$CH_4$ = 35%,$C_2 H_4$ = 5%, CO = 15%, $H_2$ = 40%, $N_2$ = 1 %, water vapours =4%, ( mol wt of air = 28.949)

$H_2 +1/2 O_2 \rightarrow H_2O$ = 0.20 $m^3$

$CH_4 +2O_2 \rightarrow CO_2 + 2H_2O$ = 0.7 $m^3$

$CO+1/2 O_2 \rightarrow CO_2$ = 0.11 $m^3$

$C_2H_4 +3O \rightarrow_2 2CO_2 + 2H_2O$ = 0.15 $m^3$

$=0.7 m^3 + 0.11m^3 + 0.2m^3 + 0.15 m^3$

Amount of oxygen required = $1.16 m^3$

Volume of air required = $1.16 m^3$

$1.16 m^3$ of air will weigh = $(1.16/22.4) \times 28.949$ = $1.499 Kg$

$\therefore$ Wt of air required = 1.499 Kg