\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
\(n_{C_2H_4Br_2}=\dfrac{56,4}{188}=0,3mol\)
\(\Rightarrow n_{etilen}=0,3mol\)
\(n_{hh}=\dfrac{11,2}{22,4}=0,5mol\)
\(\Rightarrow n_{metan}=0,5-0,3=0,2mol\)
\(\%V_{metan}=\dfrac{0,2}{0,5}\cdot100\%=40\%\)
\(\%V_{etilen}=100\%-40\%=60\%\)
