Question

# A solid compound 'x' on heating gives $C{O_2}$ gas and a residue. The residue mixed with water forms 'y'. On passing an excess of $C{O_2}$ through 'y' in water, a clear solution 'z' is obtained. On boiling 'z' compound 'x' is reformed. The compound 'x' is..???$A.\;\;\;\;\;Ca{\left( {HC{O_3}} \right)_2} \\ B.\;\;\;\;\;CaC{O_3} \\ C.\;\;\;\;\;N{a_2}C{O_3} \\ D.\;\;\;\;\;{K_2}C{O_3} \\$

Hint: We must know that the residue obtained from $Ca{\left( {HC{O_3}} \right)_2}$will finally form $CaC{O_3}$ whereas residue from $N{a_2}C{O_3}$ and ${K_2}C{O_3}$ finally forms hydroxides.

Complete step by step answer:
A solid compound X must be Calcium carbonate,
Therefore, a solid compound X is a Calcium carbonate that on heating releases carbon dioxide gas $C{O_2}$ and forms Calcium oxide as a residue.
The residue called calcium oxide is when mixed with water forms compound Y that is calcium hydroxide.
On passing an excess of carbon dioxide gas through compound Y – calcium hydroxide in water, a clear solution Z that is calcium bicarbonate is formed.
The solution becomes colorless when the excess carbon dioxide is passed through the calcium hydroxide. This occurs due to the formed calcium bicarbonate being soluble in water.
On boiling compound Z that is calcium bicarbonate, compound X – calcium carbonate is reformed.
Therefore, we can write the reaction as
Hence, the compound 'x' that is Calcium carbonate $CaC{O_3}$.
So, the correct option is C

Note:
We must remember that when less amount of carbon dioxide is passed through the calcium hydroxide solution then it becomes milky in appearance due to the formation of calcium carbonate. Calcium carbonate is then settled as a residue or precipitate when large amounts of carbon dioxide are treated.