Gọi $n_{CuO}=x(mol);n_{Fe_2O_3}=y(mol)$

$CuO+H_2SO_4\to CuSO_4+H_2O\\\quad x\quad\quad\quad x\quad\quad\quad\quad x$

$Fe_2O_3+3H_2SO_4\to Fe_2(SO_4)_3+3H_2O\\\quad y\quad\quad\quad\quad 3y\quad\quad\quad\quad y$

a/ $m_{hh}=80x+160y=96g$ 

Tỉ lệ số mol 2 muối là $2:1$

$→\dfrac{x}{2}=\dfrac{y}{1}$

$→$ Ta có hệ phương trình: $\begin{cases}80x+160y=96\\\dfrac{x}{2}=\dfrac{y}{1}\end{cases}$

$↔\begin{cases}x=0,6(mol)\\y=0,3(mol)\end{cases}$ hay $\begin{cases}n_{CuO}=0,6(mol)\\n_{Fe_2O_3}=0,3(mol)\end{cases}$

$→\begin{cases}m_{CuO}=48g\\m_{Fe_2O_3}=48g\end{cases}$

$→\begin{cases}\%m_{CuO}=\dfrac{48}{96}=50\%\\m_{Fe_2O_3}=100\%-50\%=50\%\end{cases}$

b/ Sửa: $d=1,2g/ml$

$n_{H_2SO_4}=x+3y=0,6+3.0,3=1,5(mol)\\V_{dd\,H_2SO_4}=\dfrac{1,5}{2}=0,75l=75ml$

$m_{dd\,H_2SO_4}=75.1,2=90g$

Vậy $m_{dd\,H_2SO_4}=90g$