Question

Please show your work:

Benzene gas (C6H6)is burned in a continuous flow reactor with 85% theoretical dry air. What is the mole fraction of CO in the products?

Answer: [6.8%]

Answer 1

The combustion of Benzene is

C6H6+7.5O2-------->6CO2 + 3H2O

1 mole of Benzene requires 7.5moles of oxygen for complete combustion.

moles of oxygen required = 7.5 moles. since air contains 21%O2 and 79%N2, moles of air = 7.5/0.21=35.71

supplied air is = 0.85*35.71=30.35

Oxygen in this air = 30.35*0.21=6.35

this leads to incomplete combustion as C6H6+4.5O2----> 6CO+ 3H2O

Products contains ( assuming all the C6H6 got combusted to CO)

CO= 6 moles, H2O= 3 moles, O2 ( unreacted)= 6.35-4.5= 1.85 moles, N2= 30.35*0.79=24

total moles of products = 6+3+1.85+24= 34.85

%CO in the products = 100*6/34.85 =17.21%