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Preparations and Uses of Caustic Soda

Element A whi...

Question

Element A which belongs to group 12 and period 4 is a bluish white metal. A reacts with conc. $H_{2} S O_{4}$ and gives B with the liberation of $S O_{2}$ . With $N a O H$ . A gives compound C. Identify A, B and C. Explain the reaction.

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Solution

$A \to Z i n c (Z n)$
$B \to Z i n c s u l p h a t e (Z n S O_{4})$
$C \to S o d i u m Z i n c a t e (N a_{2} Z n O_{2})$
Reaction:
Zinc reacts with hot con. $H_{2} S O_{4}$ formating $Z n S O_{4}$
$Z n (A) + 2 H_{2} S O_{4} \to Z n S O_{4} (B) + S O_{2} ↑ + 2 H_{2} O$
Zinc dissolves in hot NaOH solution formating soluble zincate ion.
$Z n + 2 N a O H \to N a_{2} Z n O_{2} S o d i u m Z i n c a t e (C) + H_{2}$

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