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NH(4)CI on heating with NaOH liberates...

`NH_(4)CI` on heating with NaOH liberates

A

`NaCl`

B

`NH_(3)`

C

`HCl`

D

`NaOCl`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the question regarding the reaction of ammonium chloride (NH₄Cl) with sodium hydroxide (NaOH) upon heating, we can follow these steps: ### Step-by-Step Solution: 1. **Identify the Reactants**: The reactants in this reaction are ammonium chloride (NH₄Cl) and sodium hydroxide (NaOH). 2. **Write the Reaction**: When NH₄Cl is heated with NaOH, the reaction can be represented as: \[ NH₄Cl + NaOH \rightarrow NaCl + NH₃ + H₂O \] 3. **Analyze the Products**: - **NaCl**: Sodium chloride is formed as a solid. - **NH₃**: Ammonia is liberated as a gas. - **H₂O**: Water is produced as a liquid. 4. **Identify the Gas Liberated**: The question specifically asks what is liberated when NH₄Cl is heated with NaOH. The gas that is liberated in this reaction is ammonia (NH₃). 5. **Conclusion**: Therefore, when NH₄Cl is heated with NaOH, it liberates ammonia (NH₃). ### Final Answer: The compound liberated when NH₄Cl is heated with NaOH is **ammonia (NH₃)**. ---
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