Question

Compound A reacts with $NH _4 Cl$ and forms a compound B.

Compound B reacts with $H _2 O$ and excess of $CO _2$ 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.

• A. $CaCl _2, NH _4^{+},\left( NH _4\right)_2 CO _3$
• B. $CaCl _2, NH _3, NH _4 HCO _3$
• C. $Ca ( OH )_2, NH _3, NH _4 HCO _3$
• D. $Ca ( OH )_2, NH _4^{+},\left( NH _4\right)_2 CO _3$

Answer 1

The Correct Option is C

To solve this problem, we need to identify the compounds \(A\), \(B\), and \(C\) based on the given chemical reactions.

1. The problem states that compound \(A\) reacts with \(NH_4Cl\) to form compound \(B\).
   • If we consider \(A\) to be \(Ca(OH)_2\) (calcium hydroxide), this compound can react with ammonium chloride \((NH_4Cl)\) to release ammonia \((NH_3)\), a potential compound \(B\).
2. Compound \(B\) now reacts with water and an excess of carbon dioxide to produce compound \(C\).
   • Ammonia \((NH_3)\) reacting with water \((H_2O)\) and carbon dioxide \((CO_2)\) forms ammonium bicarbonate \((NH_4HCO_3)\), which can be identified as compound \(C\).
3. When compound \(C\), which is ammonium bicarbonate \((NH_4HCO_3)\), is passed through or reacts with a saturated sodium chloride solution, sodium hydrogen carbonate \((NaHCO_3)\) can be formed.

Based on these reactions, we can deduce the following assignments:

• \(A\) is \(Ca(OH)_2\) (calcium hydroxide).
• \(B\) is \(NH_3\) (ammonia).
• \(C\) is \(NH_4HCO_3\) (ammonium bicarbonate).

Thus, the correct answer is:

\(Ca(OH)_2, NH_3, NH_4HCO_3\)

This solution is chosen based on the fact that calcium hydroxide reacting with ammonium chloride can indeed release ammonia gas, and excess carbon dioxide with ammonia readily forms ammonium bicarbonate, which is consistent throughout the problem statement.