Compound A reacts with NH₄Cl and forms a compound B. Compound B reacts with H₂O and excess of CO₂ to form compound C which on passing through or reaction with saturated NaCl solution forms sodium hydrogen carbonate. Compound A, B and C, are respectively :

Solution

Chemistry ma’am · 4 steps

This is the Solvay process run backwards: start from NaHCO₃ and ask what made it.

1. Find C first. Sodium hydrogen carbonate forms when NH₄HCO₃ meets saturated brine. NaHCO₃ is only sparingly soluble, so it precipitates: NH₄HCO₃ + NaCl → NaHCO₃ v + NH₄Cl. So C is NH₄HCO₃.
2. Find B. NH₄HCO₃ comes from ammonia, water and excess CO₂: NH₃ + H₂O + CO₂ → NH₄HCO₃. With excess CO₂ you get the hydrogen carbonate, not the carbonate. So B is NH₃.
3. Find A. Ammonia is driven out of an ammonium salt by a base. Slaked lime does this in the Solvay plant to recover ammonia.

   Ca(OH)₂ + 2NH₄Cl → CaCl₂ + 2NH₃ ↑ + 2H₂O

4. So. A is Ca(OH)₂, B is NH₃ and C is NH₄HCO₃, which is option (d).

The idea to keep. In the Solvay process, CO₂ in excess with ammonia gives NH₄HCO₃, brine turns it into NaHCO₃, and Ca(OH)₂ frees NH₃ from NH₄Cl for reuse.

Answer from the NTA final answer key. Solution by PrepWiser.