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JEE Main & Advanced Sample Paper JEE Main Sample Paper-35

  • question_answer
    Sodium carbonate is manufactured by Solvay process, the products that are recycled are:

    A)  \[C{{O}_{2}}\] and \[N{{H}_{3}}\]           

    B)  \[C{{O}_{2}}\] and \[N{{H}_{4}}Cl\]

    C)  \[NaCl,\,CaO\]                

    D)  \[CaC{{l}_{2}},CaO\]

    Correct Answer: A

    Solution :

    (i) \[N{{H}_{3}}\,C{{O}_{2}}+{{H}_{2}}O\xrightarrow{\,}\,N{{H}_{4}}HC{{O}_{3}}\]
    (ii) \[N{{H}_{4}}\,HC{{O}_{3}}+NaCl\xrightarrow{\,}\,\underset{sodium\,\,carbonate}{\mathop{NaHC{{O}_{3}}+N{{H}_{4}}Cl}}\,\]
    (iii) \[2NaHC{{O}_{3}}\,\xrightarrow{\Delta }\,\underset{sodium\,\,carbonate}{\mathop{N{{a}_{2}}C{{O}_{3}}\,+{{H}_{2}}O+C{{O}_{2}}}}\,\]
    The \[C{{O}_{2}}\] obtained in this step can be recycled. The \[C{{O}_{2}}\] required in reaction is prepared by heating \[CaC{{O}_{3}}\].             \[CaC{{O}_{3}}\xrightarrow{\Delta }CaO+C{{O}_{2}}\]             \[CaO+{{H}_{2}}O\xrightarrow{\,}\,Ca{{(OH)}_{2}}\] \[Ca{{(OH)}_{2}}\] is used to decompose \[N{{H}_{4}}Cl\] The \[N{{H}_{3}}\] thus obtained can be recycled. \[\therefore \,\,C{{O}_{2}}\] and \[N{{H}_{3}}\] are the products which can be recycled in Solvay process.


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