Question 2.9: To ensure complete combustion, 20 per cent excess air is sup...

Basis: 100 mol gas, as the analysis is volume percentage.
Reactions: CH_{4}+ 2O_{2}\longrightarrow CO_{2}+2H_{2}O
C_{2}H_{6} + 3\frac{1}{2}O_{2}\longrightarrow 2CO_{2}+ 3H_{2}O
Stoichiometric mols O_{2}required = 95\times 2+ 5\times 3\frac{1}{2}=\underline{{\underline{207.5}} }
With 20 per cent excess, mols O_{2} required = 207.5\times \frac{120 }{10}=\underline{{\underline{249 } }}
Mols air (21 per cent O_{2}) = 249\times \frac{100}{21}=1185.7
Air per mol fuel = \frac{1185.7}{100}=\underline{\underline{11.86 mol} }