CameraIcon

SearchIcon

MyQuestionIcon

2

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Laws of Mass Conservation

Equal weights...

Question

Equal weights of mercury and iodine are allowed to react completely to form a mixture of mercurous and mercuric iodide leaving none of the reactants. The ratio by mole of $H g_{2} I_{2}$ and $H g I_{2}$ formed (Hg= $200$ , I= $127$ ) is :

A

1 : 2.7

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

B

2.7 : 1

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

C

1 : 3

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

D

3 : 1

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

Open in App

Solution

The correct option is D $1 : 2.7$
The molecular weights of $H g$ and iodine are $200$ g/mol and $254$ g/mol respectively.
To produce $1$ mole of $H g_{2} I_{2}$ , $400$ g of Hg and $254$ g of iodine are required.
To produce $x$ mole of $H g I_{2}$ , $200 x$ g of mercury and $254 x$ g of iodine are required.
Total mass of mercury required is $200 x + 400$ .
Total mass of iodine required is $254 x + 254$ .
The mass of mercury required is equal to the mass of iodine required.
Hence, $200 x + 400 = 254 x + 254$ .
Hence, $x = 2.7$ .
The ratio of the number of moles of $H g_{2} I_{2}$ and $H g I_{2}$ produced is $1 : 2.7$ .

Suggest Corrections

thumbs-up

0

Similar questions

Q. Equal weights of mercury and

l_{2}

are allowed to react completely to form a mixture of mercurous and mercuric iodide leaving none of the reactants. Calculate the ratio of the weights of

H g_{2} l_{2}

H g l_{2}

Q. Equal weights of Hg and iodine arc allowed to react completely to form a mixture of mercuric iodide and mercuric iodide. Calculate the ratio of the masses of mercuric and mercuric iodides formed. (Hg = 201, I= 117)

Q. Two substance,

P_{4}

O_{2}

are allowed to react completely to form mixture of

P_{4} O_{6}

P_{4} O_{10}

leaving none of the reactants. The composition of final mixture if

P_{4}

O_{2}

P_{4} + 3 O_{2} \to P_{4} O_{6}

P_{4} + 5 O_{2} \to P_{4} O_{10}

.

1

P_{4}

4

O_{2}

3

P_{4}

11

O_{2}

3

P_{4}

13

O_{2}

Q. A mixture of

254 g

142 g

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

I C l

I c l_{3}

. How many moles of each product are formed?

(I = 127, C l = 35.5)

.

H g

I_{2}

are allowed to react to form

H g_{2} I_{2}

(assume no other side products to be formed). 50 g of

H g

is exposed to excess

I_{2}

H g_{2} I_{2}

is obtained. Calculate the percentage yield of the reaction.
(Molar mass of

H g

I

are 200 g/mol and 127 g/mol respectively)

2 H g + I_{2} \to H g_{2} I_{2}

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Laws of Chemical Combination

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Laws of Mass Conservation

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon