• # question_answer 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        [CBSE PMT 2004] A) $N{{a}_{2}}C{{O}_{3}}$ B) ${{K}_{2}}C{{O}_{3}}$ C) $Ca{{(HC{{O}_{3}})}_{2}}$ D) $CaC{{O}_{3}}$

Solution :

The given compound x must be $CaC{{O}_{3}}$. It can be explained by following reactions, $\underset{\text{(x)}}{\mathop{CaC{{O}_{3}}}}\,\xrightarrow{\Delta }\underset{\text{Residue}}{\mathop{CaO}}\,+C{{O}_{2}}\uparrow$; $CaO+{{H}_{2}}O\to \underset{\text{(y)}}{\mathop{Ca{{(OH)}_{2}}}}\,$ $Ca{{(OH)}_{2}}+C{{O}_{2}}+{{H}_{2}}O\to \underset{\text{z}}{\mathop{Ca(HC{{O}_{3}})}}\,$ $Ca{{(HC{{O}_{3}})}_{2}}\xrightarrow{\Delta }\underset{\text{(x)}}{\mathop{CaC{{O}_{3}}}}\,+C{{O}_{2}}\uparrow +{{H}_{2}}O$

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