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Preparations and Uses of Caustic Soda

When Zn is ...

Question

When $Z n$ is treated with excess of $N a O H,$ the product obtained is :

Z n (O H)_{2}

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Z n O H

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N a_{2} Z n O_{2}

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none of these

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Solution

The correct option is C $N a_{2} Z n O_{2}$
When $Z n$ reacts with excess of $N a O H$ , the following reaction takes place:

$Z n + N a O H \to N a_{2} Z n O_{2} + H_{2}$

Hence, option $C$ is correct.

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