\(n_{HCl}=\frac{60.36,5\%}{36,5}=0,6\left(mol\right)\)
\(FeS+2HCl\rightarrow FeCl_2+H_2S\)
\(Na_2S+2HCl\rightarrow2NaCl+H_2S\)
Gọi a là mol FeS; b là mol Na2S
\(\left\{{}\begin{matrix}88a+78b=25,4\\2a+2b=0,6\rightarrow\end{matrix}\right.\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\)
\(\%_{FeS}=\frac{0,2.88.100}{25,4}=69,29\%\)
\(\%_{Na2S}=100\%-69,29\%=30,71\%\)
\(n_{FeCl2}=0,2\left(mol\right);n_{NaCl}=0,2\left(mol\right)\)
\(\rightarrow n_{Cl}=2n_{FeCl2}+n_{NaCl}=0,6\left(mol\right)\)
\(PTHH:Pb+2Cl\rightarrow PbCl_2\)
\(\rightarrow n_{PbCl2}=0,3\left(mol\right)\)
\(\rightarrow m+PbCl2=83,4\left(g\right)\)
