$n_{O_2}=\dfrac{3,2}{32}=0,1(mol)$
Đặt CTTQ ankan là $C_nH_{2n+2}$
$C_nH_{2n+2}+\dfrac{3n+1}{2}O_2\to nCO_2+(n+1)H_2O$
$\to n_A=\dfrac{0,1.2}{3n+2}=\dfrac{0,2}{3n+1}(mol)$
$\to M_A=\dfrac{7,2(3n+1)}{0,2}=36(3n+1)=14n+2$
$\to n=-0,36$ (vô lí, bạn xem lại đề)