$n_{C_4H_{10}}=\dfrac{5,8}{58}=0,1(mol)$
Quy đổi hỗn hợp anken gồm $CH_2$
$m_{\text{brom tăng}}=m_{\text{anken}}=m_{CH_2}$
$\to n_{CH_2}=\dfrac{2,016}{14}=0,144(mol)$
Bảo toàn $C$:
$n_{CO_2}=n_{C(Y)}=4n_{C_4H_{10}}-n_{CH_2}=0,256(mol)$
Bảo toàn $H$:
$n_{H_2O}=5n_{C_4H_{10}}-n_{CH_2}=0,356(mol)$
$\to V=22,4(0,256+0,356)=13,7088l$