Question

\( \mathrm{Al_2O_3} \) was leached with alkali to get \( X \). The solution of \( X \), on passing the gas \( Y \), forms \( Z \). \( X, Y \) and \( Z \) respectively are :

• A. X = Na[Al(OH)_4], Y = SO_2, Z = Al_2O_3
• B. X = Al(OH)_3, Y = SO_2, Z = Al_2O_3 . xH_2O
• C. X = Al(OH)_3, Y = CO_2, Z = Al_2O_3
• D. X = Na[Al(OH)_4], Y = CO_2, Z = Al_2O_3 . xH_2O

Hint:

This is Baeyer's Process. $CO_2$ is used specifically because it is a weak acid that precipitates hydrated alumina without introducing new metallic impurities.

Answer 1

The Correct Option is D

Step 1: Leaching of Bauxite: $Al_2O_3 + NaOH \to Na[Al(OH)_4]$ (X).
Step 2: Neutralization with $CO_2$ (Y): $2Na[Al(OH)_4] + CO_2 \to Al_2O_3 \cdot xH_2O$ (Z) $+ 2NaHCO_3$.
Step 3: Hydrated alumina (Z) is then filtered and heated to give pure alumina.