a, \(n_{CH_4}+n_{C_2H_2}=\dfrac{0,28}{22,4}=0,0125\left(mol\right)\left(1\right)\)
PT: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
\(2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\)
Theo PT: \(n_{O_2}=2n_{CH_4}+\dfrac{5}{2}n_{C_2H_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{CH_4}=0,0025\left(mol\right)\\n_{C_2H_2}=0,01\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,0025.22,4}{0,28}.100\%=20\%\\\%V_{C_2H_2}=80\%\end{matrix}\right.\)