Gọi \(n_{CH_4}=a\left(mol\right)\) và \(n_{C_2H_6}=b\left(mol\right)\)
\(n_{hh}=\dfrac{11,1555}{22,4}=0,5\)
\(\Rightarrow a+b=0,5\left(1\right)\)
\(n_{CO_2}=\dfrac{16,1135}{22,4}=0,72mol\)
Bảo toàn C: \(n_{CO_2}=n_{CH_4}+2n_{C_2H_6}=a+2b=0,72\left(2\right)\)
Từ (1) và (2)\(\Rightarrow\left\{{}\begin{matrix}a=0,28mol\\b=0,22mol\end{matrix}\right.\)
\(\%V_{CH_4}=\dfrac{0,28}{0,28+0,22}\cdot100\%=56\%\)
\(\%V_{C_2H_6}=100\%-56\%=44\%\)