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Stoichiometric Calculations

254 g of I2...

Question

$254$ g of $I_{2}$ and $142$ g of ${C l}_{2}$ are made to react completely to yield a mixture of $I C l$ and ${I C l}_{3}$ . The moles of ${I C l}_{3}$ formed are :

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Solution

The balanced chemical equation for the reaction is,

$I_{2} + 2 C l_{2} \to I C l + I C l_{3}$
Thus, $1$ mole of iodine reacts with $2$ moles of chlorine to form $1$ mole of iodine trichloride.
The molar masses of iodine and chlorine are $254 g / m o l$ and $71 g / m o l$ respectively.

$254$ g of iodine corresponds to $1$ mole and $142$ g of chlorine corresponds to $2$ moles. $1$ mole of iodine requires $2$ moles of chlorine. Hence iodine and chlorine are present in stoichiometric amount. Thus, $1$ mole of iodine trichloride is formed.

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Similar questions

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 moles of

I C l

and

I C l_{3}

formed.

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

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

I_{2}

and 14.2 of

C l_{2}

are made to react completely to form

I C l

and

{I C l}_{3}

. Determine

(i) Limiting reagent(ii) No. of moles of

I C l

and

{I C l}_{3}

formed.

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.

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254 g of I2 and 142 g of Cl2 are made to react completely to yield a mixture of ICl and ICl3. The moles of ICl3 formed are :

$254$ g of $I_{2}$ and $142$ g of ${C l}_{2}$ are made to react completely to yield a mixture of $I C l$ and ${I C l}_{3}$ . The moles of ${I C l}_{3}$ formed are :