Single Choice

Potassium dichromate can be prepared by:

Aheating $$Na_{2}CrO_{4}$$ solution with $$KCl$$ solution and then cooling
Bheating powdered chromite ore with quicklime in the presence of air
Cheating $$K_{2}CrO_{4}$$ with sulphuric acid
Dboth $$A$$ and $$C$$
Correct Answer

Solution

Potassium dichromate is usually prepared by the reaction of potassium chloride on sodium dichromate.
$$Na_{2}Cr_{2}O_{7}+2KCl \rightarrow K_{2}Cr_{2}O_{7}+2NaCl$$

It can also be prepared by heating $$K_{2}CrO_{4}$$ with $$H_{2}SO_{4}$$.
$$2K_{2}CrO_{4}+2H^+ \rightarrow K_{2}Cr_{2}O_{7}+2K^++H_{2}O$$

Hence, option D is correct.


SIMILAR QUESTIONS

d - block and f - block Elements

What would happen when a solution of potassium chromate is treated with an excess of dilute nitric acid?

d - block and f - block Elements

Potassium dichromate on heating does not gives :

d - block and f - block Elements

How is sodium chromate converted into sodium dichromate in the manufacture of potassium dichromate from chromite ore?

d - block and f - block Elements

Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH on a solution of potassium dichromate?

d - block and f - block Elements

Potassium dichromate is prepared from:

d - block and f - block Elements

Which of the following reactions do not result in the preparation of potassium dichromate from chromate? (I) 4$$FeCr_2O_4$$ + 8$$Na_2CO_3$$+ $$7O_2$$$$\rightarrow$$ (II) $$Na_2CrO_4$$ + $$H_2SO_4$$ $$\rightarrow$$ (III) $$Na_2Cr_2O_7$$+ $$2KCl$$$$\rightarrow$$

d - block and f - block Elements

Which of the following compounds is used as the starting material for the preparation of potassium dichromate?

d - block and f - block Elements

Name an ore used in the preparation of potassium dichromate.

d - block and f - block Elements

The chromate ion in acidic medium changes to ..............

d - block and f - block Elements

What would happen when a solution of potassium chromate is treated with an excess of dilute nitric acid?

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