Al_{2}O_{3} was leached with alkali to get X. | vedclass

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(B) In the Bayer's process,$Al_{2}O_{3}$ is leached with $NaOH$ to form sodium aluminate,$X = Na[Al(OH)_{4}]$.Passing $CO_{2}$ gas $(Y)$ through the s

Vedclass · Accepted Answer

$X = Na[Al(OH)_{4}], Y = CO_{2}, Z = Al_{2}O_{3} \cdot xH_{2}O$

Vedclass · Answer

(B) In the Bayer's process,$Al_{2}O_{3}$ is leached with $NaOH$ to form sodium aluminate,$X = Na[Al(OH)_{4}]$.Passing $CO_{2}$ gas $(Y)$ through the solution of $X$ results in the precipitation of hydrated alumina,$Z = Al_{2}O_{3} \cdot xH_{2}O$.The chemical reactions are:$Al_{2}O_{3}(s) + 2NaOH(aq) + 3H_{2}O(l) \rightarrow 2Na[Al(OH)_{4}](aq)$$2Na[Al(OH)_{4}](aq) + CO_{2}(g) \rightarrow Al_{2}O_{3} \cdot xH_{2}O(s) + 2NaHCO_{3}(aq)$

$Al_{2}O_{3}$ was leached with alkali to get $X$. The solution of $X$ on passing of gas $Y$,forms $Z$. $X$,$Y$ and $Z$ respectively are

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