\(n_{CO2}=\dfrac{17,6}{44}=0,4\left(mol\right),n_{H2O}=\dfrac{10,8}{18}=0,6\left(mol\right)\)
Ta thấy : \(n_{CO2}< n_{H2O}\) --> A là ankan (CnH2n + 2)
Có : \(n\left(sốC\right)=\dfrac{n_{CO2}}{n_{H2O}-n_{CO2}}=\dfrac{0,4}{0,6-0,4}=2\)
\(\rightarrow CTPT:C_2H_6\)