PTHH: \(CH_4+2O_2\xrightarrow[]{t^o}CO_2+2H_2O\)
Ta có: \(n_{CH_4}=\dfrac{2,016}{22,4}=0,09\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{H_2O}=n_{CO_2}=0,18\left(mol\right)\\n_{CO_2}=0,09\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{H_2O}=0,18\cdot18=3,24\left(g\right)\\V_{CO_2}=0,09\cdot22,4=2,016\left(l\right)\\V_{kk}=\dfrac{0,18\cdot22,4}{20\%}=20,16\left(l\right)\end{matrix}\right.\)
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