Khí đi ra là CH4
\(\Rightarrow V_{etilen,propilen}=8,96-1,12=7,84\left(l\right)\)
Gọi a là mol C2H4, b là mol C3H6
\(\Rightarrow22,4a+22,4b=7,84\left(1\right)\)
m tăng= m anken
\(\Rightarrow28a+42b=11,9\left(2\right)\)
(1)(2)\(\Rightarrow\left\{{}\begin{matrix}a=0,2\\b=0,15\end{matrix}\right.\)
\(\%_{CH4}=\frac{1,12.100}{8,96}=12,5\%\)
\(\%_{C2H4}=\frac{0,2.22,4.100}{8,96}=50\%\)
\(\%_{C3H6}=100\%-\left(12,5\%+50\%\right)=37,5\%\)