Hộ mik nốt lần này thuii

..

`a)`

`n_{HCl}={200.14,6\%}/{36,5}=0,8(mol)`

Đặt `n_{Fe_2O_3}=x(mol);n_{ZnO}=y(mol)`

`->160x+81y=24,1(1)`

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

`ZnO+2HCl->ZnCl_2+H_2O`

Theo PT: `6x+2y=n_{HCl}=0,8(2)`

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

`\%m_{Fe_2O_3}={160.0,1}/{24,1}.100\%\approx 66,39\%`

`\%m_{ZnO}=100-66,39=33,61\%`

`b)`

Theo PT: `n_{FeCl_3}=2x=0,2(mol);n_{ZnCl_2}=y=0,1(mol)`

`m_{dd\ spu}=24,1+200=224,1(g)`

`C\%_{FeCl_3}={0,2.162,5}/{224,1}.100\%\approx 14,5\%`

`C\%_{ZnCl_2}={0,1.136}/{224,1}.100\%\approx 6,07\%`