\(n_{HH}=\dfrac{11,2}{22,4}=0,5mol\\ n_{Br_2}=\dfrac{112}{160}=0,7mol\\ n_{C_2H_4}=a;n_{C_2H_2}=b\\ C_2H_4+Br_2\rightarrow C_2H_4Br_2\\ C_2H_2+2Br_2\rightarrow C_2H_2Br_4\\ \Rightarrow\left\{{}\begin{matrix}a+b=0,5\\a+2b=0,7\end{matrix}\right.\\ \Rightarrow a=0,3;b=0,2\\ \%V_{C_2H_4}=\dfrac{0,3}{0,5}\cdot100\%=60\%\\ \%V_{C_2H_2}=40\%\)