Có ${{n}_{C{{H}_{4}}}}+{{n}_{{{C}_{2}}{{H}_{4}}}}=\frac{V}{22,4}=\frac{3,36}{22,4}=0,15(mol)$ (1)

Có ${{d}_{\frac{X}{{{H}_{2}}}}}=10\to \frac{{{M}_{X}}}{{{M}_{{{H}_{2}}}}}=10\to {{M}_{X}}=20$

$\to {{m}_{C{{H}_{4}}}}+{{m}_{{{C}_{2}}{{H}_{4}}}}=M\times n=20\times 0,15=3(g)$

$\to 16\times {{n}_{C{{H}_{4}}}}+28\times

{{n}_{{{C}_{2}}{{H}_{4}}}}=3$    (2)

Từ (1) và (2) $\to $ ${{n}_{C{{H}_{4}}}}=0,1(mol)$ ;  ${{n}_{{{C}_{2}}{{H}_{4}}}}=0,05(mol)$

$\to {{V}_{C{{H}_{4}}}}=2,24(l)$ ;   ${{V}_{{{C}_{2}}{{H}_{4}}}}=1,12(l)$

$\Rightarrow $ Chọn đáp án C