Question

Benzene is burned with 50% excess air. What is the flue gas composition?

(Determine: CO2, H2O, O2, N2 v/v content in flue gas)

Answer 1

Let us consider a reaction,

C6H6 + 15/2O2 ----> 6CO2 + 3H2O

1 mol of C6H6 requires = 15 /2

= 7.5 mol of O2

Volume of O2 = 7.5 * 22.4 ( STP)

  = 168 litres

Air used = 168 litres * 100 / 21

  = 800 litres ( In air, O2 = 21% & N2 =79%)

and we used 50% excess = 1.5* 800

= 1200 litres

22.4 litres of Benzene and1200 litres of Air = Total feed volume

= 1222.4 litres

1mol C6H6 produces 6 mol CO2 = 6mol * 22.4 litres/mol

= 134.4 litres CO2

and H2O produced = 3 * 22.4

= 67.2 litres H2O

and O2 consumed = 168 litres

O2 unburned = 0.5 * 168

= 84 litres

N2 unburned =79 *1200 / 100

= 948 litres

Total volume = 134.4+67.2+84+948

= 1233.6 litres

CO2 (%V/V) = 134.4 / 1233.6

= 0.10894 or 10.89%

H2O = 67.2/1233.6

= 0.05447 %

O2 = 84 / 1233.6

= 0.068%

N2 = 948 / 1233.6

= 0.7684 %