X reacts with aqueous NaOH to form Y and rele | vedclass

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(D) $1$. The reaction of $Al$ with aqueous $NaOH$ produces sodium aluminate $(NaAlO_2)$ and releases $H_2$ gas: $2Al + 2NaOH + 2H_2O \rightarrow 2NaAl

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$Al, NaAlO_2, Al(OH)_3$

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(D) $1$. The reaction of $Al$ with aqueous $NaOH$ produces sodium aluminate $(NaAlO_2)$ and releases $H_2$ gas: $2Al + 2NaOH + 2H_2O \rightarrow 2NaAlO_2 + 3H_2$. Thus,$X = Al$ and $Y = NaAlO_2$.$2$. When $CO_2$ is passed through the aqueous solution of $NaAlO_2$ $(Y)$,it forms aluminum hydroxide $(Al(OH)_3)$ as a precipitate: $2NaAlO_2 + 3H_2O + CO_2 \rightarrow 2Al(OH)_3 \downarrow + Na_2CO_3$. Here,$Z = Al(OH)_3$.$3$. Heating $Al(OH)_3$ $(Z)$ at $1200 \ ^\circ C$ results in the formation of alumina $(Al_2O_3)$: $2Al(OH)_3 \xrightarrow{1200 \ ^\circ C} Al_2O_3 + 3H_2O$.$4$. Therefore,the correct sequence is $X = Al, Y = NaAlO_2, Z = Al(OH)_3$.

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