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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