$n_Y=\dfrac{11,2}{22,4}=0,5(mol)$
Ta có $n_X:n_{H_2}=1:1$
$\to n_X=n_{H_2}=\dfrac{0,5}{2}=0,25(mol)$
$\to 0,25M_X+0,25.2=11,5$
$\to M_X=44$
Trong $X$, $\%m_A=27,27\%$
$\to \dfrac{M_A.100}{44}=27,27$
$\to M_A=12\quad(C)$
$\to M_B=\dfrac{44-M_A}{2}=16\quad(O)$
Vậy khí $X$ là $CO_2$.