\(n_{Br_2}=\dfrac{32}{160}=0,2mol\Rightarrow n_{etilen}=0,2mol\)
\(n_{hh}=\dfrac{11,2}{22,4}=0,5mol\Rightarrow n_{metan}=0,5-0,2=0,3mol\)
\(\%m_{etilen}=\dfrac{0,2\cdot28}{0,2\cdot18+0,3\cdot16}\cdot100\%=53,85\%\)
\(\%m_{metan}=100\%-53,85\%=46,15\%\)