`a)`

Đặt `n_{CuO}=x(mol);n_{FeO}=y(mol)`

`->80x+72y=11,6(1)`

`CuO+2HCl->CuCl_2+H_2O`

`FeO+2HCl->FeCl_2+H_2O`

Theo PT: `n_{HCl}=2x+2y={200.5,475\%}/{36,5}=0,3(2)`

`(1)(2)->x=0,1(mol);y=0,05(mol)`

`->\%m_{CuO}={0,1.80}/{11,6}.100\%\approx 68,97\%`

`->\%m_{FeO}=100-68,97=31,03\%`

`b)`

Theo PT: `n_{CuCl_2}=x=0,1(mol);n_{FeCl_2}=y=0,05(mol)`

`m_{dd\ spu}=11,6+200=211,6(g)`

`->C\%_{CuCl_2}={0,1.135}/{211,6}.100\%\approx 6,38\%`

        `C\%_{FeCl_2}={0,05.127}/{211,6}.100\%\approx 3\%`

`c)`

`2NaOH+CuCl_2->Cu(OH)_2+2NaCl`

`2NaOH+FeCl_2->Fe(OH)_2+2NaCl`

`Cu(OH)_2`  $\xrightarrow{t^o}$  `CuO+H_2O`

`4Fe(OH)_2+O_2`  $\xrightarrow{t^o}$  `2Fe_2O_3+4H_2O`

Theo PT: `n_{CuO}=x=0,1(mol);n_{Fe_2O_3}=0,5y=0,025(mol)`

`->m_{CR}=0,1.80+0,025.160=12(g)`