Question:
Ten grams of calcium carbonate which is only 90% pure is treated with excess hydrochloric acid. What is the mass of $CO_{2}$ gas liberated? (Atomic mass: $Ca=40$, $C=12$ & $O=16$) ________.
Ten grams of calcium carbonate which is only 90% pure is treated with excess hydrochloric acid. What is the mass of $CO_{2}$ gas liberated? (Atomic mass: $Ca=40$, $C=12$ & $O=16$) ________.
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Always calculate based on the pure weight of the reactant.
Updated On: Jun 26, 2026
- 4.4g
- 3.96g
- 2.2g
- 0.44g
- 0.22g
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The Correct Option is B
Solution and Explanation
Step 1: Concept
Determine the amount of pure reactant and use stoichiometry.
Step 2: Meaning
Pure $CaCO_3 = 90\% \text{ of } 10g = 9g$. Molar mass of $CaCO_3 = 100g/mol$; Molar mass of $CO_2 = 44g/mol$.
Step 3: Analysis
Reaction: $CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2$. 100g $CaCO_3$ gives 44g $CO_2$. Mass of $CO_2 = \frac{44}{100} \times 9$.
Step 4: Conclusion
Mass $= 3.96g$. Final Answer: (B)
Determine the amount of pure reactant and use stoichiometry.
Step 2: Meaning
Pure $CaCO_3 = 90\% \text{ of } 10g = 9g$. Molar mass of $CaCO_3 = 100g/mol$; Molar mass of $CO_2 = 44g/mol$.
Step 3: Analysis
Reaction: $CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2$. 100g $CaCO_3$ gives 44g $CO_2$. Mass of $CO_2 = \frac{44}{100} \times 9$.
Step 4: Conclusion
Mass $= 3.96g$. Final Answer: (B)
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