calculate weight and volume of air required for $2m^3$ of the gas

$H_2 + \frac{1}{2} O_2 H_2O$ = $0.25 m^3$

1vol     0.5 vol

$CH_4 +2O_2 CO_2+2H_2O = 0.6 m^3$

$CO+\frac{1}{2} O_2 CO_2 = 0.035 m^3$

$C_2H_4 +3O_2 2CO_2 +2H_2O = 0.09 m^3$

$C_2H_6+3O_2 2CO_2 + O_2 = 0.15 m^3$

$= 0.25 m^3 + 0.035m^3 + 0.6 m^3 + 0.09 m^3 + 0.175 m^3$

Amount of oxygen required = $1.15 m^3$

Volume of air required = $1.15 \times 2 = 2.30 m^3$

$2.30 m^3$ of air will weigh = $ \frac{2.30}{22.4} \times 28.949 = 2.97 Kg$

$\therefore$ Wt of air required = 2.97 Kg