Question

What is the air fuel ratio on molar basis for the complete combustion of octane $$C_8{H}_{18}$$ with just right amount of air available for combustion?

• A. 59.5
• B. 65.5
• C. 62.4
• D. 69.5

Answer 1

The correct option is A 59.5

For complete combustion of $$C_8{H}_{18}$$ with the theoretical amount of air, the products contain carbon dioxide, water and nitrogen only.

$C_8{H}_{18}+a(O_2+3.76N_2)\to bC{O}_2+c{H}_{2}O+d{N}_2$

Applying the conservation of mass principle to the carbon, hydrogen, oxygen and nitrogen respectively gives,

C : b = 8
H : 2c = 18
O = 2a = 2b + c
N = d = 3.76 a

Solving these equations

a = 12.5, b = 8, c = 9

d = 47

The balanced chemical equation is

$C_8{H}_{18}+12.5(O_2+3.76N_2)\to 8C{O}_2+9H_{2}O+47N_2$

The air fuel ratio on molar basis is

$A/F=\frac{12.5+12.5\left(3.76\right)}{1}=\frac{59.5}{1}\frac{Kmol\left(air\right)}{kmol\left(fuel\right)}$