Problem 94 - A Complete Resource Book in Chemistry for JEE Main

Question

Gas 'A' is bubbled through slaked lime when a white precipitate is formed. On prolonged bubbling the precipitate dissolved. On heating the resultant solution, the white precipitate reappeared with solution of gas 'B'. The gases 'A' and 'B' respectively are (a) \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\) (b) \(\mathrm{CO}_{2}\) and \(\mathrm{CO}_{2}\) (c) \(\mathrm{CO}\) and \(\mathrm{CO}\) (d) \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\)

Step-by-Step Solution

Answer

The gases 'A' and 'B' are both \(\text{CO}_2\), so the answer is option (b).

Step 1: Identifying the Reaction with Slaked Lime

When a gas is bubbled through slaked lime (calcium hydroxide, \(\text{Ca(OH)}_2\)) and a white precipitate forms, this commonly indicates the formation of calcium carbonate (\(\text{CaCO}_3\)). This suggests that gas 'A' is \(\text{CO}_2\), since \(\text{Ca(OH)}_2 + \text{CO}_2 \rightarrow \text{CaCO}_3 + \text{H}_2\text{O}\).

Step 2: Analyzing the Dissolution of Precipitate

On prolonged bubbling of gas 'A', the white precipitate dissolves. For \(\text{CaCO}_3\) to dissolve, the presence of an excess of \(\text{CO}_2\) forms calcium bicarbonate \((\text{Ca(HCO}_3\text{)}_2)\), which is soluble: \(\text{CaCO}_3 + \text{CO}_2 + \text{H}_2\text{O} \rightarrow \text{Ca(HCO}_3\text{)}_2\). This confirms gas 'A' is \(\text{CO}_2\).

Step 3: Observing Precipitate Reformation on Heating

When heating the solution, the calcium bicarbonate decomposes back to calcium carbonate (the white precipitate) and releases carbon dioxide gas: \(\text{Ca(HCO}_3\text{)}_2 \rightarrow \text{CaCO}_3 + \text{CO}_2 + \text{H}_2\text{O}\). With \(\text{CO}_2\) as both gas came off during reaction and is produced mirroring gas 'A', gas 'B' is also \(\text{CO}_2\).

Step 4: Conclusion and Correct Option

Since both the initial white precipitate forms due to \(\text{CO}_2\) and the gas released on heating matches \(\text{CO}_2\), the gases 'A' and 'B' are \(\text{CO}_2\) and \(\text{CO}_2\), respectively. Therefore, the correct answer is option \(b\).

Key Concepts

Calcium Carbonate Formation

When carbon dioxide (\(\text{CO}\_2\)) is bubbled through a solution of slaked lime, which is chemically known as calcium hydroxide (\(\text{Ca(OH)}\_2\)), a white precipitate forms. This white solid is calcium carbonate (\(\text{CaCO}\_3\)). The reaction can be represented by the chemical equation:

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

The formation of calcium carbonate is a reversible process, where the \(\text{CO}\_2\) reacts with the calcium ions in the slaked lime to create this compound. You will see a white, chalky appearance, confirming the presence of calcium carbonate.

Gas Reaction with Slaked Lime

Slaked lime is an important reactant in many chemical reactions involving gases. When exposed to carbon dioxide (\(\text{CO}\_2\)), slaked lime reacts to form calcium carbonate.

• Initial Reaction: As \(\text{CO}\_2\) bubbles through the \(\text{Ca(OH)}\_2\) solution, it combines with hydroxide ions to create the insoluble white precipitate of calcium carbonate.

This demonstrates that \(\text{CO}\_2\) easily reacts with bases, like slaked lime, converting them into carbonates. This principle is pivotal in chemistry, as it shows how gaseous substances can be manipulated via controlled reactions in aqueous solutions.

Calcium Bicarbonate Solubility

Upon continual exposure to carbon dioxide beyond the point of initial reaction, calcium carbonate (\(\text{CaCO}\_3\)) dissolves to form calcium bicarbonate (\(\text{Ca(HCO}\_3\text{)}_2\)). The process illustrates that with an excess of \(\text{CO}\_2\), even a solid like calcium carbonate can transform into a soluble form:

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

Calcium bicarbonate's solubility is crucial because it dictates how compounds dissolve and reform in various conditions. This phenomenon is observed in nature, especially in the formation and erosion of limestone structures.

Precipitate Reformation

Heat applied to a solution containing calcium bicarbonate initiates the decomposition back into calcium carbonate, demonstrating the reversible nature of the reaction:

• Reaction on Heating: \( \text{Ca(HCO}_3\text{)}_2 \rightarrow \text{CaCO}_3 + \text{CO}_2 + \text{H}_2\text{O} \)

This reformation of the precipitate signifies the removal of excess carbon dioxide and moisture, causing the white solid to reappear. The white precipitate signals the release of carbon dioxide gas, affirming the cyclical relationship between calcium bicarbonate and calcium carbonate when environmental conditions change, like temperature.

Gas Identification in Reactions

Identifying gases in chemical reactions requires understanding the reagents and products involved. In the provided scenario, both gases 'A' and 'B' are determined to be carbon dioxide (\(\text{CO}\_2\)). The process involves:

• First, the initial bubbling of \(\text{CO}\_2\) with slaked lime forms a precipitate of calcium carbonate.
• As more \(\text{CO}\_2\) is added, calcium bicarbonate forms, dissolving the precipitate.
• On heating, the precipitate reforms, releasing additional \(\text{CO}\_2\), confirmed as gas 'B.'

By observing these transformations, chemists can identify unknown gases based on their reactions and changing conditions in a chemical system, ensuring accurate detection and management of substances.