\(n_{hh}=\dfrac{8.96}{22.4}=0.4\left(mol\right)\)
\(n_{Br_2}=\dfrac{24}{160}=0.15\left(mol\right)\)
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
\(0.15..........0.15\)
\(n_{CH_4}=0.4-0.15=0.25\left(mol\right)\)
\(\%V_{C_2H_4}=\dfrac{0.15}{0.4}\cdot100\%=37.5\%\)
\(\%V_{CH_4}=100-37.5=62.5\%\)