a, \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
b, \(n_{CH_4}=\dfrac{32}{16}=2\left(mol\right)\)
Theo PT: \(n_{CO_2}=n_{CH_4}=2\left(mol\right)\)
\(n_{H_2O}=2n_{CH_4}=4\left(mol\right)\)
⇒ m sản phẩm = mCO2 + mH2O = 2.44 + 4.18 = 160 (g)
c, \(n_{O_2}=2n_{CH_4}=4\left(mol\right)\Rightarrow V_{O_2}=4.22,4=89,6\left(l\right)\)