PT: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
Ta có: \(n_{CH_4}=\dfrac{0,224}{22,4}=0,01\left(mol\right)\)
Theo PT: \(\left\{{}\begin{matrix}n_{CO_2}=n_{CH_4}=0,01\left(mol\right)\\n_{H_2O}=2n_{CH_4}=0,02\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\Sigma n_{hhk}=0,03\left(mol\right)\)
\(\Rightarrow V_{hhk}=0,03.22,4=0,672\left(l\right)\)
Bạn tham khảo nhé!