1. 1:08 AM Unit 1: Stoichiometry Chemistry 2202 1

2. 1:08 AM Stoichiometry Stoichiometry deals with quantities used in OR produced by a chemical reaction 2

3. 1:08 AM 3 Parts Mole Calculations (Chp. 2 & 3) Stoichiometry and Chemical Equations (Chp. 4) Solution Stoichiometry (Chp. 6) 3

4. 1:08 AM PART 1 - Mole Calculations Isotopes and Atomic Mass (pp. 43 - 46) Avogadro’s number (pp. 47 – 49) Mole Conversions (pp. 50 - 74) M, MV, N A, n, m, v, N 4

5. 1:08 AM 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. 1:08 AM 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. 1:08 AM 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. 1:08 AM 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 isotope - atoms which have the same number of protons and electrons but different numbers of neutrons 8

9. 1:08 AM Isotopes and Atomic Mass eg. 9

10. 1:08 AM Isotopes and Atomic Mass 10

11. 1:08 AM Isotopes and Atomic Mass not all isotopes are created equal 79 % 10 % 11 % 11

12. 1:08 AM Isotopes and Atomic Mass 12

13. 1:08 AM 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. 1:08 AM 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 – 4 p. 46#’s 1 – 6 p. 75 #’s 9 - 12 14

15. 1:08 AM 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. 1:08 AM 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. 1:08 AM x + y = 1 78.92x + 80.92y = 79.90 17

18. 1:08 AM Avogadro’s Number (p. 47) p. 48 18

19. 1:08 AM Avogadro’s Number The MOLE is a number used by chemists to count atoms The MOLE is the number of atoms contained in exactly 12 g of carbon-12. In honor of Amedeo Avogadro, the number of particles in 1 mol has been called Avogadro’s number. 19

20. 1:08 AM How big is Avogadro's number? An Avogadro's number of soft drink cans would cover the surface of the earth to a depth of over 200 miles. Avogadro's number of unpopped popcorn kernels spread across the USA, would cover the entire country to a depth of over 9 miles. 20

21. 1:08 AM If we were able to count atoms at the rate of 10 million per second, it would take about 2 billion years to count the atoms in one mole. How big is Avogadro's number? 21

22. 1:08 AM Avogadro’s Number 1 mole = 6.02214199 x 10 23 particles 1 mol = 6.022 x 10 23 particles N A = 6.022 x 10 23 particles/mol 22

23. 1:08 AM Avogadro’s Number 23

24. 1:08 AM Avogadro’s Number Number of molesNumber of atoms 5 mol 0.01 mol 4.65 x 10 24 atoms 8.01 x 10 21 atoms 7.72 mol 0.0133 mol 6.022 x 10 21 atoms 3.011 x 10 24 atoms 24

25. 1:08 AM 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 25

26. 1:08 AM 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 26

27. 1:08 AM 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? 27

28. 1:08 AM Avogadro’s Number 1. How many molecules of glucose are in 0.435 mol of C 6 H 12 O 6 ? How many carbon atoms? 2. Calculate the number of moles in a sample of glucose that has 3.56 x 10 22 hydrogen atoms. 28

29. 1:08 AM Avogadro’s Number pp. 51 – 53: #’s 5 – 15 p. 54: #’s 4 - 8 29

30. 1:08 AM Molar Mass The mass of one mole of a substance is called the molar mass of the substance eg. 1 mole of Pb has a mass of 207.19 g 1 mole of Ag has a mass of 107.87 g 30

31. 1:08 AM Molar Mass The symbol for molar mass is M and the unit is g/mol eg. M Pb = 207.19 g/mol M Ag = 107.87 g/mol 31

32. 1:08 AM 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 32

33. 1:08 AM 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 33

34. 1:08 AM 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 34

35. 1:08 AM 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 35

36. 1:08 AM 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 36

37. 1:08 AM Molar Mass Calculations N = n x N A mass molar mass m = n x M Avogadro’s # 37

38. 1:08 AM Molar Mass Calculations N = n x N A m = n x M 38

39. 1:08 AM 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 39

40. 1:08 AM 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 40

41. 1:08 AM Molar Mass Calculations Practice:p. 59 #’s 20 - 23 p. 60 #’s 24 - 27 41

