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NEET levelSingle correct optionDifficulty 1 of 3sb118

In the Solvay process the compound that separates out as a solid from the ammoniated brine is:

A $Na X_{2} CO X_{3}$
B $NH X_{4} Cl$
C $NaHCO X_{3}$
D $NaCl$

Answer: C

The reaction

NaCl + NH X_{3} + CO X_{2} + H X_{2} O NaHCO X_{3} + NH X_{4} Cl

gives sodium hydrogencarbonate, which has low solubility in the cold solution and precipitates. Heating it later gives

Na X_{2} CO X_{3}

.

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