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When pure carbon is burned in air, some of it oxidizes into CO2 and some to CO. The molar ratio of N2 to O2 is 7.18 and the ratio of CO to CO2 is 2.0 in the product gas. What is the percent excess air used? The exit gases contain only N2, O2, CO and CO2.(a) 50 %(b) 60 %(c) 40 %(d) 20 %The question was asked by my college professor while I was bunking the class.This interesting question is from Material Balance Numericals topic in chapter Material Balance of Bioprocess Engineering

Answer»

Right choice is (c) 40 %

Easiest explanation: The process involves TWO reactions:

R1 = C + O2 = CO2

R2 = C + 1/2 O2 = CO

As the product (output) gas composition is specified more clearly, we may USE it as starting point.

TAKE N2 in the product as nN2 = Take N2 in the product as, then O2 = 1 mol

N2 as an inert gas: input = output = 7.18 mol

O2 input with AIR = 7.18 (21/79) = 1.91 mol

Total O2 consumption = 1.91-1 = 0.91 mol

If R1 uses n1 mol O2, generating n1 mol CO2, R2 uses 0.91-n1 mol O2, generating 2

(0.91-n1) mol CO.

Since CO/CO2 in the product gas = 2, 2 (0.91-n1)/n1 = 2

n1 = 0.91/2 = 0.455 mol

Therefore, total moles of C input = 3 n1 = 1.365 mol

O2 required for COMPLETE combustion of 1.365 mol C = 1.365 mol (for R1 only)

O2 excess % = (1.91 -1365)/1.365 x 100% = 39.9% = 40% .


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