a) FeS + H2SO4 → FeSO4 + H2S (1)
ZnS + H2SO4 → ZnSO4 + H2S (2)
b) \(n_{H_2S}=\frac{3,36}{22,4}=0,15\left(mol\right)\)
Gọi số mol của FeS là x (mol)
Theo đề bài: \(n_{ZnS}=x\left(mol\right)\)
\(n_{CuS}=2x\left(mol\right)\)
Theo pT1: \(n_{H_2S}=n_{FeS}=x\left(mol\right)\)
Theo PT2: \(n_{H_2S}=n_{ZnS}=x\left(mol\right)\)
Ta có: \(x+2x=0,15\)
\(\Leftrightarrow3x=0,15\)
\(\Leftrightarrow x=0,05\left(mol\right)\)
Vậy \(n_{FeS}=0,05\left(mol\right)\Rightarrow m_{FeS}=0,05\times88=4,4\left(g\right)\)
\(n_{ZnS}=0,05\left(mol\right)\Rightarrow m_{ZnS}=0,05\times97=4,85\left(g\right)\)
\(n_{CuS}=2\times0,05=0,1\left(mol\right)\Rightarrow m_{CuS}=0,1\times96=9,6\left(g\right)\)
Vậy \(m=4,4+4,85+9,6=18,85\left(g\right)\)
