$n_{CH_4}=\dfrac{4,48}{22,4}=0,2\ (mol)$
$n_{O_2}=\dfrac{8,96}{22,4}=0,4\ (mol)$
$CH_4+2O_2\xrightarrow{t^o} CO_2+2H_2O$
Ta có: $\dfrac{n_{CH_4}}{1}=\dfrac{n_{O_2}}{2}$
$\Rightarrow$ Cả hai khí tham gia phản ứng đều hết
Ta có: $\begin{cases}n_{CO_2}=n_{CH_4}=0,2\ (mol)\\n_{H_2O}=2n_{CH_4}=2.0,2=0,4\ (mol)\end{cases}$
$\Rightarrow \begin{cases}m_{CO_2}=0,2.22,4=4,48\ (l)\\m_{H_2O}=0,4.18=7,2\ (g)\end{cases}$