\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
\(n_{Br_2}=0,2mol\)\(\Rightarrow n_{etilen}=0,2mol\)
\(n_{hh}=\dfrac{5,6}{22,4}=0,25mol\)
\(\Rightarrow n_{metan}=0,25-0,2=0,05mol\)
\(\%V_{metan}=\dfrac{0,05}{0,25}\cdot100\%=20\%\)
\(\%V_{metan}=100\%-20\%=80\%\)


