Question

A liquid fuel containing Co and H2, was burned using excess air. Its Orsat analysis of the combustion gas is 5.34% CO2, 13.23% O2 and 81.43% N2. The molal humidity of the product is 0.179 mol water vapor/mol of dry gas. Calculate percent composition by weight of the liquid fuel.

Answer 1

200 + O2 +2602 612 + + 2 + 420 let take 100 mol of Combustion gas (day) without taking H20 Vapour 5-3 u mol 0 = 13.23 molla N2=81.43 mol Ny in product aN2in fresh air Ng in fresh an a 81.43 mol O in fresh an = 21 so y 8.43 79 = 21.65 mol Co2 in product = 5.34 mol so zmol Co2 required lmolo2 5.34 mol Con required 1 x S-34 2 22.67 molar

So 02 neact with H₂ = og in frech feed - Q in product O₂ Consumed for lo 21.65 -13.23 -2.67 - 5-75 mol 5-75 mol O₂ is consumed for H₂O so H₂ present in fuel = 265.75 = 11.5 mol Co present in fuel co2 produced ²5-34 mol or liquid fuel Contain Co = 5-34 mol = 03/08 H2= Ils mol in weight Molecular weight Cot 5.34 x 28 = of Co=28 H₂ = 11.5x2= 239 & H₂ = 2 = 149.59 of x1w % Composition of Co = 149.5 = 86.67%. 149.5+23 H2 = 23 X/W = 1495+23 13:33 y