Elementary Principles 101 - 4.68 (cont'd) b. i) Theoretical...

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4- 60 4.68 (cont’d) b. i) Theoretical oxygen = (100 mol C 4 H 10 )(6.5 mol O 2 /mol C 4 H 10 ) = 650 mol O 2 n mol O mol air / 0.21 mol O mol air air 2 2 = = () ( ) 650 1 3095 100% conversion n C4H10 = 0, n O 2 = 0 n n n N2 CO2 4 10 2 4 10 2 H2O 4 10 2 4 10 2 2 2 2 mol 100 mol C H react 4 mol CO mol C H mol CO 100 mol C H react 5 mol H O mol C H mol H O 73.1% N 12.0% CO 14.9% H O == U V | W | 0 79 3095 mol 2445 400 500 . bg b g b g b g ii) 100% conversion n C4H10 = 0 20% excess n air = 1.2(3095) = 3714 mol (780 mol O 2 , 2934 mol N 2 ) Exit gas : 400 mol CO 2 500 mol H 2 O 130 mol O 2 2934 mol N 2 10.1% CO 2 12.6% H 2 O 3.3% O 2 74 0% . N 2 iii) 90% conversion n C4H10 = 10 mol C 4 H 10 (90 mol C 4 H 10 react, 585 mol O 2 consumed) 20% excess : n air = 1.2(3095) = 3714 mol (780 mol O 2 , 2483 mol N 2 ) Exit gas : 10 mol 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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