\(C_2H_2+Br_2\rightarrow C_2H_2Br_2\)
\(m_{tang}=m_{C2H2}=4,68\left(g\right)\)
\(\rightarrow n_{C2H2}=\frac{4,68}{26}=0,18\left(mol\right)\)
\(n_{hh}=\frac{6,72}{22,4}=0,3\left(mol\right)\)
\(\rightarrow C\%_{C2H2}=\frac{0,18}{0,3}.100\%=60\%\)
\(\rightarrow C\%_{CH4}=100\%-60\%=40\%\)