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`a)`

`Fe+2HCl->FeCl_2+H_2`

`FeO+2HCl->FeCl_2+H_2O`

`FeCO_3+2HCl->FeCl_2+CO_2+H_2O`

`->n_{FeCl_2}={31,75}/{127}=0,25(mol)`

Theo PT: `n_{HCl}=2n_{FeCl_2}=0,5(mol)`

`->V_{dd\ HCl}={0,5}/{0,5}=1(l)`

`b)`

Sửa `185->18,5`

`M_B=18,5.M_{H_2}=37`

`->{n_{CO_2}}/{n_{H_2}}={37-2}/{44-37}=5`

`->n_{CO_2}=5n_{H_2}`

Đặt `n_{Fe}=x(mol);n_{FeO}=y(mol);n_{FeCO_3}=z(mol)`

`->56x+72y+116z=21,6(1)`

Theo PT: `n_{H_2}=x(mol);n_{CO_2}=z(mol)`

`->5x=z(2)`

Theo PT: `n_{FeCl_2}=x+y+z=0,25(3)`

`(1)(2)(3)->x=3/{170};y={49}/{340};z=3/{34}`

`\%m_{Fe}={3/{170}.56}/{21,6}.100\%\approx 4,575\%`

`\%m_{FeO}={{49}/{340}.72}/{21,6}.100\%\approx 48,04\%`

`\%m_{FeCO_3}=100-4,575-48,04=47,385\%`