\(n_{CH_4}=\dfrac{3,2}{16}=0,2\left(mol\right)\)
\(n_{O_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)
PT: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
Xét tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{0,1}{2}\), ta được CH4 dư.
Theo PT: \(\left\{{}\begin{matrix}n_{CO_2}=\dfrac{1}{2}n_{O_2}=0,05\left(mol\right)\\n_{H_2O}=n_{O_2}=0,1\left(mol\right)\end{matrix}\right.\)
⇒ m = 0,05.44 + 0,1.18 = 4 (g)