Question

The compound A on heating gives a colourless gas and a residue that is dissolved in water to obtain B. Excess of $ C{{O}_{2}} $ ​ is bubbled through an aqueous solution of B,C is formed which is recovered in the solid form. Solid C on gentle heating gives back A. The compound A is:
(A) $ CaC{{O}_{3}} $
(B) $ N{{a}_{2}}C{{O}_{3}} $
(C) $ {{K}_{2}}C{{O}_{3}} $
(D) $ CaS{{O}_{4}}\cdot 2{{H}_{2}}O $

Answer 1

Hint :According to the question, compound X on heating a colourless gas. That colourless gas is carbon dioxide. On passing excess carbon dioxide gas, the colorless calcium bicarbonate is formed.

Complete Step By Step Answer:
 We can write the reactions according to the given question as Compound X on heating gives a colorless gas (Carbon dioxide) along with the residue (Calcium Oxide)
 $ CaC{{O}_{3}}\xrightarrow{\Delta }CaO+C\uparrow $
The residue (Calcium Oxide) is dissolved in water to obtain Y (Calcium Hydroxide)
 $ CaO+{{H}_{2}}O\to Ca{{\left( OH \right)}_{2}} $
Excess carbon dioxide is bubbled through an aqueous solution of Y (Calcium Hydroxide) to form Z (Calcium Bicarbonate). $ Ca{{\left( OH \right)}_{\text{2}}}\text{+}2\text{C}{{O}_{2}}\xrightarrow{Excess}\text{ Ca}{{\left( HC{{O}_{3}} \right)}_{2}} $
Z (Calcium Bicarbonate) on gently heating gives back X (Calcium Carbonate).
 $ Ca{{\left( HC{{O}_{3}} \right)}_{2}}\xrightarrow{\Delta }\text{ }CaC{{O}_{3}}+C{{O}_{2}}\uparrow +{{\text{H}}_{\text{2}}}O $
From the above reactions, we can see that the compound
X is $ CaC{{O}_{3}} $ Calcium carbonate;
 Y is $ Ca{{(OH)}_{2}} $ Calcium hydroxide;
 Z is $ Ca{{(HC{{O}_{3}})}_{2}} $ Calcium bicarbonate.
Therefore, Correct answer is option A, i.e. $ CaC{{O}_{3}} $ .

Note :
When excess carbon dioxide gas is passed through calcium hydroxide (compound Y), it turns milky due to the formation of the calcium carbonate (compound X) as a precipitate which is insoluble in water and then become colorless due to the formation of calcium bicarbonate (compound Z).