\(CH_4+2O_2\xrightarrow[]{t^o}CO_2+2H_2O\)
\(n_{CH_4}=\dfrac{12,395}{24,79}=0,5\left(mol\right)\)
Theo PTHH: \(n_{O_2}=2n_{CH_4}=2\cdot0,5=1\left(mol\right)\)
\(\Rightarrow V_{O_2}=1\cdot24,79=24,79\left(l\right)\)
Theo PTHH: \(n_{CO_2}=n_{CH_4}=0,5\left(mol\right)\)
\(\Rightarrow V_{CO_2}=0,5\cdot24,79=12,395\left(l\right)\)