\(n_{C_2H_4Br_2}=\dfrac{18,8}{188}=0,1mol\Rightarrow n_{etilen}=0,1mol\)
\(n_{hh}=\dfrac{22,4}{22,4}=1mol\Rightarrow n_{metan}=1-0,1=0,9mol\)
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
\(\%V_{C_2H_4}=\dfrac{0,1}{1}\cdot100\%=10\%\)
\(\%V_{CH_4}=100\%-10\%=90\%\)