.0 g of Al 4 C 3 (M = 143.95 g/mol) Find volume of CH 4 #2n = 0.174 mol Al 4 C 3 #3n = 0.522 mol CH 4 (MV = 22.4 L/mol) #4m = 11.7 L CH 4 9:47 PM123

124. How many moles of aluminum chloride can be produced from the reaction of chlorine and 10.8 mol of aluminum ? Cl 2(g) + Al (s) → AlCl 3(s) 9:47 PM124

125. How many moles of magnesium are needed to react with 27 g of iodine to form magnesium iodide? 9:47 PM125

126. How many grams of nitrogen are needed to react with 14.0 mol of oxygen to produce nitrogen dioxide ? N 2(g) + O 2(g) → NO 2(g) 9:47 PM126

127. 9:47 PM Mole Calculations (p. 121 #14) 2.34 g? L #2n = 0.05086 mol NO 2 #3n = 0.01272 mol O 2 (MV = 22.4 L/mol) #4n = (0.01272)(22.4) = 0.285 L O 2 127

128. 9:47 PM Limiting Reactant (p. 128) 1. 10.0 g of Li requires _______ of Br 2 2. 15.0 g of Br 2 requires _______ of Li 3. 10.0 g of Li produces _______ of LiBr 4. 15.0 g of Br 2 produces ______ of LiBr This problem has _____ g of excess _____ and will produce ______ g of LiBr 9:47 PM128

129. 9:47 PM Limiting Reactant (p. 128) 10.0 g of Li reacts with 15.0 g of Br 2. Calculate the mass of LiBr produced. 129

130. 9:47 PM Limiting Reactant (p. 128) The Limiting Reactant (LR) OR Limiting Reagent (LR) is the substance that is completely used in a chemical reaction. The Excess Reactant is the reactant that is left over after a reaction is complete. 130

131. 9:47 PM Limiting Reactant (p. 128) eg. 2.00 g of NaI reacts with 2.00 g of Pb(NO 3 ) 2. Determine the LR and calculate the amount of PbI 2 produced.  write a balanced equation  find n for each reactant (Step #2)  find moles produced by each reactant (Step #3) 131

132. Pb(NO 3 ) 2 + 2 NaI → 2 NaNO 3 + PbI 2 2.00 g n Pb(NO 3 ) 2 = 2.00 g 331.21 g/mol = 0.006038 mol n NaI = 2.00 g 149.89 g/mol = 0.013343 mol 9:47 PM132

133. n PbI2 = 0.006038 mol Pb(NO 3 ) 2 x 1 mol PbI 2 1 mol Pb(NO 3 ) 2 = 0.006038 mol PbI 2 n PbI2 = 0.013343 mol NaI x 1 mol PbI 2 2 mol NaI = 0.006672 mol PbI 2 m PbI2 = 0.006038 mol x 460.99 g/mol = 2.78 g PbI 2 9:47 PM133

134. What mass of calcium carbonate will be produced when 20.0 g of calcium phosphate reacts with 15.0 g of sodium carbonate? (14.2 g) 9:47 PM134

135. What mass of barium hydroxide will be produced when 10.0 g of barium nitrate reacts with 30.0 g of sodium hydroxide? (6.56 g) Ba(NO 3 ) 2 + 2 NaOH → 2 NaNO 3 + Ba(OH) 2 9:47 PM135 n Ba(NO 3 ) 2 = 10.0 g 261.35 g/mol = 0.03826 mol n NaOH = 30.0 g 40.00 g/mol = 0.750 mol n Ba(OH) 2 = 0.03826 mol Ba(NO 3 ) 2 x 1 mol Ba(OH) 2 1 mol Ba(NO 3 ) 2 = 0.03826 mol Ba(OH) 2 n Ba(OH) 2 = 0.750 mol NaOH x 1 mol Ba(OH) 2 2 mol NaOH = 0.373 mol Ba(OH) 2

136. What volume of hydrogen gas at STP will be produced when 10.0 g of zinc metal reacts with 20.0 g of hydrogen chloride? Zn + 2 HCl → H 2 + ZnCl 2 9:47 PM136

137. Ba(NO 3 ) 2 + 2 NaOH → 2 NaNO 3 +Ba(OH) 2 9:47 PM137

138. 9:47 PM138

139. 9:47 PM Law of Conservation of Mass ( p. 118) In a chemical reaction, the total mass of reactants always equals the total mass of products. eg.2 Na 3 N → 6 Na + N 2 When 500.00 g of Na 3 N decomposes 323.20 g of N 2 is produced. How much Na is produced in this decomposition? 139

140. 9:47 PM Law of Conservation of Mass ( p. 118) eg. To produce 90.1 g of water, what mass of hydrogen gas is needed to react with 80.0 g of oxygen? eg.If 3.55 g of chlorine reacts with exactly 2.29 g of sodium, what mass of NaCl will be produced? 140

141. 9:47 PM The theoretical yield is the amount of product that we calculate using stoichiometry The actual yield is the amount of product obtained from a chemical reaction Percent yield (p. 137) 141

142. 9:47 PM Percent yield (p. 137) 142

143. DEMO: silver nitrate + copper Equation: Mass AgNO 3 = Mass Cu = 9:47 PM143

144. DEMO: silver nitrate + copper Mass of filter paper and precipitate = Mass of empty filter paper = Mass of precipitate = 9:47 PM144