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Sulphuric Acid

Conc. H2SO4...

Question

Conc. $H_{2} S O_{4}$ can not be used to prepare pure $H B r$ from $N a B r$ as it reacts slowly with $N a B r$ .

True

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False

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Solution

The correct option is B False
$H B r$ cannot be prepared from $N a B r$ by using conc. $H_{2} S O_{4}$ because $H B r$ is strong reducing agent and reduces $H_{2} S O_{4}$ to $S O_{2}$ and get itself oxidised to $B r_{2}$ .
$N a B r + H_{2} S O_{4} \to N a H S O_{4} + H B r$
$H_{2} S O_{4} + 2 H B r \to S O_{2} + B r_{2} + 2 H_{2} O$

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H_{2} S O_{4}

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F e + B r_{2} \to F e B r_{2}

F e B r_{2} + B r_{2} \to F e_{3} B r_{8}

F e_{3} B r_{8} + N a_{2} C O_{3} \to N a B r + C O_{2} + F e_{3} O_{4}

F e

4.12 \times 10^{3} k g N a B r

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