\(CaS+FeO+4HCl\rightarrow FeCl_2+CaCl_2+H_2O+H_2S\uparrow\)
.0,025...0,025.....0,1......................................................................
\(n_{HCl}=C_M.V=0,5.0,2=0,1\left(mol\right)\)
Ta có : \(m_A=m_{CaS}+m_{FeO}=0,025.72+0,025.72=3,6\left(g\right)\)
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