\(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_20\)
\(n_{CH_{\text{4}}}=\frac{8,96}{22,4}=0,4\left(mol\right)\)
theo PTHH: \(n_{O_2}=2n_{CH_4}=2.0,4=0,8\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,8.22,4=17.92l\)
b,theo PTHH: \(n_{CO_2}=n_{CH_4}=0,4mol\)
\(\Rightarrow m_{CO_2}=44.0,4=17.6g\)