PTHH: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
\(CO_2+Ba\left(OH\right)_2\rightarrow BaCO_3\downarrow+H_2O\)
Ta có: \(n_{H_2O}=\dfrac{7,2}{18}=0,4\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{CO_2}=n_{BaCO_3}=0,2\left(mol\right)=n_{CH_4}\\n_{O_2}=0,4\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{BaCO_3}=0,2\cdot197=39,4\left(g\right)\\V_{CH_4}=0,2\cdot22,4=4,48\left(l\right)\\V_{kk}=\dfrac{0,4\cdot22,4}{20\%}=44,8\left(l\right)\end{matrix}\right.\)