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: $Ca=40$, $C=12$ & $O=16$) ________.

• A. 4.4g
• B. 3.96g
• C. 2.2g
• D. 0.44g
• E. 0.22g

Hint:

Always calculate based on the pure weight of the reactant.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
This is a stoichiometry problem involving a chemical reaction with an impure reactant. We need to first calculate the mass of the pure reactant that actually participates in the reaction, then use the balanced chemical equation to find the mass of the product formed.
Step 2: Key Formula or Approach
1. Calculate the mass of pure Calcium Carbonate (CaCO\(_3\)).
2. Write the balanced chemical equation for the reaction.
3. Calculate the molar masses of CaCO\(_3\) and Carbon Dioxide (CO\(_2\)).
4. Convert the mass of pure CaCO\(_3\) to moles.
5. Use the mole ratio from the balanced equation to find the moles of CO\(_2\) produced.
6. Convert the moles of CO\(_2\) to mass.
Step 3: Detailed Explanation
1. Calculate the mass of pure CaCO\(_3\).
- Total mass of the sample = 10 g.
- Purity = 90%.
- Mass of pure CaCO\(_3\) = \(10 \text{ g} \times \frac{90}{100} = 9 \text{ g}\).
2. Balanced Chemical Equation.
Calcium carbonate reacts with hydrochloric acid to produce calcium chloride, water, and carbon dioxide.
\[ \text{CaCO}_3(s) + 2\text{HCl}(aq) \to \text{CaCl}_2(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g) \] The equation is balanced. The mole ratio between CaCO\(_3\) and CO\(_2\) is 1:1.
3. Calculate Molar Masses.
- Molar mass of CaCO\(_3\) = Ca + C + 3(O) = 40 + 12 + 3(16) = 40 + 12 + 48 = 100 g/mol.
- Molar mass of CO\(_2\) = C + 2(O) = 12 + 2(16) = 12 + 32 = 44 g/mol.
4. Calculate moles of pure CaCO\(_3\).
\[ \text{Moles of CaCO}_3 = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{9 \text{ g}}{100 \text{ g/mol}} = 0.09 \text{ mol} \] 5. Calculate moles of CO\(_2\) produced.
From the balanced equation, 1 mole of CaCO\(_3\) produces 1 mole of CO\(_2\).
Therefore, 0.09 moles of CaCO\(_3\) will produce 0.09 moles of CO\(_2\).
6. Calculate the mass of CO\(_2\).
\[ \text{Mass of CO}_2 = \text{Moles} \times \text{Molar Mass} \] \[ \text{Mass of CO}_2 = 0.09 \text{ mol} \times 44 \text{ g/mol} \] \[ \text{Mass of CO}_2 = 3.96 \text{ g} \] Step 4: Final Answer
The mass of CO\(_2\) gas liberated is 3.96g.