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 \(X\)?

• A. \(CaCl_2\)
• B. \(CaCO_3\)
• C. \(Ca(OH)_2\)
• D. \(Ca(HCO_3)_2\)

Hint:

Passing \(CO_2\) through lime water forms milky \(CaCO_3\), which is the standard test for carbon dioxide gas.

Answer 1

The Correct Option is B

Step 1: Write the reaction of CaCO3 with dilute HCl.
Calcium carbonate reacts with dilute hydrochloric acid to give three products: \[ CaCO_3 + 2HCl \rightarrow CaCl_2 + CO_2 + H_2O \] The products are \( CaCl_2 \), \( CO_2 \), and \( H_2O \).
Step 2: Identify slaked lime.
Slaked lime is \( Ca(OH)_2 \) (calcium hydroxide). It is a white powder formed by adding water to quicklime.
Step 3: Decide which product reacts with slaked lime.
Of the three products, only \( CO_2 \) is a gas that reacts with bases. \( CaCl_2 \) and \( H_2O \) do not react meaningfully with \( Ca(OH)_2 \) under these conditions.
Step 4: Write the reaction of CO2 with slaked lime.
Carbon dioxide passed into slaked lime gives a white precipitate: \[ CO_2 + Ca(OH)_2 \rightarrow CaCO_3 + H_2O \]
Step 5: Identify compound X.
The product X is \( CaCO_3 \) (calcium carbonate). This is the same compound we started with, which nicely completes the cycle.
Step 6: Match with the options.
X is \( CaCO_3 \), which is option 2. \[ \boxed{CaCO_3} \]