\(\begin{array}{l}
a)\\
CuO+2HCl\to CuCl_2+H_2O\\
Fe_2O_3+6HCl\to 2FeCl_3+3H_2O\\
b)\\
Dat\,n_{CuO}=x(mol);n_{Fe_2O_3}=y(mol)\\
\to 80x+160y=32(1)\\
n_{HCl}=\frac{200.18,25\%}{36,5}=1(mol)\\
Theo\,PT:\,2x+6y=n_{HCl}=1(2)\\
(1)(2)\to x=0,2(mol);y=0,1(mol)\\
\to m_{CuO}=0,2.80=16(g);m_{Fe_2O_3}=32-16=16(g)\\
c)\\
Theo\,PT:\,n_{CuCl_2}=x=0,2(mol);n_{FeCl_3}=2y=0,2(mol)\\
m_{dd\,spu}=32+200=232(g)\\
\to\begin{cases} C\%_{CuSO_4}=\frac{0,2.135}{232}.100\%≈11,64\%\\ C\%_{FeCl_3}=\frac{0,2.162,5}{232}.100\%≈14\%\end{cases}
\end{array}\)

`n_{HCl}=(200.18,25\%)/(36,5)=1(mol)`

Gọi `n_{CuO}=a;n_{Fe_2O_3}=b`

Ta có : `80a+160b=32(1)`

`a,`

PTHH:

`CuO+2HCl->CuCl_2+H_2O(1)`

`Fe_2O_3+6HCl->2FeCl_3+3H_2O(2)`

`b,`

Từ PTHH:

`2a+6b=n_{HCl}=1(2)`

Từ (1) và (2) ta suy ra :

`a=0,2;b=0,1`

`m_{CuO}=0,2.80=16(g)`

`m_{Fe_2O_3}=0,1.160=16(g)`

`c,`

(đề : tính nồng độ % các chất có trong dung dịch sau phản ứng)

Theo PTHH :

`n_{CuCl_2}=n_{CuO}=0,2(mol)`

`=>m_{CuCl_2}=0,2.135=27(g)`

`n_{FeCl_3}=2n_{Fe_2O_3}=2.0,1=0,2(mol)`

`=>m_{FeCl_3}=0,2.162,5=32,5(g)`

`m_{dd (sau p//ứ)}=32+200=232(g)`

`C\%_{CuCl_2}=(27)/(232).100\%≈11,64\%`

`C\%_{FeCl_3}=(32,5)/(232).100\%≈14,01\%`