Gọi \(\left\{{}\begin{matrix}n_{C_2H_2}=x\\n_{CH_4}=y\end{matrix}\right.\)
\(n_{hh}=\dfrac{5,6}{22,4}=0,25mol\)
Ta có:
\(\left\{{}\begin{matrix}x+y=0,25\\26x+16y=5,5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,15\\y=0,1\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}\%V_{C_2H_2}=\dfrac{0,15}{0,25}.100=60\%\\\%V_{CH_4}=100\%-60\%=40\%\end{matrix}\right.\)