\(n_{Br_2}=0,05.2=0,1\left(mol\right)\)
PTHH: C2H2 + 2Br2 ---> C2H2Br4
0,05 0,1
\(n_{hhkhí}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\\ \rightarrow\left\{{}\begin{matrix}\%V_{C_2H_2}=\dfrac{0,05}{0,5}=10\%\\\%V_{CH_4}=100\%-10\%=90\%\end{matrix}\right.\)