Dung dịch X gồm AlCl3 a mol/l và Al2(SO4)3 b mol/l. Cho 400ml dung dịch X

Đáp án B.

$\text{X}\left\{\begin{array}{l}{\text{AlCl}}_{\text{3}}:0,4a\\ {\text{Al}}_{\text{2}}{\left(S{O}_4\right)}_3:0,4b\end{array}\right.\stackrel{{\text{BaCl}}_{\text{2}}}{\to }\downarrow {\text{BaSO}}_{\text{4}}:0,144$

BT số mol gốc $S{O}_4^{2-} n_{A{l}_2{\left(S{O}_4\right)}_3}=\frac{1}{3}{n}_{BaS{O}_4}=0,048mol$

$$\begin{gathered} \to b=0,12 \\ \text{X}\left\{\begin{array}{l}{\text{AlCl}}_{\text{3}}:0,4a\\ {\text{Al}}_{\text{2}}{\left(S{O}_4\right)}_3:0,4b\end{array}\right.\stackrel{\text{NaOH:0,612}}{\to }\downarrow {\text{Al(OH)}}_{\text{3}}:0,108 \\ {n}_{{\text{Al}}^{\text{3+}}}=n_{{\text{AlCl}}_{\text{3}}}+2n_{{\text{Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}}=0,4a+0,048.2=0,4a+0,096 \end{gathered}$$

$n_{\text{NaOH}}>n_{{\text{Al(OH)}}_{\text{3}}}\to$ kết tủa tạo thành đã tan một phần trong NaOH.

$A{l}^{3+}+3O{H}^{-}\to Al{\left(OH\right)}_3$

x        3x           x

$Al{\left(OH\right)}_3+O{H}^{-}\to Al{O}_2^{-}+2H_{2}O$

y              y

$$\begin{gathered} \to \left\{\begin{array}{l}x-y=0,108\\ 3x+y=0,612\end{array}\right.\to \left\{\begin{array}{l}x=0,18\\ y=0,672\end{array}\right. \\ \to 0,4a+0,096=0,18\to a=0,21 \\ \to \frac{a}{b}=1,75. \end{gathered}$$