2H2O \(\underrightarrow{to}\) 2H2 + O2 (1)
\(m_{H_2O}=1000\times1=1000\left(g\right)\)
\(\Rightarrow n_{H_2O}=\frac{1000}{18}=\frac{500}{9}\left(mol\right)\)
a) Theo pT1: \(n_{H_2}=n_{H_2O}=\frac{500}{9}\left(mol\right)\)
\(\Rightarrow V_{H_2}=\frac{500}{9}\times22,4=1244,44\left(l\right)\)
Theo pT1: \(n_{O_2}=\frac{1}{2}n_{H_2O}=\frac{1}{2}\times\frac{500}{9}=\frac{250}{9}\left(mol\right)\)
\(\Rightarrow V_{O_2}=\frac{250}{9}\times22,4=622,22\left(l\right)\)
b) 3O2 + 2H2S \(\underrightarrow{to}\) 2H2O + 2SO2 (2)
\(m_{O_2}=\frac{250}{9}\times32=888,89\left(g\right)\)
Theo pT2: \(n_{H_2S}=\frac{2}{3}n_{O_2}=\frac{2}{3}\times\frac{250}{9}=\frac{500}{27}\left(mol\right)\)
\(\Rightarrow m_{H_2S}=\frac{500}{27}\times34=629,63\left(g\right)\)
Theo ĐL BTKL ta có:
\(m_{sp}=m_{O_2}+m_{H_2S}=888,89+629,63=1518,52\left(g\right)\)
