Question

If $20 \mathrm{g}$ of $\mathrm{CaCO}_3$ is treated with $100 \mathrm{mL}$ of $20%$ HCl solution, the amount of $\mathrm{CO}_2$ produced is

• A. $22.4 \mathrm{L}$
• B. $8.80 \mathrm{g}$
• C. $4.40 \mathrm{g}$
• D. $2.24 \mathrm{L}$

Hint:

Identify the limiting reagent to find the amount of product formed.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The reaction is $\mathrm{CaCO_3 + 2HCl \rightarrow CaCl_2 + CO_2 + H_2O}$. Identify the limiting reagent.
Step 2: Detailed Explanation:
Molar mass of $\mathrm{CaCO_3} = 100$ g/mol. $20$ g = $0.2$ mol. HCl: $100$ mL of $20%$ solution = $20$ g HCl = $\frac{20}{36.5} \approx 0.548$ mol. Limiting reagent is $\mathrm{CaCO_3}$. $0.2$ mol $\mathrm{CaCO_3}$ produces $0.2$ mol $\mathrm{CO_2} = 0.2 \times 44 = 8.8$ g.
Step 3: Final Answer:
The amount of $\mathrm{CO}_2$ produced is $8.80$ g.