JEE Main & Advanced Chemistry The s-Block Elements Question Bank Alkali Metals

  • 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) \[CaC{{O}_{3}}\]

    Correct Answer: D

    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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