Chapter 9

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HEINS09-095-117v4.qxd 12/30/06 1:58 PM Page 106 - <strong>Chapter</strong> 9 - (b) Reaction between 12.0 g of C 3 H 6 and 25.0 g of O 2 . Convert each amount to moles of H 2 O 112.0 g C 3 H 6 2¢ 1 mol 42.08 g ≤¢6 mol H 2O 2 mol C 3 H 6 ≤ = 0.856 mol H 2 O 125.0 g O 2 2¢ 1 mol 32.00 g ≤¢6 mol H 2O 9 mol O 2 ≤ = 0.521 mol H 2 O O 2 is the limiting reactant. The yield is 0.521 mol H 2 O. (c) Reaction between 5.0 mol of C 3 H 6 and 15.0 mol of O 2 . Convert each to moles of CO 2 (5.0 mol C 3 H 6 )¢ 6 mol CO 2 2 mol C 3 H 6 ≤ = 15 mol CO 2 (15.0 mol O 2 )¢ 6 mol CO 2 ≤ = 10 mol CO 9 mol O 2 2 Since is the limiting reactant. C 3 H 6 will be left unreacted. O 2 27. 28. X 8 + 12 O 2 ¡ 8 XO 3 The conversion is: g O 2 ¡ mol O 2 ¡ mol X 8 (120.0 g O 2 2a 1 mol 32.00 g b ¢ 1 mol X 8 12 mol O 2 ≤ = 0.3125 mol X 8 80.0 g X 8 = 0.3125 mol X 8 80.0 g 0.3125 mol = 256 g>mol X 8 g 256 mol g molar mass X = = 32.0 8 mol Using the periodic table we find that the element with 32.0 g>mol is sulfur. X + 2 HCl ¡ XCl 2 + H 2 The conversion is: g H 2 ¡ mol H 2 ¡ mol X 12.42 g H 2 2a 1 mol 2.016 g b ¢ 1 mol X 1 mol H 2 ≤ = 1.20 mol X 78.5 g X = 1.20 molX 78.5 g = 65.4 g>mol 1.20 mol Using the periodic table we find that the element with atomic mass 65.4 is zinc. - 106 -
HEINS09-095-117v4.qxd 12/30/06 1:58 PM Page 107 - <strong>Chapter</strong> 9 - 29. Limiting reactant calculation and percentage yield 2 Al + 3 Br 2 ¡ 2 AlBr 3 Reaction between 25.0 g Al and 100. g Br 2 Calculate the grams of AlBr 3 from each reactant. 125.0 g Al2a 1 mol 26.98 g b ¢ 2 mol AlBr 3 ≤ a 266.7 g b = 247 g AlBr 2 mol Al mol 3 (100. g Br 2 )a 1 mol 159.8 g b ¢ 2 mol AlBr 3 ≤ a 266.7 g b = 111 g AlBr 3 mol Br 2 mol 3 Br 2 is limiting; 111 g AlBr 3 is the theoretical yield of product. actual yield 64.2 g Percent yield = a b11002 = a b11002 = 57.8% theoretical yield 111 g 30. Percent yield calculation Fe(s) + CuSO 4 (aq) ¡ Cu(s) + FeSO 4 (aq) 1400. g CuSO 4 2a 1 mol 1 mol Cu b ¢ ≤ a 63.55 g b = 159 g Cu (theoretical yield) 159.6 g 1 mol CuSO 4 mol actual yield 151 g % yield = a b11002 = a b11002 = 95.0% yield of Cu theoretical yield 159 g 31. The balanced equation is 3 C + 2 SO 2 ¡ CS 2 + 2 CO 2 Calculate the g C needed to produce 950 g CS 2 taking into account that the yield of CS 2 is 86.0%. First calculate the theoretical yield of CS 2 . 950 g CS 2 0.860 = 1.1 * 10 3 g CS 2 1theoretical yield2 Now calculate the grams of coke needed to produce 1.1 * 10 3 g CS 2 . 11.1 * 10 3 g CS 2 2a 1 mol 76.15 g b ¢ 3 mol C ≤ a 12.01 g b = 5.2 * 10 2 g C 1 mol CS 2 mol 32. The balanced equation is CaC 2 + 2 H 2 O ¡ C 2 H 2 + Ca1OH2 2 First calculate the grams of pure CaC 2 in the sample from the amount of C 2 H 2 produced. 10.540 mol C 2 H 2 2¢ 1 mol CaC 2 ≤ a 64.10 g CaC 2 b = 34.6 g of pure CaC 1 mol C 2 H 2 mol CaC 2 in the sample 2 - 107 -
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Chapter 9

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