a, PT: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
Ta có: \(n_{CH_4}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
Theo PT: \(n_{O_2}=2n_{CH_4}=0,5\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,5.22,4=11,2\left(l\right)\)
b, PT: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
Ta có: \(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{2}{3}n_{Fe}=0,2\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,2.22,4=4,48\left(l\right)\)
c, PT: \(2H_2+O_2\underrightarrow{t^o}2H_2O\)
Ta có: \(n_{H_2}=\dfrac{3}{2}=1,5\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{H_2}=0,75\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,75.22,4=16,8\left(l\right)\)
d, PT: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
Ta có: \(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{5}{4}n_P=0,25\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,25.22,4=5,6\left(l\right)\)
