Write the balanced chemical equations for the following reactions.
Copper sulphate on treatment with potassium iodide precipitates cuprous iodide $(C u I)$ , liberates iodine gas and also forms potassium sulphate.

2 C u S O_{4} + 4 K I \to C u_{2} I_{2} + 2 K_{2} S O_{4} + 2 I_{2}

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2 C u S O_{4} + 2 K I \to 2 C u I + I_{2} + 2 K_{2} S O_{4}

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C u S O_{4} + 4 K I \to C u I + 2 I_{2} + 2 K_{2} S O_{4}

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2 C u S O_{4} + 4 K I \to 2 C u I + I_{2} + 2 K_{2} S O_{4}

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Solution

The correct option is D $2 C u S O_{4} + 4 K I \to 2 C u I + I_{2} + 2 K_{2} S O_{4}$
$2 C u S O_{4} + 4 K I ⟶ 2 C u I + I_{2} + 2 K_{2} S O_{4}$

No. of atom on reactant side

No. of atom on product side

$C u$

$2$

$2$

$S$

$2$

$2$

$O$

$8$

$8$

$K$

$4$

$4$

$I$

$4$

$4$

Hence, option D is correct.

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