Question

Q-
n-factor of HCl in the given reaction is
K,Cr,O, + HCI → KCl + Crçiz + Cl2 + H2O
M
(1)
112)
(3) 1
(4) 6
I know the answer,i just want to know how to get the answer ?

Answer 1

Answer:

Question

Answers

Related Questions

The n-factor of HCl in the equation:

K2Cr2O7+14HCl→2KCl+2CrCl3+3Cl2+7H2O

A.23

B.113

C.73

D.37

Answer

VerifiedVerified

72.7k+ views

Hint: The n factor is also known as the valence factor and its value varies depending upon the compound being considered. For oxidizing agents and reducing agents, the n factor is equal to the number of electrons transferred by one mole of the oxidizing agent or reducing agent.

Complete step by step answer:

In the given reaction,

K2Cr2O7+14HCl→2KCl+2CrCl3+3Cl2+7H2O

Potassium dichromate is converted to chromium chloride.

In dichromate ion, chromium has an oxidation number of +6 , while in chromium chloride its oxidation number is +3 . As a result, we can say that three electrons are gained per mole of chromium atom.

Since in the given reaction, two moles of chromium atom are involved the number of electrons transferred will be =2×3=6

Since the dichromate molecule gains 6 electrons, thus, the other reactant, that is, hydrochloric acid must lose 6 electrons.

The reaction involves 14 moles of HCl.

Thus, we can say that 14 moles of HCl lose 6electrons.

So, 1mole of HCl loses 614electrons.

Thus, the n factor of HCl is 37.