Chapter 9

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HEINS09-095-117v4.qxd 12/30/06 1:58 PM Page 108 - <strong>Chapter</strong> 9 - Now calculate the percent in the impure sample. 33. No. There are not enough screwdrivers, wrenches or pliers. 2400 screwdrivers, 3600 wrenches and 1200 pliers are needed for 600 tool sets. 34. A subscript is used to indicate the number of atoms in a formula. It cannot be changed without changing the identity of the substance. Coefficients are used only to balance atoms in chemical equations. They may be changed as needed to achieve a balanced equation. 35. Consider the reaction A ¡ 2B and assume that you have 1 gram of A. This does not guarantee that you will produce 1 gram of B because A and B have different molar masses. One gram of A does not contain the same number of molecules as 1 gram of B. However, 1 mole of A does have the same number of molecules as one mole of B. (Remember, 1 mole = 6.022 * 10 23 molecules always.) If you determine the number of moles in one gram of A and multiply by 2 to get the number of moles of B then from that you can determine the grams of B using its molar mass. Equations are written in terms of moles not grams. 36. (a) CaC 2 - 108 - ¢ 34.6 g CaC 2 44.5 g sample ≤11002 = 77.8% CaC 2 in the impure sample 4 KO 2 + 2 H 2 O + 4 CO 2 ¡ 4 KHCO 3 + 3 O 2 Á ¢ 0.85 g CO 2 ≤ a 1 mol min 44.01 g b ¢ 4 mol KO 2 ≤ = 0.019 mol KO 2 4 mol CO 2 min ¢ 0.019 mol KO 2 b110.0 min2 = 0.19 mol KO min 2 (b) 37. (a) The conversion is: g CO 2 min ¡ mol CO 2 min ¡ mol O 2 min ¢ 0.85 g CO 2 ≤ a 1 mol min 44.01 g b ¢ 3 mol O 2 ≤ a 32.00 g 4 mol CO 2 mol 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 = 380 g C 1 mol C 6 H 12 O 6 mol 2 H 5 OH ba ¡ g O 2 min ¡ g O 2 hr 60.0 min b = 28 g O 2 1.0 hr hr 1750 g C 6 H 12 O 6 2a 1 mol 180.2 g b ¢ 2 mol CO 2 ≤ a 44.01 g b = 370 g CO 1 mol C 6 H 12 O 6 mol 2 Alternate Solution: 750 g C 2 H 6 O 6 - 380 g C 2 H 5 OH = 370 g CO 2 by the conservation of mass method. (b) 1380 g C 2 H 5 OH2a 1 mL 0.79 g b = 480 mL C 2H 5 OH
HEINS09-095-117v4.qxd 12/30/06 1:58 PM Page 109 - <strong>Chapter</strong> 9 - 38. 2 CH 3 OH + 3 O 2 ¡ 2 CO 2 + 4 H 2 O The conversion is: mL CH 3 OH ¡ g CH 3 OH ¡ mol CH 3 OH ¡ mol O 2 ¡ g O 2 160.0 mL CH 3 OH2a 0.72 g mL b ¢ 1 mol 32.04 g ≤¢ 3 mol O 2 2 mol CH 3 OH ≤ a 32.00 g b = 65 g O mol 2 39. The balanced equation is The conversion is: 7 H 2 O 2 + N 2 H 4 ¡ 2 HNO 3 + 8 H 2 O (a) 175 kg N 2 H 4 2a 1000 g 1 kg b ¢ 1 mol 32.05 g ≤¢2 mol HNO 3 ≤ a 63.02 g b = 2.9 * 10 5 g HNO 1 mol N 2 H 4 mol 3 (b) (c) 1250 L H 2 O 2 2a 1000 mL 1 L b ¢ 1.41 g 1 mol ≤¢ 1 mL 34.02 g ≤ a 8 mol H 2O ba 18.02 g 7 mol H 2 O 2 mol 1725 g H 2 O 2 2a 1 mol 34.02 g b ¢ 1 mol N 2H 4 ≤ a 32.05 g b = 97.6 g N 7 mol H 2 O 2 mol 2 H 4 b = 2.1 * 10 5 g H 2 O (d) Reaction between 750 g of N 2 H 2 and 125 g of H 2 O 2 . Convert each amount to grams of H 2 O. 1750 g N 2 H 4 2a 1 mol 32.05 g b ¢ 8 mol H 2O ≤ a 18.02 g b = 3.4 * 10 3 g H 1 mol N 2 H 4 mol 2 O 1125 g H 2 O 2 2a 1 mol 34.02 g b ¢ 8 mol H 2O ≤ a 18.02 g b = 75.7 g H 7 mol H 2 O 2 mol 2 O 75.7 g H 2 O can be produced. (e) Since H 2 O 2 is the limiting reactant, N 2 H 4 is in excess. 1125 g H 2 O 2 2a 1 mol 34.02 g b ¢ 1 mol N 2H 4 ≤ a 32.05 g b = 16.8 g N 7 mol H 2 O 2 mol 2 H 4 reacted 750 g N 2 H 4 given - 16.8 g N 2 H 4 used = 730 g N 2 H 4 remaining - 109 -
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Chapter 9

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