$25.4$ g of iodine and $14.2$ g of chlorine are made to react completely to yield a mixture of $I C l$ and $I C l_{3}$ . Calculate the number of moles of $I C l$ and $I C l_{3}$ formed.

0.1 mole, 0.1 mole

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0.1 mole, 0.2 mole

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0.5 mole, 0.5 mole

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0.2 mole, 0.2 mole

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Solution

The correct option is A 0.1 mole, 0.1 mole
$I_{2} + 2 C l_{2} \to I C l + I C l_{3}$

Molar mass of $I_{2}$ $= 127$

Molar mass of $C l_{2}$ $= 70.9$

Moles of $I_{2}$ used = $\frac{25.4}{127}$ = 0.2 moles of $I_{2}$

Moles of $C l_{2}$ used = $\frac{14.2}{70.9}$ = 0.2 moles of $C l_{2}$

Since the stoichiometry indicates that each mole of $I_{2}$ reacts with 2 moles of $C l_{2}$ to give 1 mole each of $I C l$ and $I C l_{3}$ .
Chlorine is limiting reagent, in this case.

Hence 0.2-mole chlorine will react with 0.1 moles of iodine to form 0.1 mole each of the product.

Hence, the correct option is $A$

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25.4 g of iodine and 14.2 g of chlorine are made to react completely to yield a mixture of ICl and $I C l_{3}$ . Calculate the number of moles of ICl and $I C l_{3}$ is formed

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