\(n_{CH_4}=\dfrac{3,2}{16}=0,2\left(mol\right)\)
a. PTHH: CH4 + 2O2 ---to---> CO2 + 2H2O
Theo PT: \(n_{CO_2}=n_{CH_4}=0,2\left(mol\right)\)
=> \(V_{CO_2}=0,2.22,4=4,48\left(lít\right)\)
b. Theo PT: \(n_{H_2O}=2.n_{CH_4}=2.0,2=0,4\left(mol\right)\)
=> \(m_{H_2O}=0,4.18=7,2\left(g\right)\)
PTHH: \(CH_4+2O_2\xrightarrow[]{t^o}CO_2+2H_2O\)
Ta có: \(n_{CH_4}=\dfrac{3,2}{16}=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{O_2}=0,4\left(mol\right)=n_{H_2O}\\n_{CO_2}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{O_2}=0,4\cdot22,4=8,96\left(l\right)\\V_{CO_2}=0,2\cdot22,4=4,48\left(l\right)\\m_{H_2O}=0,4\cdot18=7,2\left(g\right)\end{matrix}\right.\)
