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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 11

  • question_answer
    A solid 'X' on heating gives \[C{{O}_{2}}\]and a residue. The residue with \[{{H}_{2}}O\] form 'Y'. On passing an excess of \[{{H}_{2}}O\]through 'Y in \[{{H}_{2}}O\], a clear solution of 'Z' is obtained. On boiling 'Z; 'X'  is reformed. 'X' is

    A) \[Ca{{(HC{{O}_{3}})}_{2}}\]

    B) \[CaC{{O}_{3}}\]

    C) \[N{{a}_{2}}C{{O}_{3}}\]      

    D) \[{{K}_{2}}C{{O}_{3}}\]

    Correct Answer: B

    Solution :

    [b]: \[\underset{(X)}{\mathop{CaC{{O}_{3}}}}\,\,\overset{Heat}{\mathop{\rightleftharpoons }}\,\,\,CaO+C{{O}_{2}}\] \[CaO+{{H}_{2}}O\xrightarrow[{}]{{}}Ca{{(OH)}_{2}}\] \[\underset{(Y)}{\mathop{Ca{{(OH)}_{2}}}}\,+C{{O}_{2}}+{{H}_{2}}O\xrightarrow[{}]{{}}\underset{(Z)}{\mathop{Ca{{(HC{{O}_{3}})}_{2}}}}\,\] \[\underset{(Z)}{\mathop{Ca{{(HC{{O}_{3}})}_{2}}}}\,\xrightarrow[{}]{boil}\underset{(X)}{\mathop{CaC{{O}_{3}}+}}\,C{{O}_{2}}+{{H}_{2}}O\]


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