$n_{CH_4}=\dfrac{1,12}{22,4}=0,05\ (mol)$
$CH_4+2O_2\xrightarrow{\ t^o\ } CO_2+2H_2O$
$\Rightarrow \begin{cases}n_{CO_2}=n_{CH_4}=0,05\ (mol)\\n_{O_2}=2n_{CH_4}=2.0,05=0,1\ (mol)\end{cases}$
$\Rightarrow \begin{cases}V_{O_2}=0,1.22,4=2,24\ (l)\\V_{CO_2}=0,05.22,4=1,12\ (l)\end{cases}$