Question

When pure carbon is burnt in air some of it is oxidized to CO2 and some to CO. If the molar ratio of N2 to O2 is 7.18 and the molar ratio of CO to CO2 is 2, what is the percentage excess air used? Exit gases contain only N2 , O2 , CO and CO2.

Answer 1

Basis : A moles of Carbon

let X moles be converted to CO2

XC +O2---XCO2

(A-X)C+{(A-X)/2}O₂---> (A-X) CO

given molar ratio of CO2 to CO= X/(A-X)=2

X= 2A-2X

3X =2A X= 2A/3

So 0.66A moles got converted to CO2 and rest 1-0.66=0.34A moles got converted to CO

total moles of air =0.66A+0.17A= 0.83A

moles of air =0.83A/0.21=3.95A

let x= % excess moles of air suppled

Air supplied =3.95A*(1+x/100)

N2= 3.95A*(1+0.01x)*0.79=3.1205A*(1+0.01x)

O2 supplied =3.95A*(1+x/100)*0.21

O2 remaining= 0.83A*x/100 =0.0083Ax

given N2/O2= 7.18,    3.1205*(1+0.01x)/x= 7.18

3.1205+3.1205*0.01x= 7.18x

x=2.29%

Excess air suppleid = 2.29%

N2 in the air is 79%, N2= (1+0.01y)3.95*0.79

O2 in the air is 21%, O2= (1+0.01y)*3.95*0.21