Elementary Principles 166 - 5.83 Basis: 10 mL C5 H 10 l...

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5- 54 5.83 Basis: 10 mL C H charged to reactor 51 0 l bg CH 15 2 O5 C O 5 H O 0 2 2 2 +→+ a. n 10.0 mL C H l 0.745 g 1 mol mL 70.13 g 0.1062 mol C H 1 0 0 == Stoichiometric air: n 0.1062 mol C H 7.5 mol O 1 mol air 1 mol C H 0.21 mol O 3.79 mol air 2 0 2 21 0 2 P nRT V 3.79 mol 0.08314 L bar 300K 11.2 L mol K 8.44 bars o = (We neglect the C H 0 that may be present in the gas phase due to evaporation) Initial gauge pressure bar 1 bar bar = = 844 744 .. b. n 0.1062 mol C H 5 mol CO 1 mol C H 0.531 mol CO n 0.531 mol CO 1 mol H O 1 mol CO 0.531 mol H O n 0.79 3.79 2.99 mol N 4.052 mol product gas 3 0 2 0 2 4 22 2 2 52 U V | | | W | | | CO y = 0.531/ 4.052 = 0.131 mol CO / mol, T = 304.2 K P = 72.9 atm H O y = 0.531/ 4.052 = 0.131 mol H O / mol, T = 647.4 K P = 218.3 atm N y = 2.99 / 4.052 = 0.738 mol N / mol, T =126.2 K P = 33.5 atm 23 2 c c 24 2 c c 25 2 c c :
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This note was uploaded on 02/27/2012 for the course CHEMICAL E 312 taught by Professor Cheung during the Fall '11 term at The University of Akron.

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