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When a nitrate is boiled with aluminum t...

When a nitrate is boiled with aluminum turnings and concentrated NaOH solution, the gas liberated is

A

`NO_2`

B

`NH_3`

C

NO

D

`N_2O`

Text Solution

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The correct Answer is:
To solve the question of what gas is liberated when a nitrate is boiled with aluminum turnings and concentrated NaOH solution, we can follow these steps: ### Step-by-Step Solution: 1. **Identify the Reactants**: - We have a nitrate (for example, sodium nitrate, NaNO3). - We also have aluminum turnings (Al). - The third reactant is concentrated sodium hydroxide (NaOH). 2. **Understand the Reaction**: - When aluminum reacts with sodium nitrate in the presence of concentrated NaOH, a redox reaction occurs. - Aluminum is a strong reducing agent and will reduce the nitrate ion (NO3-) to ammonia (NH3). 3. **Write the Reaction**: - The general reaction can be represented as: \[ 3 \text{NaNO}_3 + 8 \text{Al} + 5 \text{NaOH} \rightarrow 3 \text{NaAlO}_2 + 3 \text{NH}_3 + 2 \text{H}_2O \] - In this reaction, sodium nitrate is reduced to ammonia, and aluminum is oxidized. 4. **Identify the Gas Produced**: - The gas that is liberated during this reaction is ammonia (NH3). 5. **Conclusion**: - Therefore, when a nitrate is boiled with aluminum turnings and concentrated NaOH solution, the gas liberated is ammonia. ### Final Answer: The gas liberated is **ammonia (NH3)**.
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