11th Class Chemistry The s-Block Elements

  • question_answer 74)   In the Solvay process, can we obtain sodium carbonate directly by treating the solution containing \[{{(N{{H}_{4}})}_{2}}C{{O}_{3}}\] with sodium chloride? Explain.

    Answer:

      In solvay process, when \[C{{O}_{2}}\] is passed through ammonical sodium chloride solution, the sparingly soluble \[NaHC{{O}_{3}}\] is formed which on heating gives\[N{{a}_{2}}C{{O}_{3}}\]. \[NaCl+N{{H}_{3}}+{{H}_{2}}O+C{{O}_{2}}\to NaHC{{O}_{3}}+N{{H}_{4}}Cl\]\[2NaHC{{O}_{3}}\xrightarrow{Heat}N{{a}_{2}}C{{O}_{3}}+{{H}_{2}}O+C{{O}_{2}}\] The solution which reacts with \[NaCl\] consists \[N{{H}_{4}}HC{{O}_{3}}\] and not \[(N{{H}_{4}})C{{O}_{3}}.\] \[N{{H}_{3}}+{{H}_{2}}O+C{{O}_{2}}\to N{{H}_{4}}HC{{O}_{3}}\] Thus, in solvay process, Na2C03 cannot be obtained directly.

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