2H2+O2->2H2O
30m3=30000l;20m3=20000l
\(n_{H_2}=\frac{30000}{22,4}=1339,285714\left(mol\right)\)
\(n_{O_2}=\frac{20000}{22,4}=892,8571429\left(mol\right)\)
Ta có: \(\frac{1339,285714}{2}< \frac{892,8571429}{1}\)
=> Oxi dư
\(n_{O_2}\)dư: 892,8571429-1339,285714.\(\frac{1}{2}\)=223,2142859(mol)
V O2 dư: 223,2142859.22,4\(\simeq\)5000l\(\simeq\)5m3
b) \(V_{H_2O}=1339,285714.22,4\simeq30000l=30m^3\)
\(m_{H_2O}=1339,285714.18=24107,14285g=24,10714285kg\)