When cone. $H_{2} {SO}_{4}$ was treated with $K_{4} [Fe {(CN)}_{6}]$ , CO gas was evolved. By mistake, somebody used dilute $H_{2} {SO}_{4}$ instead of cone. $H_{2} {SO}_{4}$ then the gas evolved was :

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$N_{2}$

$HCN$

${CO}_{2}$

$CO$

answer is B.

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Detailed Solution

These are the responses:

$\begin{array}{l} K_{4} Fe (CN)_{6} + 6 H_{2} {SO}_{4} + 6 H_{2} O \to 2 K_{2} {SO}_{4} + {FeSO}_{4} + 3 {({NH}_{4})}_{2} {SO}_{4} + 6 CO \\ 3 K_{4} [Fe (CN)_{6}] + 6 H_{2} {SO}_{4} \to 12 HCN + {Fe}_{2} [Fe (CN)_{6}] + 6 K_{2} {SO}_{4} \end{array}$

HCN is the gas that developed.

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