Question:
\(CaCO_3\) was reacted with dil HCl to form three products. One of the products formed, when passed into slaked lime gave \(X\). What is \(X\)?
\(CaCO_3\) was reacted with dil HCl to form three products. One of the products formed, when passed into slaked lime gave \(X\). What is \(X\)?
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Passing \(CO_2\) through lime water forms milky \(CaCO_3\), which is the standard test for carbon dioxide gas.
Updated On: Jun 25, 2026
- \(CaCl_2\)
- \(CaCO_3\)
- \(Ca(OH)_2\)
- \(Ca(HCO_3)_2\)
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The Correct Option is B
Solution and Explanation
Step 1: Write the reaction of \(CaCO_3\) with dilute HCl.
\[ CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O \] The three products formed are: \[ CaCl_2,\ CO_2,\ H_2O \]
Step 2: Identify the product passed into slaked lime.
Slaked lime is \[ Ca(OH)_2 \] Carbon dioxide reacts with slaked lime as: \[ CO_2+Ca(OH)_2\rightarrow CaCO_3+H_2O \]
Step 3: Identify \(X\).
The white precipitate formed is calcium carbonate: \[ CaCO_3 \]
Step 4: Final conclusion.
Hence, \[ \boxed{CaCO_3} \] is \(X\).
\[ CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O \] The three products formed are: \[ CaCl_2,\ CO_2,\ H_2O \]
Step 2: Identify the product passed into slaked lime.
Slaked lime is \[ Ca(OH)_2 \] Carbon dioxide reacts with slaked lime as: \[ CO_2+Ca(OH)_2\rightarrow CaCO_3+H_2O \]
Step 3: Identify \(X\).
The white precipitate formed is calcium carbonate: \[ CaCO_3 \]
Step 4: Final conclusion.
Hence, \[ \boxed{CaCO_3} \] is \(X\).
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