`a)`

`n_{MgSO_4}={12}/{120}=0,1(mol)`

`MgSO_4+2NaOH->Mg(OH)_2+Na_2SO_4`

`0,1->0,2->0,1->0,1(mol)`

`m_{dd\ NaOH}={0,2.40}/{10\%}=80(g)`

`b)`

`m_{dd\ sau}=12+80-0,1.58=86,2(g)`

`C\%_{Na_2SO_4}={0,1.142}/{86,2}.100\%=16,47\%`

`c)`

`Na_2SO_4+Ba(NO_3)_2->BaSO_4+2NaNO_3`

`0,1->0,1->0,1(mol)`

`m_{BaSO_4}=0,1.233=23,3(g)`

nMgSO₄=12/120 =0,1(mol)
PTHH : MgSO₄+2NaOH---->Mg(OH)₂ +Na₂SO₄ (1)
                            0,1           0,2              0,1              0,1    /mol
từ ptrinh 1 => nNaOH=0,2(mol)
                  =>mNaOH=0,2x40 =8(g)
C%=m_ct : mdd x100%
10%=8: mdd x100% => mdd_NaOH =80(g)
từ ptrinh 1 => nNa₂SO₄ =0,1(mol)
                  =>mNa₂SO₄= 0,1 x 142 = 14,2(g)
Bảo toàn khối lượng:
 mdd sau pứ = mMgSO₄ +mNaOH +mNa₂SO₄ - mMg(OH)₂
=> 12 + 80 +  14,2 – (0,1x58) =100,4(g)
=>%mNa₂SO₄ = 14,2/100,4 x100% = xấp xỉ 14,14%
PTHH :  Na₂SO₄ + Ba(NO₃)₂---> BaSO₄ +2NaNO3(2)
                   0,1                                  0,1                  /mol
Từ ptrinh (2) => nBaSO₄ =0,1 (mol)
                      =>mBaSO₄=0,1x233 = 23,3(g)
         
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