Question

A non-metal which reacts with concentrated sulphuric acid to form two gases which turn lime water milky is ______.

• A. Sulphur
• B. Carbon
• C. Oxygen
• D. Nitrogen

Hint:

While both gases turn lime water milky, SO_2 can be distinguished because it has a pungent smell of burning sulphur and turns acidified potassium dichromate paper green!

Answer 1

The Correct Option is B

The question requires identifying a non-metal that reacts with concentrated sulphuric acid to form two gases, which subsequently turn lime water milky. Let's analyze the options:

1. Sulphur: When sulphur reacts with sulphuric acid, it usually forms sulphur dioxide gas. However, this reaction does not produce two gases that specifically turn lime water (a test for carbon dioxide) milky.
2. Carbon: When carbon reacts with concentrated sulphuric acid, it forms carbon dioxide (\( \text{CO}_2 \)) and carbon monoxide (\( \text{CO} \)) gases. The carbon dioxide formed turns lime water milky due to the formation of calcium carbonate (\( \text{CaCO}_3 \)), which is insoluble in water.
3. Oxygen: Oxygen does not generally react with concentrated sulphuric acid in a manner that produces gases which turn lime water milky.
4. Nitrogen: Nitrogen is an inert gas and does not typically react with concentrated sulphuric acid under normal conditions to produce any gases that affect lime water.

Therefore, the correct answer is Carbon. This reaction can be summarized as follows:

The reaction between carbon and concentrated sulphuric acid can be represented by the equations:

• \(C + 2\text{H}_2\text{SO}_4 \rightarrow \text{CO}_2 + 2\text{SO}_2 + 2\text{H}_2\text{O}\) (simplified to show formation of gases)

The carbon dioxide gas produced turns lime water milky due to the following reaction:

• \(\text{Ca(OH)}_2 + \text{CO}_2 \rightarrow \text{CaCO}_3 + \text{H}_2\text{O}\)

This results in the formation of calcium carbonate (\( \text{CaCO}_3 \)), which is a white precipitate, thus turning the lime water milky.