\(n_{hh}=\dfrac{11,2}{22,4}=0,5mol\)
Gọi \(\left\{{}\begin{matrix}n_{C_2H_4}=x\left(mol\right)\\n_{C_2H_2}=y\left(mol\right)\end{matrix}\right.\)
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=0,5\\x+2y=0,7\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,3\\y=0,2\end{matrix}\right.\)
\(\%V_{C_2H_4}=\dfrac{0,3}{0,3+0,2}\cdot100\%=60\%\)
\(\%V_{C_2H_2}=100\%-60\%=40\%\)
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