Chapter 9

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HEINS09-095-117v4.qxd 12/30/06 1:58 PM Page 112 - <strong>Chapter</strong> 9 - 44. Composition of a mixture of and KCl. In the mixture only KCl reacts with AgNO 3 . KCl(aq) + AgNO 3 (aq) ¡ AgCl(s) + KNO 3 (aq) 74.55 g KCl 14.33 g AgCl2a b = 2.25 g KCl in the mixture 143.7 g AgCl KNO 3 - 112 - 10.00 g mixture - 2.25 g KCl = 7.75 g KNO 3 2.25 g KCl a b11002 = 22.5% KCl 10.00 g mixture a 7.75 g KNO 3 10.00 g mixture b11002 = 77.5% KNO 3 45. The balanced equation is Zn + 2 HCl ¡ ZnCl 2 + H 2 180.0 g Zn - 35 g Zn = 145 g Zn reacted with HCl (a) (b) (c) 1145 g Zn2a 1 mol 65.39 g ba1 mol H 2 1 mol Zn b a 2.016 g b = 4.47 g H mol 2 produced 1145 g Zn2a 1 mol mol HCl ba2 65.39 g 1 mol Zn ba36.46 g b = 162 g HCl reacted mol 1180.0 g Zn2a 1 mol mol HCl ba2 65.39 g 1 mol Zn ba36.46 g b = 201 g HCl reacts mol 201 g - 162 g = 39 g more HCl needed to react wih the 180.0 g Zn 46. Fe(s) + CuSO 4 (aq) 2.0 mol 3.0 mol ¡ Cu(s) FeSO 4 (aq) (a) 2.0 mol Fe react with 2.0 mol CuSO 4 to yield 2.0 mol Cu and 2.0 mol FeSO 4 . 1.0 mol CuSO 4 is unreacted. At the completion of the reaction, there will be 2.0 mol Cu, 2.0 mol FeSO 4 , and 1.0 mol CuSO 4 . (b) Determine which reactant is limiting and then calculate the g FeSO 4 produced from that reactant. 120.0 g Fe2a 1 mol mol Cu ba1 55.85 g 1 mol Fe ba63.55 g b = 22.8 g Cu mol 140.0 g CuSO 4 2a 1 mol 1 mol Cu ba ba 63.55 g b = 15.9 g Cu 159.6 g 1 mol CuSO 4 mol Since CuSO 4 produces less Cu, it is the limiting reactant. Determine the mass of FeSO 4 produced from 40.0 g CuSO 4 . +
HEINS09-095-117v4.qxd 12/30/06 1:58 PM Page 113 - <strong>Chapter</strong> 9 - 140.0 g CuSO 4 2a 1 mol 159.6 g b ¢ 1 mol FeSO 4 ≤ a 151.9 g b = 38.1 g FeSO 1 mol CuSO 4 mol 4 produced Calculate the mass of unreacted Fe. 140.0 g CuSO 4 2a 1 mol 159.6 g b ¢ 1 mol Fe ≤ a 55.85 g b = 14.0 g Fe will react 1 mol CuSO 4 mol Unreacted Fe = 20.0 g - 14.0 g = 6.0 g. Therefore, at the completion of the reaction, 15.9 g Cu, 38.1 g FeSO 4 , 6.0 g Fe, and no CuSO 4 remain. 47. Limiting reactant calculation CO(g) + 2 H 2 (g) ¡ CH 3 OH(l) Reaction between 40.0 g CO and 10.0 g H 2 : determine the limiting reactant by calculating the amount of CH 3 OH that would be formed from each reactant. 140.0 g CO2a 1 mol 28.01 g b ¢ 1 mol CH 3OH 1 mol CO ≤ a 32.04 g b = 45.8 g CH mol 3 OH 110.0 g H 2 2a 1 mol 2.016 g b ¢ 1 mol CH 3OH ≤ a 32.04 g b = 79.5 g CH 2 mol H 2 mol 3 OH CO is limiting; H 2 is in excess; 45.8 g CH 3 OH will be produced. Calculate the mass of unreacted H 2 : 140.0 g CO2a 1 mol 28.01 g b ¢ 2 mol H 2 1 mol CO ≤ a 2.016 g b = 5.76 g H mol 2 react 10.0 g H 2 - 5.76 g H 2 = 4.2 g H 2 remain unreacted 48. The balanced equation is C 6 H 12 O 6 ¡ 2 C 2 H 5 OH + 2 CO 2 (a) First calculate the theoretical yield. 1750 g C 6 H 12 O 6 2a 1 mol 180.2 g b ¢ 2 mol C 2H 5 OH ≤ a 46.07 g b 1 mol C 6 H 12 O 6 mol Then take 84.6% of the theoretical yield to obtain the actual yield. actual yield = 1theoretical yield2184.62 100 = 3.2 * 10 2 g C 2 H 5 OH - 113 - = 3.8 * 10 2 g C 2 H 5 OH (theoretical yield) = 13.8 * 102 g C 2 H 5 OH2184.62 100
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Chapter 9

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