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$\text{Gọi x, y, z lần lượt là số mol }$ $Fe_{2}O_{3},MgO,CuO$ 

\(PTHH:Fe_3O_4+4H_2⇌3Fe+4H_2O\left(1\right)\)

________x______________3x_________mol

\(MgO+H_2\rightarrow Mg+H_2O\left(2\right)\)

y____________y________mol$

$\(CuO+H_2\rightarrow Cu+H_2O\left(3\right)\)$

z___________z _________mol

$Từ (1),(2),(3) :$

\(\Rightarrow\left\{{}\begin{matrix}232x+40y+80z=66,8\left(+\right)\\168x+40t+64z=52,4\left(++\right)\end{matrix}\right.\)

\(Fe_3O_4+8HCl\rightarrow FeCl_2+2FeCl_2+4H_2O\left(4\right)\)

x_______8x____________ mol

\(MgO+2HCl\rightarrow MgCl_2+H_2O\left(5\right)\)

y _______2y _____________mol

\(CuO+2HCl\rightarrow CuCl_2+H_2O\left(6\right)\)

z___2z________________________ mol

\(n_{HCl}=1,1\left(mol\right)\)

$Từ 4,5,6 :$

\(\left\{{}\begin{matrix}x+y+z=0,325\\8x+2y+2z=1,1\end{matrix}\right.\)

\(\frac{8x+2y+2z}{x+y+z}=\frac{1,1}{0,325}\)

$\text{Ta có tỉ lệ: }$

\(\Leftrightarrow8x+2y+2z=\frac{44}{13x}+\frac{44}{13y}+\frac{44}{13z}\)

\(\Rightarrow\frac{60}{13x}-\frac{18}{13y}-\frac{18}{13z}=0\left(+++\right)\)

$Từ (+), (++), (+++):$

\(\Rightarrow\left\{{}\begin{matrix}x=0,15\left(mol\right)\\y=0,2\left(mol\right)\\z=0,3\left(mol\right)\end{matrix}\right.\)