42. 1:08 AM The Mole #4 answers 1.a) 0.038 mol 2.a) 17.4 g3.a) 5631 g b) 3.75 mol b) 1560. g b) 3.73 g c) 0.276 mol c) 528 g c) 10.02 g d) 23 mol d) 97513 g d) 1662.9 g e) 0.0575 mol e) 9328 g e) 981.2 g 4.a) 1.82 molc) 0.02133 mole) 0.0000573 mol b) 13 mold) 4.3423 mol 42

43. 1:08 AM 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 Particle–Mole-Mass Conversions 43

44. 1:08 AM = 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 Particle–Mole-Mass Conversions 44

45. 1:08 AM 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 Particle–Mole-Mass Conversions 45

46. 1:08 AM = 77,881 mol m = n x M = 77881 mol x 118.69 g/mol = 9.24 x 10 6 g Particle–Mole-Mass Conversions 46

47. 1:08 AM 1.Calculate the mass of 4.80 x 10 24 water molecules. 2.How many grams are in 2.53 x 10 26 CH 4 molecules? 3.How many atoms are in 68.0 g of Sn? 4.Calculate the number of molecules in 105 g of C 6 H 12 O 6. Particle–Mole-Mass Conversions 47

48. Particles (N)Moles (n)Mass (m) 4.80 x 10 24 H 2 O molecules 2.53 x 10 26 CH 4 molecules 68.0 g of Sn 105 g of C 6 H 12 O 6 Particle–Mole-Mass Conversions 1:08 AM48

49. 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 1:08 AM49

50. particles (N) Moles (n) Mass (m) 4.18 x 10 22 Al atoms 2.45 g C 8 H 18 0.145 mol S 2 F 4 Particle–Mole-Mass Conversions 1:08 AM50

51. 1:08 AM Practice:p. 63 #’s 28 - 33 p. 64 #’s 34 – 37 p. 76 # 15 moles (n) mass (m) particles (N) x M ÷ M ÷ N A x N A Particle–Mole-Mass Conversions 51

52. 1:08 AM 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 Particle–Mole-Mass Conversions 1:08 AM52

53. 1:08 AM = 0.02653 mol glucose N = n x N A = 0.02653 X 6.022 x 10 23 = 8.99 x 10 23 molecules Particle–Mole-Mass Conversions 1:08 AM53

54. 1:08 AM Molar Mass Calculations Practice:p. 63 #’s 28 - 33 p. 64 #’s 34 – 37 p. 76 # 15 54

55. 1:08 AM 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 55

56. 1:08 AM 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) 56

57. 1:08 AM 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. 57

58. 1:08 AM 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 58

59. 1:08 AM Molar Mass Calculations N = n x N A m = n x M v = n x V STP given volume in Litres 59

60. 1:08 AM Molar Mass Calculations moles (n) mass (m) particles (N) x M ÷ M ÷ N A x N A volume (v) x V STP ÷ V STP 1:08 AM60

61. 1:08 AM Molar Volume p. 73 #’s 38 – 43 p. 74 #’s 1 – 4 p. 76 #’s 26, 27 61

62. 1:08 AM62

63. 1:08 AM Percent Composition (p. 79) The mass percent of a compound is the mass of each element in a compound expressed as a percent of the total mass of the compound. 63

64. 1:08 AM 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 64

65. 1:08 AM Percent Composition mass percent may be found from the molar mass of a compound. eg. Find the percentage composition for CH 4 65

66. 1:08 AM Percent Composition M = 12.01 g/mol + 4(1.01 g/mol) = 12.01 g/mol + 4.04 g/mol = 16.05 g/mol p. 85 #’s 5 - 8 66

67. p. 85 #’s 5 - 8 p. 86 #’s 1, 3 – 6 p. 107 #’s 6 – 10 1:08 AM67

68. 1:08 AM Empirical Formulas An empirical formula gives the simplest ratio of elements in a compound. A molecular formula shows the actual number of atoms in a molecule of a compound. Ionic compounds are always written as empirical formulas 68

69. 1:08 AM Empirical Formulas CompoundMolecular Formula Empirical Formula butaneC 4 H 10 glucoseC 6 H 12 O 6 waterH2OH2O benzeneC6H6C6H6 C2H5C2H5 CH 2 O H2OH2O CH 69

