Question

What are the products formed in sequence when excess of \(CO_2\) is passed in slaked lime ?

• A. \(CaCO_3, Ca(HCO_3)_2\)
• B. \(Ca(HCO_3)_2, CaCO_3\)
• C. \(CaO, CaCO_3\)
• D. \(CaO, Ca(HCO_3)_2\)

Hint:

Passing \(CO_2\) through lime water is the standard test for identifying carbon dioxide gas. Remember: milky white = \(CaCO_3\); clear = \(Ca(HCO_3)_2\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Slaked lime is calcium hydroxide, \(Ca(OH)_2\). The reaction with carbon dioxide is a two-step process depending on the amount of \(CO_2\) available.
Step 2: Detailed Explanation:
1. Step 1 (Limited \(CO_2\)): When \(CO_2\) is passed through slaked lime, it reacts to form calcium carbonate, which is insoluble and makes the solution appear milky.
\[ Ca(OH)_2 + CO_2 \to CaCO_3(s) + H_2O(l) \]
2. Step 2 (Excess \(CO_2\)): If the passage of \(CO_2\) continues, the calcium carbonate reacts with water and more \(CO_2\) to form calcium bicarbonate. This compound is soluble in water, so the milkiness disappears.
\[ CaCO_3 + H_2O + CO_2 \to Ca(HCO_3)_2(aq) \]
Therefore, the sequence of products is first \(CaCO_3\) then \(Ca(HCO_3)_2\).
Step 3: Final Answer:
The products formed in sequence are \(CaCO_3\) and \(Ca(HCO_3)_2\).