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

Van't Hoff Factor

25.4g of I2 a...

Question

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.

0.1,0.1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.2,0.2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.1,0.2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.2,0.1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A
0.1,0.1

$I_{2} (_{\frac{25.4}{254}})$ + 2 $C l_{2} (_{\frac{14.2}{71}})$ $\to$ $I C l (_{0})$ + $I C l_{3} (_{0})$

1 2 1 1 as per equation

0.1 0.2 0 0 moles in the beginning

0 0 0.1 0.1 moles after reaction

Suggest Corrections

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

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

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

)

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 :

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.

Join BYJU'S Learning Program

25.4g of I2 and 14.2g of Cl2 are made to react completely to yield a mixture of ICl and ICl3 .Calculate moles of ICl and ICl3 formed.

The correct option is A 0.1,0.1 I2(25.4254) + 2Cl2(14.271) → ICl(0) + ICl3(0) 1 2 1 1 as per equation 0.1 0.2 0 0 moles in the beginning 0 0 0.1 0.1 moles after reaction

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.

The correct option is A
0.1,0.1

$I_{2} (_{\frac{25.4}{254}})$ + 2 $C l_{2} (_{\frac{14.2}{71}})$ $\to$ $I C l (_{0})$ + $I C l_{3} (_{0})$

1 2 1 1 as per equation

0.1 0.2 0 0 moles in the beginning

0 0 0.1 0.1 moles after reaction