70. 1:08 AM Empirical Formulas (p.87) 70

71. 1:08 AM Empirical Formulas The empirical formula of a compound may be determined by using the % composition of a given compound. 71

72. 1:08 AM 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 72

73. 1:08 AM 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 73

74. 1:08 AM 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. 74

75. 1:08 AM p. 90 75

76. 1:08 AM Empirical Formulas eg. A compound contains 89.91% C and 10.08 % H by mass. Determine the empirical formula of the compound. p. 91 #’s 13 – 16 p. 94 #’s 2-4, 6, 7 Answers on p. 109 76

77. 1:08 AM MgO Lab 77

78. 1:08 AM Molecular Formulas The molecular formula of a compound is a multiple of the empirical formula. See p. 95 78

79. 1:08 AM 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? 79

80. 1:08 AM Molecular Formulas p. 97 #’s 17 – 20 p. 107, 108 #’s 11 - 14 80

81. 1:08 AM 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 81

82. 1:08 AM 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. 1:08 AM 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 83

84. m water = 42.15 – 34.94 = 7.21 g H 2 O m LiCl = 34.94 – 26.35 = 8.59 g LiCl 1:08 AM84

85. 1:08 AM eg.Na 2 CO 3 ● xH 2 O crucible= 15.96 g crucible + hydrate = 22.19 g crucible + anhydrous compound = 19.67 g 85

86. 1:08 AM 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 86

87. 1:08 AM Formula of a hydrate mass of empty beaker mass of beaker & hydrate mass of beaker & anhydrous compound 87

88. 1:08 AM Review – Chp. 3 p. 86 #’s 1, 3 – 6 p. 94 #’s 1 – 7 p. 103 # 23 pp. 107–109 #’s 5 – 22 88

89. 1:08 AM 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 89

90. 1:08 AM 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 90

91. 1:08 AM91 Test Monday – Dec. 7 1.Formula of a hydrate - lab activity - p. 103 # 24 2.% composition - given mass p. 82 #’s 1 – 4 - given formula p. 85 #’s 5 - 8 p. 86 #’s 3 & 4 p. 103 # 23 p. 107 # 5 3.EF and MF - What is an empirical formula - Finding EF from % composition pp. 89, 91 #’s 9 - 16 (watch for.5 or.333) - Finding MF from EF and molar mass p. 97 #’s 17 – 20 p. 108 #13 4.Mole calculations – Chp. 2

92. 1:08 AM 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. 92

93. 1:08 AM Stoichiometry – p. 111 Clubhouse sandwich recipe 93

94. 1:08 AM Clubhouse sandwich recipe Slices of Toast Slices of Turkey Strips of Bacon # of Sandwiches 12 27 66 100 Fill in the missing quantities: 94

95. 1:08 AM Mole Ratios A mole ratio is a mathematical expression that shows the relative amounts of two species involved in a chemical change. 95

96. 1:08 AM Mole Ratios A mole ratio Comes from a balanced chemical equation Shows the relative amounts of the reactants/products in moles Looks like 96

97. 1:08 AM N 2(g) + 3 H 2 (g) → 2 NH 3 (g) 20 66 140 81 1:08 AM97

98. 1:08 AM C 3 H 8 + 5 O 2 → 3 CO 2 + 4 H 2 O How many moles of CO 2 are produced when 31.5 mol of O 2 react? 98

99. 1:08 AM C 3 H 8 + 5 O 2 → 3 CO 2 + 4 H 2 O How many moles of H 2 O are produced when 1.35 mol of O 2 react? 1:08 AM99

100. 1:08 AM C 3 H 8 + 5 O 2 → 3 CO 2 + 4 H 2 O How many moles of C 3 H 8 are needed to react with 0.369 mol of O 2 ? 1:08 AM100

101. 1:08 AM C 5 H 12 + O 2 → CO 2 + H 2 O How many moles of CO 2 are produced when 6.35 mol of O 2 react? 1:08 AM101

102. 1:08 AM Al (s) + Br 2(l) → AlBr 3(s) How many moles of Br 2 are needed to produce 0.315 mol of AlBr 3 ? p. 115 #’s 4 – 7 p. 117 #’s 8 - 10 1:08 AM102

