a) \(Mg+2HCl--.MgCl2+H2\)
x-------------------------x(mol)
\(Fe2O3+6HCl-->2FeCl3+3H2O\)
y----------------------------2y(mol)
Theo bài ta có hpt
\(\left\{{}\begin{matrix}24x+160y=43,6\\95x+325y=95,5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,15\\y=0,25\end{matrix}\right.\)
\(\%m_{Mg}=\frac{0,15.24}{43,6}.100\%=8,26\%\)
\(\%m_{Fe2O3}=100-8,96=91,04\%\)