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Mole Concept

25.4 g of I 2...

Question

$25.4 g$ of $I_{2}$ and $14.2 g$ of $C l_{2}$ 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 $I C l_{3}$ . (Given: Molar mass of $I$ = $127$ $g / m o l$ )

0.2, 0.2

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0.1, 0.1

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0.1, 0.3

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0.3, 0.1

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Solution

The correct option is B $0.1, 0.1$
$I_{2} + 2 C l_{2} \to I C l + I C l_{3}$
Number of moles of $I_{2} = \frac{25.4 g}{254 g/mol} = 0.1 mol$
1 mol of $I_{2}$ produces 1 mol of $I C l$ and 1 mol of $I C l_{3}$
0.1 mol of $I_{2}$ produces 0.1 mol of $I C l$ and 0.1 mol of $I C l_{3}$

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Similar questions

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

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are made to react completely to yield a mixture of

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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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25.4 g of I2 and 14.2 g of Cl2 are made to react completely to yield a mixture of ICl and ICl3. Calculate the number of moles of ICl and ICl3. (Given: Molar mass of I= 127 g/mol)

The correct option is B 0.1, 0.1I2+2Cl2→ICl+ICl3 Number of moles of I2=25.4 g254 g/mol=0.1 mol 1 mol of I2 produces 1 mol of ICl and 1 mol of ICl3 0.1 mol of I2 produces 0.1 mol of ICl and 0.1 mol of ICl3

$25.4 g$ of $I_{2}$ and $14.2 g$ of $C l_{2}$ 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 $I C l_{3}$ . (Given: Molar mass of $I$ = $127$ $g / m o l$ )

The correct option is B $0.1, 0.1$
$I_{2} + 2 C l_{2} \to I C l + I C l_{3}$
Number of moles of $I_{2} = \frac{25.4 g}{254 g/mol} = 0.1 mol$
1 mol of $I_{2}$ produces 1 mol of $I C l$ and 1 mol of $I C l_{3}$
0.1 mol of $I_{2}$ produces 0.1 mol of $I C l$ and 0.1 mol of $I C l_{3}$