Question:
Which of the following compounds is formed when chlorine reacts with dry slake lime \(\text{Ca(OH)}_2\) ?
Which of the following compounds is formed when chlorine reacts with dry slake lime \(\text{Ca(OH)}_2\) ?
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Chlorine and dry slaked lime give bleaching powder, CaOCl2.
Updated On: Oct 1, 2026
- \(\text{CaOCl}_2\)
- \(\text{CaO}_2\text{Cl}_2\)
- \(\text{CaOCl}\)
- \(\text{CaCl}_2\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
Chlorine reacts with dry slaked lime to form bleaching powder. It is a mixed salt of hypochlorite and chloride.
Step 2: Detailed Explanation:
\[ \text{Ca(OH)}_2 + \text{Cl}_2 \rightarrow \text{CaOCl}_2 + \text{H}_2\text{O} \]
The product has the formula \(\text{CaOCl}_2\), often written \(\text{Ca(OCl)Cl}\). The other formulae \(\text{CaO}_2\text{Cl}_2\) and \(\text{CaOCl}\) are not the product of this reaction. \(\text{CaCl}_2\) alone would form only if the hypochlorite part were absent.
Step 3: Final Answer:
The compound formed is \(\text{CaOCl}_2\), option (A).
Final Answer:
Bleaching powder, CaOCl2, forms. \[ \boxed{\text{(A) }\text{CaOCl}_2} \]
Chlorine reacts with dry slaked lime to form bleaching powder. It is a mixed salt of hypochlorite and chloride.
Step 2: Detailed Explanation:
\[ \text{Ca(OH)}_2 + \text{Cl}_2 \rightarrow \text{CaOCl}_2 + \text{H}_2\text{O} \]
The product has the formula \(\text{CaOCl}_2\), often written \(\text{Ca(OCl)Cl}\). The other formulae \(\text{CaO}_2\text{Cl}_2\) and \(\text{CaOCl}\) are not the product of this reaction. \(\text{CaCl}_2\) alone would form only if the hypochlorite part were absent.
Step 3: Final Answer:
The compound formed is \(\text{CaOCl}_2\), option (A).
Final Answer:
Bleaching powder, CaOCl2, forms. \[ \boxed{\text{(A) }\text{CaOCl}_2} \]
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