\(Mg+2HCl\rightarrow MgCl_2+H_2\)
x______2x_______x_____ x
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
y____2y______y_________ y
\(\left\{{}\begin{matrix}24x+56y=8\\x+y=\frac{2,28}{22,4}=0,2\end{matrix}\right.\rightarrow x=y=0,1\)
\(\%m_{Mg}=\frac{0,1.24}{8}.100\%=30\%,\%m_{Fe}=100\%-30\%=70\%\)
\(n_{HC}=2.0,1+2.0,1=0,4\left(mol\right)\)
\(m_{dd_{HCl}}=\frac{0,4.36,5}{14,6\%}=100\left(g\right)\)
\(C\%_{MgCl2}=\frac{0,1.\left(24+71\right)}{8+100-0,1.2}.100\%=8,83\%\)
\(C\%_{FeCl_2}=\frac{0,1.\left(56+71\right)}{8+100-0,2.2}.100\%=11,8\%\)
\(Mg+X_2\rightarrow MgX_2\)
\(n_{Mg}=\frac{1,2}{24}=0,05\left(mol\right)\)
\(n_{MgX2}=\frac{4,75}{24+2X}\)
Ta có nMg=nMgX2
\(\Leftrightarrow0,05=\frac{4,75}{24+2X}\)
\(\rightarrow X=35,5\left(Cl\right)\)
