`a)`

`Fe+2HCl->FeCl_2+H_2`

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

`b)`

`n_{H_2}={4,48}/{22,4}=0,2(mol)`

BTKL: `m+m_{dd\ HCl}=m_{dd\ sau}+m_{H_2}`

`->m=368,8+0,2.2-350=19,2(g)`

Theo pt:

`n_{Fe}=n_{FeCl_2}=n_{H_2}=0,2(mol)`

`n_{Fe_2O_3}={19,2-0,2.56}/{160}=0,05(mol)`

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

`n_{HCl\ pu}=2n_{Fe}+6n_{Fe_2O_3}=0,7={350.7,3\%}/{36,5}(mol)`

`->HCl` hết

`C\%_{FeCl_2}={0,2.127}/{368,8}.100\%=6,89\%`

`C\%_{FeCl_3}={0,1.162,5}/{368,8}.100\%=4,41\%`