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

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

Step 2: Calculating the number of moles of $I C l$ and $I C l_{3}$
The formula for calculation of the number of moles
$n = \frac{m}{M}$
where $n =$ number of moles
$m =$ given mass of the compound
$M =$ molar mass of the compound

Molar mass of $I_{2} = 127 \times 2 = 254 g$
Molar mass of $C l_{2} = 2 \times 35.5 = 71 g$

$n_{(I_{2})} = \frac{25.4 g}{254 g} = 0.1$

$n_{(C l_{2})} = \frac{14.2 g}{71 g} = 0.2$

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25.4g of $I_{2}$ and 14.2g of $C l_{2}$ are made to react completely to yield a mixture of $I C l$ and $I C l_{3}$ .Calculate moles of $I C l$ and $I C l_{3}$ formed.

25.4 g of iodine and 7.1 g chlorine are made to react to give a mixture of ICl and $I C l_{3}$ . Number of moles of ICl and $I C l_{3}$ formed are respectively ?