$\overline{M}_{X}=9,4.2=18,8$
Đặt $n_{CH_4}=x(mol); n_{C_2H_6}=y(mol)$
$\to x+y=1$
Mặt khác: $16x+30y=1.18,8=18,8$
Giải hệ: $x=0,8; y=0,2$
$CH_4+2O_2\xrightarrow{{t^o}} CO_2+2H_2O$
$C_2H_6+\dfrac{7}{2}O_2\xrightarrow{{t^o}} 2CO_2+3H_2O$
$\to n_{O_2}=\dfrac{7}{2}.0,2+0,8.2=2,3(mol)$
$\to V=2,3.22,4=51,52l$