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Metal Oxide and Hydroxide

Excess of N...

Question

Excess of $N a O H$ reacts with $Z n$ to form:

Z n {(O H)}_{2}

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Z n O

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Z n H_{2}

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N a_{2} Z n O_{2}

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Solution

The correct option is C $N a_{2} Z n O_{2}$
Excess of sodium hydroxide reacts with zinc to form sodium zincate along with liberation of hydrogen.

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

Option D is correct.

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