\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
Ta có : \(V_{C2H4}=V_{hh}-V_{C2H6}=7,437-3,7185=3,7185\left(l\right)\)
\(\rightarrow n_{C2H4}=\dfrac{3,7185}{24,79}=0,15\left(mol\right)\)
\(\rightarrow\left\{{}\begin{matrix}\%m_{C2H4}=\dfrac{0,15.28}{\dfrac{7,437}{24,79}.\left(30+28\right)}.100\%=24,14\%\\\%m_{C2H6}=100\%-24,14\%=75,86\%\end{matrix}\right.\)