Ta có:
\(n_{CO2}=\frac{22,4}{22,4}=1\left(mol\right)\)
\(n_{H2O}=\frac{12,6}{18}=0,7\left(mol\right)\)
\(n_{CO2}>n_{H2O}\Rightarrow M,N:Ankin\)
\(n_{ankin}=1-0,7=0,3\left(mol\right)\)
CTHH là CnH2n-2
\(n=\frac{1}{0,3}=3,33\)
\(\Rightarrow\left\{{}\begin{matrix}M:C_3H_4\\N:C_4H_6\end{matrix}\right.\)