Gọi \(\left\{{}\begin{matrix}n_{Mg}:x\left(mol\right)\\n_{Fe}:y\left(mol\right)\end{matrix}\right.\)
\(Mg+S\rightarrow MgS\)
\(Fe+S\rightarrow FeS\)
\(\Rightarrow\left\{{}\begin{matrix}n_{MgS}=n_{Mg}=x\left(mol\right)\\n_{FeS}=n_{Fe}=y\left(mol\right)\end{matrix}\right.\)
Giải hệ PT:
\(\left\{{}\begin{matrix}24x+56y=13,6\\56x+88y=23,3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)
\(\Rightarrow m_{Mg}=0,1.24=2,4\left(g\right)\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\frac{2,4}{13,6}.100\%=17,65\%\\\%m_{Fe}=100\%-17,65\%=82,35\%\end{matrix}\right.\)
\(m_{MgS}=0,1.56=5,6\left(g\right)\Rightarrow m_{FeS}=17,6\left(g\right)\)
Hòa tan lượng muối trên
\(MgS+2HCl\rightarrow MgCl_2+H_2S\)
\(FeS+2HCl\rightarrow FeCl_2+H_2S\)
\(n_{HCl}=0,1.2+0,2.2=0,6\left(mol\right)\)
\(\Rightarrow V_{HCl}=\frac{0,6}{1}=0,6\left(l\right)\)