103. 1:08 AM Mole to Mole Stoichiometry 1. How many moles of copper would be produced if 20.5 mol of copper (II) oxide decomposes? 1:08 AM103

104. 1:08 AM Mass to Mole Stoichiometry 1. How many moles of water are produced when 20.6 g of CH 4 burns? 1:08 AM104

105. 1:08 AM Mass to Mole Stoichiometry 2. How many moles of nitrogen gas are needed to produce 6.75 g of NH 3 in a reaction with hydrogen gas? 105

106. 1:08 AM Mass to Mole Stoichiometry 3. How many moles of silver would be produced if 10.0 g of AgNO 3 reacts with copper metal? 106

107. 1:08 AM Mass to Mole Stoichiometry 4. How many moles of CO 2 are produced when 10.6 g of C 3 H 8 burns? 107

108. 1:08 AM 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. 1:08 AM Mole to Mass Stoichiometry 2. Calculate the mass of copper produced if 20.5 mol of CuO decomposes? 1:08 AM109

110. 1:08 AM 1. What mass of water would be produced when 5.45 g of C 3 H 8 burns? 1:08 AM110 Mass to Mass Stoichiometry

111. 1:08 AM Four step stoichiometry 1. Write a balanced chemical equation 2. Calculate moles given 3. Mole ratio – find moles required 4. Calculate required quantity m = n x Mv = n x V stp N = n x N A 111

112. 1:08 AM Mass to Mass Stoichiometry 2. Calculate the mass of HCl needed to react with 3.56 g of Fe to produce FeCl 2. 1:08 AM112

113. 1:08 AM113

114. 1:08 AM Stoichiometry (Chp.4) eg. What mass of CO 2 gas is produced when 45.9 g of CH 4 burns ? Step #1CH 4 + 2 O 2 → CO 2 + 2 H 2 O 45.9 g ? g 114

115. 1:08 AM115

116. 1:08 AM eg. What mass of HCl is needed to react with 3.56 g of Fe to produce FeCl 2. 116

117. 1:08 AM 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 117

118. 1:08 AM 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 118

119. 1:08 AM 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 119

120. 1:08 AM 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 1:08 AM120

121. 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) 1:08 AM121

122. How many moles of magnesium are needed to react with 27 g of iodine to form magnesium iodide? 1:08 AM122

123. 1:08 AM 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 2 123

124. 1:08 AM Limiting Reactant (p. 128) 10.0 g of Li reacts with 15.0 g of Br 2. Calculate the mass of LiBr produced. 124

125. Limiting Reactant (p. 128) 1:08 AM125 2 LiBr 2 2 LiBr 10.0 15.0

126. 1:08 AM 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. 126

127. 1:08 AM 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) 127

128. Pb(NO 3 ) 2 + 2 NaI → 2 NaNO 3 + PbI 2 n Pb(NO)3 = 2.00 g 331.21 g/mol = 0.006038 mol n NaI = 2.00 g 149.89 g/mol = 0.013343 mol 1:08 AM128

129. 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 1:08 AM129

130. 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) 3 Na 2 CO 3 + Ca 3 (PO 4 ) 2 → 3 CaCO 3 + 2 Na 3 PO 4 1:08 AM130

131. 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 + NaOH → Ba(OH) 2 + NaNO 3 = 0.03826 mol Ba(NO 3 ) 2 = 0.750 mol NaOH 1:08 AM131

132. Using Ba(NO 3 ) 2 = 0.03826 mol Ba(OH) 2 Using NaOH = 0.375 mol Ba(OH) 2 1:08 AM132

133. 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 1:08 AM133

134. 1:08 AM134

135. 1:08 AM135

136. 1:08 AM 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? 136

137. 1:08 AM 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? 137

138. 1:08 AM 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) 138

139. 1:08 AM Percent yield (p. 137) 139 p. 139 #’s 31, 32, & 33

140. DEMO: silver nitrate + copper Equation: Mass AgNO 3 = Mass Cu = 1:08 AM140

141. DEMO: silver nitrate + copper Mass of filter paper and precipitate = Mass of empty filter paper = Mass of precipitate = 1:08 AM141