When Zn is treated with excess of NaOH,the pr | vedclass

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Question

(C) When $Zn$ reacts with an excess of $NaOH$ solution,it forms sodium zincate and releases hydrogen gas. The reaction is: $Zn + 2NaOH + 2H_2O \longri

Vedclass · Accepted Answer

$Na_{2}ZnO_{2}$

Vedclass · Answer

(C) When $Zn$ reacts with an excess of $NaOH$ solution,it forms sodium zincate and releases hydrogen gas. The reaction is: $Zn + 2NaOH + 2H_2O \longrightarrow Na_2[Zn(OH)_4] + H_2$ This can also be represented as: $Zn + 2NaOH \longrightarrow Na_2ZnO_2 + H_2$ Thus,the product obtained is sodium zincate $(Na_2ZnO_2)$.

When $Zn$ is treated with excess of $NaOH$,the product obtained is

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