Sửa đề: 44,8 lít -> 4,48 lít
\(n_H=2.\dfrac{4,48}{22,4}=0,4\left(mol\right);n_C=n_{CaCO_3}=\dfrac{30}{100}=0,3\left(mol\right)\)
\(n_X=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
Trong 1 mol X có: \(\left\{{}\begin{matrix}n_C=\dfrac{0,3}{0,1}=3\\n_H=\dfrac{0,4}{0,1}=4\end{matrix}\right.\)
=> X là `C_3H_4`