Question

A solid compound X on heating gives ${\text{C}}{{\text{O}}_{\text{2}}}$ gas and a residue. The residue mixed with water forms Y. On passing an excess of ${\text{C}}{{\text{O}}_{\text{2}}}$ through Y in water a clear solution Z is obtained. On boiling Z, compound X is reformed. The compound X is:
A.${\text{Ca}}{\left( {{\text{HC}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}$
B.${\text{CaC}}{{\text{O}}_{\text{3}}}$
C.${\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_{\text{3}}}$
D.${{\text{K}}_2}{\text{C}}{{\text{O}}_{\text{3}}}$

Answer 1

All metal carbonates undergo one common reaction which is thermal decomposition. When a metal carbonate is heated, it undergoes breakdown to form the corresponding metal oxide and carbon dioxide gas.
The reactive metals form very stable carbonates and so they require a very high temperature to decompose. On the other hand, the unreactive metals form unstable carbonates and so they decompose at a relatively low temperature.

Complete step by step answer:
Calcium bicarbonate, ${\text{Ca}}{\left( {{\text{HC}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}$ on heating decomposes into calcium carbonate, carbon dioxide and water. The reaction is:
${\text{Ca}}{\left( {{\text{HC}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}\xrightarrow{\Delta }{\text{CaC}}{{\text{O}}_{\text{3}}}{\text{ + C}}{{\text{O}}_{\text{2}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}$
So, the option A is not correct.
Calcium carbonate is a metal carbonate and so on heating it will give carbon dioxide gas and calcium oxide as residue.
${\text{CaC}}{{\text{O}}_{\text{3}}}\left( {\text{s}} \right)\xrightarrow{\Delta }{\text{CaO}}\left( {\text{s}} \right){\text{ + C}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right)$
The residue of calcium oxide when mixed with water forms calcium hydroxide which is Y.
${\text{CaO}}\left( {\text{s}} \right){\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}\left( {\text{l}} \right) \to {\text{Ca}}{\left( {{\text{OH}}} \right)_{\text{2}}}\left( {{\text{aq}}} \right)$
When Y or calcium hydroxide is passed with an excess of carbon dioxide in water, a clear solution of calcium bicarbonate is obtained which is Z.
${\text{Ca}}{\left( {{\text{OH}}} \right)_{\text{2}}}\left( {{\text{aq}}} \right) + {\text{C}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right) \to {\text{Ca}}{\left( {{\text{HC}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}\left( {{\text{aq}}} \right)$
The solution is clear and colorless because calcium bicarbonate is soluble in water. But if less amount of carbon dioxide is passed through calcium hydroxide, then the solution becomes milky due to the formation of calcium carbonate which is a precipitate and hence is insoluble in water.
On boiling Z or calcium bicarbonate, calcium carbonate is reformed.
${\text{Ca}}{\left( {{\text{HC}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}\left( {{\text{aq}}} \right)\xrightarrow{\Delta }{\text{CaC}}{{\text{O}}_{\text{3}}}\left( {\text{s}} \right){\text{ + C}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right){\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}\left( {\text{l}} \right)$
Therefore, X is calcium carbonate. So option B is the correct answer.
${\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_{\text{3}}}$ or sodium carbonate is a very stable carbonate and it decomposes into carbon dioxide and sodium oxide.
${\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_{\text{3}}}\left( {\text{s}} \right)\xrightarrow{\Delta }{\text{N}}{{\text{a}}_{\text{2}}}{\text{O}}\left( {\text{s}} \right){\text{ + C}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right)$
Sodium oxide also reacts with water to give sodium hydroxide which when passed through excess carbon dioxide gives back sodium carbonate directly. So, option C is also wrong.
${{\text{K}}_2}{\text{C}}{{\text{O}}_{\text{3}}}$ or potassium carbonate also decomposes into carbon dioxide and potassium oxide.
${{\text{K}}_2}{\text{C}}{{\text{O}}_{\text{3}}}\left( {\text{s}} \right)\xrightarrow{\Delta }{{\text{K}}_{\text{2}}}{\text{O}}\left( {\text{s}} \right){\text{ + C}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right)$
Potassium oxide gives potassium hydroxide with water which when passed through excess carbon dioxide gives back potassium carbonate directly. So, option D is also wrong.

So option B is the correct answer.

Note:
It is possible to measure the stability of the metal carbonates by this thermal decomposition reaction of the carbonates. The carbon dioxide gas produced is bubbled through lime water. The faster the lime water becomes milky, the higher the rate of decomposition of the carbonate, indicating that the carbonate is less stable.