\(n_{CH_4}=\dfrac{24,79}{24,79}=1\left(mol\right)\)
PT: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
\(n_{CO_2}=n_{CH_4}=1\left(mol\right)\Rightarrow V_{CO_2}=1.24,79=24,79\left(l\right)\)
\(n_{CH_4}=\dfrac{24,79}{24,79}=1mol\\ CH_4+2O_2\xrightarrow[]{t^0}CO_2+2H_2O\\ n_{CO_2}=n_{CH_4}=1mol\\ V_{CO_2}=1.24,79=24,79l\)