a,Ta có
\(\left\{{}\begin{matrix}n_{O2}=0,275\left(mol\right)\\n_{CO2}=0,15\left(mol\right)\end{matrix}\right.\)
Bảo toàn O: \(2n_{O2}=2n_{CO2}+n_{H2O}\)
\(\Rightarrow n_{H2O}=0,25\left(mol\right)\)
\(n_{CO2}< n_{H2O}\Rightarrow\) Ankan CnH2n+2
\(\Rightarrow\frac{0,15}{n}=\frac{0,25}{n+1}\)
\(\Leftrightarrow n=1,5\)Vậy 2 ankan là CH4,C2H6
b,
\(n_C=n_{CO2}=0,15\left(mol\right)\)
\(\Rightarrow n_H=2n_{H2O}=0,5\left(mol\right)\)
\(\Rightarrow m=m_C+m_H=2,3\left(g\right)\)