Gọi số mol CH4, C2H6 là a, b
=> a + b = \(\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
\(n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
PTHH: CH4 + 2O2 --to--> CO2 + 2H2O
______a--------------------->a
2C2H6 + 7O2 --to--> 4CO2 + 6H2O
_b--------------------->2b
=> a + 2b = 0,25
=> a =0,15 ; b =0,05
=> \(\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,15}{0,2}.100\%=75\%\\\%V_{C_2H_6}=\dfrac{0,05}{0,2}.100\%=25\%\end{matrix}\right.\)