Question

A sample of Impure solid calcium carbonate, CaCo_3, of mass 12,5 g reacted with dilute hydrochloric acid, HCl (aq), according to the following balanced equation:
CaCo_3 (s) +2 HCl (aq) - >CaCl_2 (aq) +CO_2 (g) +H_2O
Calculate the percentage purity of the calcium carbonate

Answer 1

To calculate the percentage purity of the impure calcium carbonate sample, we can follow these steps:

1. **Determine the molar mass of calcium carbonate, CaCO₃**:
- Calcium (Ca): 40.08 g/mol
- Carbon (C): 12.01 g/mol
- Oxygen (O): 16.00 g/mol × 3 = 48.00 g/mol
- Molar mass of CaCO₃ = 40.08 + 12.01 + 48.00 = 100.09 g/mol

2. **Calculate the moles of CaCO₃ in the impure sample**:
- Mass of the impure sample = 12.5 g
- To find the moles of CaCO₃, we use the formula:
\[
\text{Moles of CaCO₃} = \frac{\text{mass}}{\text{molar mass}} = \frac{12.5 \, \text{g}}{100.09 \, \text{g/mol}} \approx 0.1247 \, \text{mol}
\]

3. **Using the balanced equation to find the moles of HCl required**:
According to the balanced equation:
\[
\text{CaCO}_3 (s) + 2 \text{HCl} (aq) \rightarrow \text{CaCl}_2 (aq) + \text{CO}_2 (g) + \text{H}_2\text{O} (l)
\]
From the equation, 1 mole of CaCO₃ reacts with 2 moles of HCl.

4. **Calculate the moles of HCl needed for 0.1247 moles of CaCO₃**:
\[
\text{Moles of HCl} = 2 \times \text{Moles of CaCO₃} = 2 \times 0.1247 \, \text{mol} \approx 0.2494 \, \text{mol}
\]

5. **Now, calculate the mass of pure CaCO₃ that corresponds to the moles calculated**:
\[
\text{Mass of pure CaCO₃} = \text{Moles} \times \text{Molar Mass} = 0.1247 \, \text{mol} \times 100.09 \, \text{g/mol} \approx 12.5 \, \text{g}
\]

6. **Finally, calculate the percentage purity of the calcium carbonate**:
\[
\text{Percentage Purity} = \left( \frac{\text{mass of pure CaCO}_3}{\text{total mass of sample}} \right) \times 100 = \left( \frac{12.5 \, \text{g}}{12.5 \, \text{g}} \right) \times 100 = 100\%
\]

### Conclusion:
The percentage purity of the calcium carbonate in the sample is **100%**.