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Stoichiometric Calculations

254g of iodin...

Question

$254 g$ of iodine and $142 g$ of chlorine are made to react completely to give the mixture of $I C l$ and $I {C l}_{3}$ . The moles of each product formed is:

0.1

mol

I C l

and

0.1

mol

I {C l}_{3}

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0.1

mol

I C l

and

1.0

mol

I {C l}_{3}

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0.5

mol

I C l

and

0.1

mol

I {C l}_{3}

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1.5

mol

I C l

and

1.0

mol

I {C l}_{3}

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Solution

The correct option is A $0.1$ mol $I C l$ and $0.1$ mol $I {C l}_{3}$
It is based on the principle of P.O.A.C ( Principle of Atom Conservation. )
see given that $I_{2} + 2 C l_{2} ⟶ I C l + I C l_{3}$
Now we have molar mass of $I_{2}$ = 127x2=254
Molar mass of $C l_{2}$ =35.5x2=71
So,,moles of I2 used in the reaction =given mass/molar mass =25.4/254=0.1 moles of $I_{2}$
Moles of used in the reaction is =14.2/71=0.2 moles of $C l_{2}$
so and therefore from the reaction we see that 1 mol of $I_{2}$ react with 2 moles of $C l_{2}$ to give one mole of each $I C l$ and $I C l_{3}$
Interpreting the same data with the same data with the given sample we can conclude that 0.2 mol of $C l_{2}$ reacts with 0.1 mol of iodine to give 0.1 mol of each $I C l$ and $I C l_{3}$

Hence, $0.1 m o l$ $I C l$ and $0.1 m o l$ $I C l_{3}$ is the answer.

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Similar questions

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.

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 ?

25.4 g

I_{2}

and

14.2 g

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

)

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 ratio of moles of

I C l

and

{I C l}_{3}

Q. A mixture of

254 g

of iodine and

142 g

of chlorine is made to react completely to given a mixture of

I C l

and

I c l_{3}

. How many moles of each product are formed?

(I = 127, C l = 35.5)

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254g of iodine and 142g of chlorine are made to react completely to give the mixture of ICl and ICl3. The moles of each product formed is:

$254 g$ of iodine and $142 g$ of chlorine are made to react completely to give the mixture of $I C l$ and $I {C l}_{3}$ . The moles of each product formed is: