Question

Identify X, Y, and Z in the given reaction sequence.

$\mathrm{Al}+\mathrm{NaOH}\to \mathrm{X}\stackrel{\mathrm{Y}\left(\mathrm{g}\right)}{\to }\mathrm{Z}$

• A. $\mathrm{X}=\mathrm{Na}\left[\mathrm{Al}{\left(\mathrm{OH}\right)}_4\right];\mathrm{Y}={\mathrm{CO}}_2;\mathrm{Z}={\mathrm{Al}}_2{\mathrm{O}}_3.{\mathrm{xH}}_2\mathrm{O}$
• B. $\mathrm{X}=\mathrm{Na}\left[\mathrm{Al}{\left(\mathrm{OH}\right)}_4\right];\mathrm{Y}={\mathrm{SO}}_2;\mathrm{Z}={\mathrm{Al}}_2{\mathrm{O}}_3.{\mathrm{xH}}_2\mathrm{O}$
• C. $\mathrm{X}=\mathrm{Al}{\left(\mathrm{OH}\right)}_3;\mathrm{Y}={\mathrm{CO}}_2;\mathrm{Z}={\mathrm{Al}}_2{\mathrm{O}}_3$
• D. $\mathrm{X}=\mathrm{Al}{\left(\mathrm{OH}\right)}_3;\mathrm{Y}={\mathrm{SO}}_2;\mathrm{Z}={\mathrm{Al}}_2{\mathrm{O}}_3$

Answer 1

The correct option is A

$\mathrm{X}=\mathrm{Na}\left[\mathrm{Al}{\left(\mathrm{OH}\right)}_4\right];\mathrm{Y}={\mathrm{CO}}_2;\mathrm{Z}={\mathrm{Al}}_2{\mathrm{O}}_3.{\mathrm{xH}}_2\mathrm{O}$

Explanation for the correct option:

Option A: Bayer's method (Leaching method):

In this method, $\mathrm{NaOH}$ solution is taken as leachate and Red Bauxite $\left(A{l}_3{O}_3.2H_{2}O\right)$ being Amphoteric is dissolved in $\mathrm{NaOH}$ solution and $F{e}_2{O}_3$ impurity being basic remains insoluble in NaOH solution.

$$\begin{gathered} \underset{\mathrm{Aluminium}\mathrm{Oxide}}{{\mathrm{Al}}_2{\mathrm{O}}_3·2{\mathrm{H}}_2\mathrm{O}}+\underset{\mathrm{Lechate}}{2\mathrm{NaOH}}+{\mathrm{H}}_2\mathrm{O}\to \underset{\mathrm{Sodium}\mathrm{meta}\mathrm{aluminat}}{2\left({\mathrm{NaAlO}}_2+2{\mathrm{H}}_2\mathrm{O}\right)}\text{or}\underset{{\mathrm{OH}}^{-}\left(\mathrm{Base}\right)}{2\mathrm{Na}\left[\mathrm{Al}(\mathrm{OH}{)}_4\right]} \\ \stackrel{\mathrm{equilibrium}\mathrm{shift}\mathrm{forward}\uparrow \downarrow {\mathrm{CO}}_2(\mathrm{weak}\mathrm{acid})\mathrm{equilibrium}\mathrm{shift}\mathrm{forward}}{3{\mathrm{H}}_2\mathrm{O}+\underset{(\text{a -crystalline form)}}{{\mathrm{Al}}_2{\mathrm{O}}_3}\stackrel{\mathrm{Al}(\mathrm{OH}{)}_3\downarrow }{⟵}} \\ {\left[\mathrm{Al}(\mathrm{OH}{)}_4\right]}^{-}\underset{{\mathrm{OH}}^{-}\left(\text{Excess}\mathrm{NaOH}\right)}{\stackrel{H^{+}\left(\text{tewdrop HCl}\right)}{\rightleftharpoons }}\mathrm{Al}(\mathrm{OH}{)}_2\downarrow \stackrel{\text{Excess HCl}}{⟶}\underset{\text{(Soluble salt solution)}}{{\mathrm{AlCl}}_3} \end{gathered}$$

It is the process of the concentration of ore by the leaching method.

Final Answer: Hence, the correct answer is option (A).

$\mathrm{X}=\mathrm{Na}\left[\mathrm{Al}{\left(\mathrm{OH}\right)}_4\right];\mathrm{Y}={\mathrm{CO}}_2;\mathrm{Z}={\mathrm{Al}}_2{\mathrm{O}}_3.{\mathrm{xH}}_2\mathrm{O}$