`a)`

Mỗi phần `m_A={12}/2=6(g)`

P1:

`MgSO_4+BaCl_2->BaSO_4+MgCl_2`

`->n_{MgSO_4}=n_{BaCl_2}={873,75.10^{-3}}/{233}=0,00375(mol)`

P2:

`MgCl_2+2KOH->Mg(OH)_2+2KCl`

`MgSO_4+2KOH->Mg(OH)_2+K_2SO_4`

`->n_{MgCl_2}=n_{Mg(OH)_2}-n_{MgSO_4}`

`={1,3775}/{58}-0,00375=0,02(mol)`

Ban đầu:

`n_{MgCl_2}=n_{KCl.MgCl_2.xH_2O}=0,02.2=0,04(mol)`

`m_{KCl.MgCl_2.xH_2O}=12-0,00375.120.2=11,1(g)`

`->M_{KCl.MgCl_2.xH_2O}=169,5+18x={11,1}/{0,04}=277,5`

`->x=6`

`->KCl.MgCl_{2}.6H_2O`

`b)`

`48g\ A` gồm `n_{MgCl_2}=0,02.6=0,12(mol);n_{MgSO_4}=0,00375.6=0,0225(mol)`

`m_{dd\ D}=48+152=200(g)`

`C\%_{MgCl_2}={0,12.95}/{200}.100\%=5,7\%`

`C\%_{MgSO_4}={0,0225.120}/{200}.100\%=1,35\%`