Đáp án:
$\rm CH_4+2O_2 \overset{t^o}\to CO_2+H_2O \\ Ta \ có : \ n_{CH_4}=\dfrac{3,36}{22,4}=0,15 \ (mol) \\ Theo \ PT \ có : \\ n_{CO_2}=n_{CH_4}=0,15 \ (mol) \\ n_{O_2}=2 n_{CH_4}=0,15.2=0,3 \ (mol) \\ \to \\ V_{CO_2}=0,15.22,4=3,36 \ (l) \\ V_{O_2}=0,3.22,4=6,72 \ (l)$