Dung dịch X gồm $AlC{{l}_{3}}$ a mol/l và $A{{l}_{2}}{{\left(...

$\text{X}\left\{ \begin{aligned}
& \text{AlC}{{\text{l}}_{\text{3}}}:0,4a \\
& \text{A}{{\text{l}}_{\text{2}}}{{\text{(S}{{\text{O}}_{\text{4}}}\text{)}}_{\text{3}}}:0,4b \\
\end{aligned} \right.\xrightarrow{\text{BaC}{{\text{l}}_{\text{2}}}}\downarrow \text{BaS}{{\text{O}}_{\text{4}}}:0,144$
BT số mol gốc $SO_{4}^{2-}$ ${{n}_{A{{l}_{2}}{{(S{{O}_{4}})}_{3}}}}=\dfrac{1}{3}{{n}_{BaS{{O}_{4}}}}=0,048mol$
$\to b=0,12$
$\text{X}\left\{ \begin{aligned}
& \text{AlC}{{\text{l}}_{\text{3}}}:0,4a \\
& \text{A}{{\text{l}}_{\text{2}}}{{\text{(S}{{\text{O}}_{\text{4}}}\text{)}}_{\text{3}}}:0,4b \\
\end{aligned} \right.\xrightarrow{\text{NaOH:0,612}}\downarrow \text{Al(OH}{{\text{)}}_{\text{3}}}:0,108$
${{n}_{\text{A}{{\text{l}}^{\text{3+}}}}}={{n}_{\text{AlC}{{\text{l}}_{\text{3}}}}}+2{{n}_{\text{A}{{\text{l}}_{\text{2}}}{{\text{(S}{{\text{O}}_{\text{4}}}\text{)}}_{\text{3}}}}}=0,4a+0,048.2=0,4a+0,096$
${{n}_{\text{NaOH}}}>{{n}_{\text{Al(OH}{{\text{)}}_{\text{3}}}}}\to $ kết tủa tạo thành đã tan một phần trong NaOH.
$\underset{x}{\mathop{A{{l}^{3+}}}} +\underset{3x}{\mathop{3O{{H}^{-}}}} \to \underset{x}{\mathop{Al{{(OH)}_{3}}}} $
$\underset{y}{\mathop{Al{{(OH)}_{3}}}} +\underset{y}{\mathop{O{{H}^{-}}}} \to AlO_{2}^{-}+2{{H}_{2}}O$
$\to \left\{ \begin{aligned}
& x-y=0,108 \\
& 3x+y=0,612 \\
\end{aligned} \right.\to \left\{ \begin{aligned}
& x=0,18 \\
& y=0,672 \\
\end{aligned} \right.$
$\to 0,4a+0,096=0,18\to a=0,21$
$\to \dfrac{a}{b}=1,75.$