\(PTHH:C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
Ta có:
\(m_{C2H2}=23,4\left(g\right)\Rightarrow n_{C2H2}=0,9\left(mol\right)\)
\(m_{C2H6}=41,4-23,4=18\left(g\right)\Rightarrow n_{C2H6}=0,6\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%V_{C2H2}=\frac{0,9}{0,9+0,6}.100\%=60\%\\\%V_{C2H6}=100\%-60\%=40\%\end{matrix}\right.\)