One mole of acidfied potassium dichromate on reaction with excess $K I$ will liberate ………….. mole(s) of $I_{2}$ .

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Solution

The correct option is A 3
When potassium iodide reacts with acidic solution of potassium dichromate, potassium dichromate liberates $I_{2}$ from $K I .$
$K_{2} C r_{2} O_{7} + 7 H_{2} S O_{4} + 6 K I \to 4 K_{2} S O_{4} + C r_{2} (S O_{4})_{3} + 3 I_{2} + 7 H_{2} O$
Thus, One mole of acidfied potassium dichromate on reaction with excess $K I$ will liberate three mole(s) of $I_{2}$ .

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