Practice_EXAM_II

Example517thisquestionistakenfromp213584

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Unformatted text preview: 195 Examples 5.11 and 5.12; 10th ed. p.p. 194‐196. This question is taken from p. 217 (5.54); 9th ed. p.p. 190‐192. This question is taken from p. 212 (5.54). From mass of glucose, we can calculate the theoretical yield of carbon dioxide in mole; from 293 K and 0.984 atm, we can calculate the actual yield of carbon dioxide in mole. Thus, the theoretical yield of carbon dioxide = 2 x mole of glucose = 2 x (5.97/12x6+1x12+16x6) = 2 X (5.97/180) = 0.0663 mole. From PV = nRT n = PV/RT we can calculate the actual yield = 0.984x1.44/(0.082x293) = 0.05898 = 0.0590 mole. So the percent yield = (actual yield / theoretical yield) x 100% = (0.0590/0.0663)x100% = 88.95 %. 17. What is the total pressure (in atm) of the mixture when a 2.5‐L flask at 15 oC contains a mixture if nitrogen, helium, and neon gases at partial pressure of 0.32 atm for nitrogen, 0.15 atm for helium, and 0.42 atm for neon? (a) 0.49 (b) 0.51 (c) 0.64 (d) 0.73 (e) 0.89 Hint: Total pressure is the sum of all partial pressures of component gases. To use this formula, all the gases in the same container can not react to each other (i.e. non‐reactive gases). 3 11th ed. p.p. 195‐198; 10th ed. p.p. 196‐199. This question is taken from p. 218 (5.64 (...
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