\(C_{\overline{n}}H_{2\overline{n}}+\frac{3\overline{n}}{2}O_2\rightarrow\overline{n}CO_2+\overline{n}H_2O\)
\(n_A=0,4\rightarrow n_{H_2O}=n_{CO_2}=0,4.\overline{n}\)
\(m_{b2}-m_{b1}=39\Leftrightarrow m_{CO_2}-m_{H_2O}=39\Leftrightarrow44.0,4.\overline{n}-18.0,4.\overline{n}=39\Rightarrow\overline{n}=3,75\)
\(\left(3< 3,75< 4\right)\Rightarrow2anken:\)\(C_3H_6;C_4H_8\)
Gọi x, y là số mol \(C_3H_6;C_4H_8\)
\(\Rightarrow\left\{\begin{matrix}x+y=0,4\\\left(44-18\right)\left(3x+4y\right)=39\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x=0,1\\y=0,3\end{matrix}\right.\Rightarrow\)%VC3H6=25%; %VC4H8 =75%