a, \(n_{CO_2}=\dfrac{17,6}{44}=0,4\left(mol\right)\)
CH4 + 2O2 -----to---> CO2 + 2H2O
x x
C2H4 + 3O2 -----to---> 2CO2 + 2H2O
y 2y
Ta có hệ pt: \(\left\{{}\begin{matrix}16x+28y=6\\x+2y=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)
\(\%m_{CH_4}=\dfrac{0,2.16.100\%}{6}=53,33\%;\%m_{C_2H_4}=100\%-53,33\%=46,67\%\)
b, \(\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,2.22,4.100\%}{\left(0,2+0,1\right).22,4}=66,67\%\\\%V_{C_2H_4}=100\%-66,67\%=33,33\%\end{matrix}\right.\)