Question

The gas liberated when metallic sodium reacts with ethanol

• A. \(O_2\)
• B. \(H_2\)
• C. \(CO_2\)
• D. \(CO\)

Answer 1

The Correct Option is B

To determine the gas liberated when metallic sodium reacts with ethanol, we need to analyze the chemical reaction involved.

1. Understanding the Reaction:
Metallic sodium (Na) is a highly reactive alkali metal. When it reacts with ethanol ($\text{C}_2\text{H}_5\text{OH}$), it displaces hydrogen from the hydroxyl group ($-\text{OH}$) of ethanol. The general reaction between an alkali metal and an alcohol can be represented as:

$$ \text{Metal} + \text{Alcohol} \rightarrow \text{Metal Alkoxide} + \text{Hydrogen Gas} $$
For sodium reacting with ethanol:
$$ 2\text{Na} + 2\text{C}_2\text{H}_5\text{OH} \rightarrow 2\text{C}_2\text{H}_5\text{ONa} + \text{H}_2 $$

2. Identifying the Products:
- Sodium ethoxide ($\text{C}_2\text{H}_5\text{ONa}$): This is the metal alkoxide formed.
- Hydrogen gas ($\text{H}_2$): This is the gas liberated during the reaction.

3. Conclusion:
The gas liberated when metallic sodium reacts with ethanol is hydrogen gas ($\text{H}_2$).

Final Answer: $ {\text{H}_2} $