When $8.3 g$ copper sulphate reacts with excess of potassium iodine then the amount of iodine liberated is :

42.3 g

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24.3 g

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4.23 g

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2.43 g

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Solution

The correct option is C $4.23 g$
$2 C u S O_{4} \cdot 5 H_{2} O 498 g + 4 K I \to C u_{2} I_{2} + 2 K_{2} S O_{4} + I_{2} 254 g + 10 H_{2} O$
$498 g$ of $C u S O_{4}$ liberated $I_{2} = 254 g$
$8.3 g$ of $C u S O_{4}$ liberated $I_{2} = \frac{254}{498} \times 8.3$
$= 4.23 g$

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