$m_{H_2O}=6,43g$
$\to n_{H_2O}=\dfrac{6,43}{18}=0,357(mol)$
$m_{CO_2}=9,82g$
$\to n_{CO_2}=\dfrac{9,82}{44}=0,223(mol)$
$\to n_{hh}=n_{H_2O}-n_{CO_2}=0,134(mol)$
Số $C$ trung bình: $\dfrac{n_{CO_2}}{n_{hh}}=1,66$
Vậy 2 ankan là $CH_4$ (x mol), $C_2H_6$ (y mol)
$\to x+y=0,134$
Bảo toàn $C$: $x+2y=0,223$
Giải hệ: $x=0,045; y=0,089$
$\%V_{CH_4}=\dfrac{0,045.100}{0,134}=33,58\%$
$\%V_{C_2H_6}=66,42\%$