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Question

(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

Vedclass · Accepted Answer

$Al, NaAlO_2, Al(OH)_3$

Vedclass · Answer

(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 $X = Al, Y = NaAlO_2, Z = Al(OH)_3$.

Similar Questions

What is the correct formula for Borazole?

$B(OH)_3 + NaOH \to NaBO_2 + Na[B(OH)_4] + H_2O$
How can this reaction be made to proceed in the forward direction?

The incorrect statement regarding the above reactions is:
$Al \xrightarrow{HCl(aq.)} 'X' + \text{Gas } 'P'$
$Al \xrightarrow{NaOH(aq.) + H_2O} 'Y' + \text{Gas } 'Q'$

Which one of the following is the correct statement?

Which of the following statements is correct regarding boron hydrides?

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