The reaction which proceeds in the forward direction is:

F e_{3} O_{4} + 6 H C l \to 2 F e C l_{3} + 3 H_{2} O

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N H_{3} + H_{2} O + N a C l \to N H_{4} C l + N a O H

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S n C l_{4} + H g_{2} C l_{2} \to S n C l_{2} + 2 H g C l_{2}

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2 C u I + I_{2} + 4 K^{+} \to 2 C u^{2 +} + 4 K I

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Solution

The correct option is A $F e_{3} O_{4} + 6 H C l \to 2 F e C l_{3} + 3 H_{2} O$
$F e C l_{3}$ and $H_{2} O$ do not react to give back $F e_{3} O_{4}$ and $H C l$ .

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Similar questions

2 F e C l_{3} + S n C l_{2} \to 2 F e C l_{2} + S n C l_{4}

2 F e C l_{3} + S n C l_{2} \to 2 F e C l_{2} + S n C l_{4}

2 F e C l_{3} + S n C l_{2} \to 2 F e C l_{2} + S n C l_{4}

N H_{4} C l + N a O H ⇌ H_{2} O + N H_{3} + N a C l

S n C l_{2} + 2 F e C l_{3} \to S n C l_{4} + 2 F e C l_{2}

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