$n_{O_2}=\dfrac{11,2}{22,4}=0,5(mol)$
Đặt CTPT ankan là $C_nH_{2n+2}$
$C_nH_{2n+2}+ \dfrac{3n+1}{2}O_2\buildrel{{t^o}}\over\to nCO_2+(n+1)H_2O$
$\Rightarrow n_{\text{ankan}}=\dfrac{0,5.2}{3n+1}=\dfrac{1}{3n+1}(mol)$
$\Rightarrow M_{\text{ankan}}=8,8(3n+1)=14n+2$
$\Leftrightarrow n=-0,5$ (vô lí, bạn xem lại đề)