Adding ethanoic acid to sodium carbonate yields which gaseous product that turns lime-water milky?

The correct option is Carbon dioxide.

Explanation

Carboxylic acids react with metal carbonates and metal hydrogen carbonates to produce a corresponding salt, water, and carbon dioxide gas. This reaction is a standard laboratory test used to identify the presence of a carboxylic group ($-COOH$).

Detailed Analysis:

• Reaction Mechanism: When ethanoic acid ($CH_3COOH$) is added to sodium carbonate ($Na_2CO_3$), a double displacement reaction occurs followed by decomposition, resulting in brisk effervescence.
• Products: The reaction yields sodium ethanoate ($CH_3COONa$), water ($H_2O$), and carbon dioxide ($CO_2$).
• Chemical Equation:
  $2CH_3COOH + Na_2CO_3 \rightarrow 2CH_3COONa + H_2O + CO_2 \uparrow$
• Identification Test: The evolved gas is identified as carbon dioxide because it turns lime water (calcium hydroxide solution) milky. This milkiness is caused by the formation of insoluble calcium carbonate ($CaCO_3$).
• Incorrect Options: Hydrogen is evolved when acids react with metals, not carbonates. Oxygen and Ammonia are not products of the reaction between simple carboxylic acids and metal carbonates.

Key Takeaway:

The reaction between an acid and a metal carbonate always evolves carbon dioxide gas, which is chemically confirmed by the lime-water test.