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

The question asks about the sequence of products formed when carbon dioxide (\(CO_2\)) is passed in excess through slaked lime (\(Ca(OH)_2\)). Understanding the chemical reactions that occur in this scenario is essential to identify the correct sequence of products.

1. Slaked lime, chemically known as calcium hydroxide (\(Ca(OH)_2\)), reacts with carbon dioxide (\(CO_2\)) to form calcium carbonate (\(CaCO_3\)) as a precipitate. The balanced chemical equation is:
   Ca(OH)_2 + CO_2 \rightarrow CaCO_3 + H_2O
2. When an excess of carbon dioxide is passed through the mixture, the initially formed calcium carbonate (\(CaCO_3\)) reacts with the excess \(CO_2\) and water to form calcium bicarbonate \((Ca(HCO_3)_2)\), which is soluble in water. The balanced equation for this reaction is:
   CaCO_3 + CO_2 + H_2O \rightarrow Ca(HCO_3)_2

The sequence of reactions above indicates that when excess \(CO_2\) is introduced, the products formed are first calcium carbonate and then calcium bicarbonate. Therefore, the correct sequence of products is \(CaCO_3\) and \(Ca(HCO_3)_2\).

Hence, the correct answer is: \(CaCO_3, Ca(HCO_3)_2\).