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A solution of a colourles salt H on boil...

A solution of a colourles salt H on boiling with excess of NaOH produces a non-flammable gas and the evolution of gas ceases after sometime. Upon addition of Zn dust to the same solution, the evolution of gas restarts. The colourless salt (S) 'H' is (are)

A

`NH_(4)NO_(3)`

B

`NH_(4)NO_(2)`

C

`NH_(4)Cl`

D

`(NH_(4))_(2)SO_(4)`

Text Solution

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The correct Answer is:
To solve the problem, we need to identify the colorless salt (H) based on the given reactions with NaOH and Zn dust. Let's break down the steps: ### Step 1: Understand the Reaction with NaOH When a colorless salt (H) is boiled with excess NaOH, it produces a non-flammable gas. The likely candidates for salt H are ammonium salts, as they release ammonia (NH3) when treated with NaOH. ### Step 2: Identify the Non-Flammable Gas The non-flammable gas produced in the reaction is ammonia (NH3). This is consistent with the behavior of ammonium salts, such as ammonium nitrate (NH4NO3) or ammonium nitrite (NH4NO2), when they react with NaOH. ### Step 3: Write the Reaction Equation For ammonium nitrate (NH4NO3): \[ \text{NH}_4\text{NO}_3 + \text{NaOH} \rightarrow \text{NH}_3 + \text{NaNO}_3 + \text{H}_2\text{O} \] For ammonium nitrite (NH4NO2): \[ \text{NH}_4\text{NO}_2 + \text{NaOH} \rightarrow \text{NH}_3 + \text{NaNO}_2 + \text{H}_2\text{O} \] In both cases, ammonia gas is produced. ### Step 4: Analyze the Cessation of Gas Evolution The evolution of gas ceases after some time because all the ammonium ions have reacted with NaOH to form ammonia. ### Step 5: Introduction of Zinc Dust When zinc dust is added to the solution containing NaOH and the products (NaNO3 or NaNO2), the reaction can regenerate ammonia: \[ \text{NaNO}_3 + \text{Zn} + \text{NaOH} \rightarrow \text{Na}_2\text{ZnO}_2 + \text{NH}_3 + \text{H}_2\text{O} \] This shows that zinc can react with the nitrate or nitrite to produce ammonia again, restarting the gas evolution. ### Step 6: Conclusion Based on the above reactions and analysis, the colorless salt (H) can be either ammonium nitrate (NH4NO3) or ammonium nitrite (NH4NO2). Therefore, the answer is: - The colorless salt (H) is (are) NH4NO3 and NH4NO2.
To solve the problem, we need to identify the colorless salt (H) based on the given reactions with NaOH and Zn dust. Let's break down the steps: ### Step 1: Understand the Reaction with NaOH When a colorless salt (H) is boiled with excess NaOH, it produces a non-flammable gas. The likely candidates for salt H are ammonium salts, as they release ammonia (NH3) when treated with NaOH. ### Step 2: Identify the Non-Flammable Gas The non-flammable gas produced in the reaction is ammonia (NH3). This is consistent with the behavior of ammonium salts, such as ammonium nitrate (NH4NO3) or ammonium nitrite (NH4NO2), when they react with NaOH. ...
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