$$\begin{gathered} nA{l}^{3+}=0,2 mol; nN{a}^{+}=0,1 mol; nSO{4}^{2-}=0,4 mol; \\ nB{a}^{2+}=0,375 mol; nO{H}^{-}=0,75 mol \\ A{l}^{3+}+3O{H}^{-}->Al\left(OH\right)3 \\ Al\left(OH\right)3+O{H}^{-}->AlO{2}^{-}+2H2O \\ B{a}^{2+}+SO{4}^{2-}->BaSO4 \end{gathered}$$

nBaSO4=nBa(2+)=0,375 mol (do nBa2+ < nSO4 (2-))

=>mBaSO4=87,375 g

$$\begin{gathered} nAl\left(OH\right)3=nA{l}^{3+}=0,2 mol \\ =>nO{H}^{-} (tạo Al\left(OH\right)3)=0,6 mol \\ =>nO{H}^{-} tạo AlO{2}^{-}=0,75-0,6=0,15 mol \\ =>nAl\left(OH\right)3 bị hòa \tan =0,15 mol=>nAl\left(OH\right)3 dư=0,05 mol \\ =>mAL\left(OH\right)3=3,9g \\ =>m kết tủa=87,375+3,9=91,275 g \end{gathered}$$

V sau=0,1+0,15=0,25 (lít)

$B: \left\{\begin{array}{l}AlO{2}^{-}:0,15 mol=>CM=0,6 M\\ SO{4}^{2-} dư: 0,4-0,375=0,025 mol=>CM=0,1 M\\ N{a}^{+}: 0,1 mol=>CM=0,4 M\end{array}\right.$