a,
$n_X=\dfrac{2,24}{22,4}=0,1(mol)$
$n_{CO_2}=\dfrac{8,8}{44}=0,2(mol)$
Gọi CTTQ ankan X là $C_nH_{2n+2}$
Bảo toàn $C$: $n.n_X=n_{CO_2}$
$\to n=2(C_2H_6)$
b,
Bảo toàn $H$: $n_{H_2O}=3n_X=0,3(mol)$
Bảo toàn $O$:
$n_{O_2}=\dfrac{2n_{CO_2}+n_{H_2O}}{2}=0,35(mol)$
$\to V_{O_2}=7,84l$