Trong hỗn hợp X, gọi \(n_{H_2S} = a(mol) ; n_S = b(mol)\)
\(2H_2S + 3O_2 \xrightarrow{t^o} 2SO_2 + 2H_2O\\
S + O_2 \xrightarrow{t^o} SO_2\)
Theo PTHH :
\(n_{O_2} = 1,5a + b = \dfrac{8,96}{22,4} = 0,4(mol)\\
n_{SO_2} = a + b = \dfrac{7,84}{22,4} = 0,35(mol)\\
\Rightarrow a = 0,1(mol) ; b = 0,25(mol)\)
Suy ra :
\(\%m_{H_2S} = \dfrac{0,1.34}{0,1.34 + 0,25.32}.100\% = 29,82\%\\
\%m_{S} = 100\% -29,82\% = 70,18\%\)