Gọi \(n_{H_2} = a(mol) ; n_{H_2S} = b(mol)\)
Ta có :
\(n_D = a + b = \dfrac{8,96}{22,4} = 0,4(mol)\\
m_D = 2a + 34b = 0,4.2.13 = 10,4(gam)\)
Suy ra a = 0,1 ; b = 0,3
\(Zn + 2HCl \to ZnCl_2 + H_2\\
ZnS + 2HCl \to ZnCl_2 + H_2S\)
Suy ra :
\(n_{Zn} = n_{H_2} +n_{H_2S} = 0,4(mol)\\
n_S = 0,3 + \dfrac{1,68}{32} = 0,3525(mol)\)
Suy ra : \(m_X = 0,4.65 + 0,3525.32=37,28(gam)\)
\(Zn + S \xrightarrow{t^o} ZnS\\
H = \dfrac{0,3}{0,3525}.100\%=85,1\%\)