You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Limiting Reagent or Reactant

25.4 g of I...

Question

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.

Open in App

Solution

$I_{2} + 2 C l_{2} ⇌ I C l + I C l_{3}$
mol of $I_{2} = \frac{25.4}{254} = 0.1 m o l$
mol of $C l_{2} = \frac{14.2}{71} = 0.2 m o l$
$∴$ both are in Stoichiometry ratio and 0.1 mol of both $I C l$ and $I C l_{3}$ are formed.

Suggest Corrections

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.

25.4

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

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

)

25.4g 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

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}

Join BYJU'S Learning Program

25.4 g of I2 and 14.2 of Cl2 are made to react completely to form ICl and ICl3. Determine(i) Limiting reagent(ii) No. of moles of ICl and ICl3 formed.

I2+2Cl2⇌ICl+ICl3 mol of I2=25.4254=0.1mol mol of Cl2=14.271=0.2mol ∴ both are in Stoichiometry ratio and 0.1 mol of both ICl and ICl3 are formed.

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.

$I_{2} + 2 C l_{2} ⇌ I C l + I C l_{3}$
mol of $I_{2} = \frac{25.4}{254} = 0.1 m o l$
mol of $C l_{2} = \frac{14.2}{71} = 0.2 m o l$
$∴$ both are in Stoichiometry ratio and 0.1 mol of both $I C l$ and $I C l_{3}$ are formed.