Solve the following

Question:

$\mathrm{Al}_{2} \mathrm{O}_{3}$ was leached with alkali to get $\mathrm{X}$. The solution of $\mathrm{X}$ on passing of gas $\mathrm{Y}$ , forms Z. X, Y and $Z$ respectively are :

  1. $\mathrm{X}=\mathrm{Na}\left[\mathrm{Al}(\mathrm{OH})_{4}\right], \mathrm{Y}=\mathrm{CO}_{2}, \mathrm{Z}=\mathrm{Al}_{2} \mathrm{O}_{3} \cdot \mathrm{xH}_{2} \mathrm{O}$

  2. $\mathrm{X}=\mathrm{Na}\left[\mathrm{Al}(\mathrm{OH})_{4}\right], \mathrm{Y}=\mathrm{SO}_{2}, \mathrm{Z}=\mathrm{Al}_{2} \mathrm{O}_{3}$

  3. $\mathrm{X}=\mathrm{Al}(\mathrm{OH})_{3}, \mathrm{Y}=\mathrm{SO}_{2}, \mathrm{Z}=\mathrm{Al}_{2} \mathrm{O}_{3} \cdot \mathrm{xH}_{2} \mathrm{O}$

  4. $\mathrm{X}=\mathrm{Al}(\mathrm{OH})_{3}, \mathrm{Y}=\mathrm{CO}_{2}, \mathrm{Z}=\mathrm{Al}_{2} \mathrm{O}_{3}$

Correct Option: 1

Solution:

(1) $\mathrm{Al}_{2} \mathrm{O}_{3}+\mathrm{NaOH} \longrightarrow \mathrm{Na}\left[\mathrm{Al}(\mathrm{OH})_{4}\right]$

(2) $\mathrm{Na}\left[\mathrm{Al}(\mathrm{OH})_{4}\right] \underset{\text { "Y" }}{\stackrel{\mathrm{CO}_{2}}{\longrightarrow}} \mathrm{Al}(\mathrm{OH})_{3}$ or $\mathrm{Al}_{2} \mathrm{O}_{3}, \mathrm{xH}_{2} \mathrm{O}$

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