\(n_{hh}=\dfrac{6.72}{22.4}=0.3\left(mol\right)\)
\(n_{Br_2}=n_{C_2H_4}=\dfrac{2}{160}=0.0125\left(mol\right)\)
\(\Rightarrow n_{CH_4}=0.3-0.0125=0.2875\left(mol\right)\)
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
\(\%V_{CH_4}=\dfrac{0.2875}{0.3}\cdot100\%=95.83\%\)
\(\%V_{C_2H_4}=100\%-95.83\%=4.17\%\)