Question

A synthesis gas is made by cracking and partially oxidizing butane (C4H10) in the presence of steam and air. The product synthesis gas has the volume percent composition as follow: 3.5% CO2, 2.3% C2H4, 23.2% CO, 39.3% H2, 11.6% CH4, 1.7% C2H6, and 18.4% N2. How many kilograms of air are used per kilogram of butane?

Answer 1

Volume composition = Mol Composition

Assume a basis of 100 mol;

Total mol of C = CO2 + C2H4 + CO + CH4 = 3.5+2.3+23.2+11.6+1.7 = 42.3

if this is butane then

C4H10 is the formula

1 mol of C4H10 = 4 mol of C

then

42.3 mol of C = 42.3/4 = 10.575 mol of butane

mass of butane = mol*MW = 10.575*58.12 = 614.619 g of Butane

0.614619 kg of butane

the ratio is

C4:10H

or 10/4

for H atoms = 10/4*42.3 mol of C = 105.75 mol of H

mol of N2 = 18.4

then, go backwards,

1 mol of air = 0.79 mol of N2

then 18.4 mol of N2 --> 18.4/0.79 = 23.29 mol of air

mass of air = mol*MW = 23.29*29 = 675.41 g of air = 0.67541 kg of air

then

kg of air per kg of butane = 0.67541/0.614619 = 1.09890 kg air/kg butane