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).