Đáp án:
`C%_{FeCl_2}=0,63%`
`C%_{FeCl_3}=1,61%`
`C%_{HCl dư}=12,99%`
Giải thích các bước giải:
`Fe_3O_4 + 8HCl -> FeCl_2 + 2FeCl_3 + 4H_2O`
`m_{HCl}=400.14,6%=58,4(gam)`
`=> n_{HCl}=(58,4)/(36,5)=1,6(mol)`
`n_{Fe_3O_4}=(4,64)/(232)=0,02(mol)`
Ta có tỷ lệ:
`n_{Fe_3O_4}(=(0,02)/1) < n_{HCl}(=(1,6)/8)`
`=> Fe_3O_4` hết; `HCl` dư
`m_{dd sau pứ}=4,64+400=404,64(gam)`
Theo pthh:
`n_{FeCl_2}=n_{Fe_3O_4}=0,02(mol)`
`-> m_{FeCl_2}=0,02.127=2,54(gam)`
`=> C%_{FeCl_2}=(2,54)/(404,64).100%=0,63%`
`n_{FeCl_3}=2.n_{Fe_3O_4}=2.0,02=0,04(mol)`
`-> m_{FeCl_3}=0,04.162,5=6,5(gam)`
`=> C%_{FeCl_3}=(6,5)/(404,64).100%=1,61%`
`n_{HCl pứ}=8.n_{Fe_3O_4}=8.0,02=0,16(mol)`
`-> n_{HCl dư}=1,6-0,16=1,44(mol)`
`-> m_{HCl dư}=1,44.36,5=52,56`
`=> C%_{HCl dư}=(52,56)/(404,64).100%=12,99%`