\(n_{CH_4}=\dfrac{5.6}{22.4}=0.25\left(mol\right)\)
\(n_{O_2}=\dfrac{6.72}{22.4}=0.3\left(mol\right)\)
\(CH_4+2O_2\underrightarrow{^{^{t^0}}}CO_2+2H_2O\)
Lập tỉ lệ : \(\dfrac{0.25}{1}>\dfrac{0.3}{2}\Rightarrow CH_4dư\)
\(V_{CO_2}=\dfrac{0.3}{2}\cdot22.4=3.36\left(l\right)\left(l\right)\)
\(V_{CH_4\left(dư\right)}=\left(0.25-\dfrac{0.3}{2}\right)\cdot22.4=2.24\left(l\right)\)