a)\(Mg+2HCl-->MgCl2+H2\)
x-------------------------------x(mol)
\(Fe2O3+6HCl--.2FeCl3+3H2O\)
y-------------------------------2y(mol)
Theo bài ta có hpt
\(\left\{{}\begin{matrix}24x+160y=18,4\\95x+325y=42\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)
\(\%m_{Mg}=\frac{24.0,1}{18,4}.100\%=13,04\%\)
\(\%m_{Fe2O3}=100-13,04=86,96\%\)
b) \(n_{HCl}=2n_{Mg}=0,2\left(mol\right)\)
\(n_{HCl}=6n_{Fe2O3}=0,6\left(mol\right)\)
\(\sum n_{HCl}=0,6+0,2=0,8\left(mol\right)\)
\(m_{HCl}=0,8.36,5=29,2\left(g\right)\)
\(mdd_{HCl}=\frac{29,2.100}{7,3}=400\left(g\right)\)
\(V_{HCl}=\frac{400}{1,2}=333,33\left(ml\right)\)
