Ta có:
\(n_{Br2}=\frac{24}{160}=0,15\left(mol\right)\)
\(PTHH:C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
\(\Rightarrow n_{C2H4}=n_{Br2}=0,15\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}V_{C2H4}=0,15.22,4=3,36\left(l\right)\\V_{CH4}=4,48-3,36=1,12\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%V_{C2H4}=\frac{3,36}{4,48}.100\%=75\%\\\%V_{CH4}=100\%-75\%=25\%\end{matrix}\right.\)