$n_{CO_2}=n_{BaCO_3\downarrow}=\dfrac{108,35}{197}=0,55(mol)$
$m_{\text{giảm}}=m_{BaCO_3}-m_{CO_2}-m_{H_2O}$
$\to n_{H_2O}=\dfrac{108,35-0,55.44-74,25}{18}=0,55(mol)$
$n_{CO_2}=n_{H_2O}\to 2$ hidrocacbon là anken $C_nH_{2n}$
$n_{hh}=\dfrac{0,56}{22,4}=0,25(mol)$
$\to n=\dfrac{n_{CO_2}}{n_{hh}}=2,2$
Vậy 2 anken là $C_2H_4$ (x mol), $C_3H_6$ (y mol)
$\to x+y=0,25$
Bảo toàn $C$: $2x+3y=0,55$
$\to x=0,2; y=0,05$
$\%V_{C_2H_4}=\dfrac{0,2.100}{0,25}=80\%$
$\%V_{C_3H_6}=20\%$