A mass of 0.4510 g of $$ \mathrm{CaCO}_{3} $$ sample was transferred into a 250 mL . Erlenmeyer fask. Then 50.00 mL of 0.2275 M HCl was pipetted into the sample and the solution was boiled for two minutes. The excess acid was titrated with 0.1158 M NaOH . The average volume of NaOH to reach the end-point was 24.19 mL .

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1. Determine the moles of $$\mathrm{HCl}$$ originally added:

$$
n_{\mathrm{HCl, initial}} = V \times M = 0.05000\,\text{L} \times 0.2275\,\text{M} = 0.011375\,\text{mol}
$$

2. Calculate the moles of $$\mathrm{NaOH}$$ used in the titration of the excess acid:

$$
n_{\mathrm{NaOH}} = 0.02419\,\text{L} \times 0.1158\,\text{M} \approx 0.002799\,\text{mol}
$$

Since the titration neutralizes the excess $$\mathrm{HCl}$$, the moles of excess $$\mathrm{HCl}$$ are also $$0.002799\,\text{mol}$$.

3. Find the moles of $$\mathrm{HCl}$$ that reacted with the $$\mathrm{CaCO}_{3}$$ sample:

$$
n_{\mathrm{HCl, reacted}} = n_{\mathrm{HCl, initial}} - n_{\mathrm{HCl, excess}} = 0.011375\,\text{mol} - 0.002799\,\text{mol} \approx 0.008576\,\text{mol}
$$

4. Use the reaction stoichiometry between $$\mathrm{CaCO}_{3}$$ and $$\mathrm{HCl}$$. The balanced reaction is:

$$
\mathrm{CaCO}_{3} + 2\,\mathrm{HCl} \rightarrow \mathrm{CaCl}_{2} + \mathrm{H}_{2}\mathrm{O} + \mathrm{CO}_{2}
$$

Since $$1$$ mole of $$\mathrm{CaCO}_{3}$$ reacts with $$2$$ moles of $$\mathrm{HCl}$$, the moles of $$\mathrm{CaCO}_{3}$$ that reacted are:

$$
n_{\mathrm{CaCO}_{3}} = \frac{n_{\mathrm{HCl, reacted}}}{2} = \frac{0.008576\,\text{mol}}{2} \approx 0.004288\,\text{mol}
$$

5. Calculate the mass of pure $$\mathrm{CaCO}_{3}$$ represented by these moles. The molar mass of $$\mathrm{CaCO}_{3}$$ is approximately:

$$
M_{\mathrm{CaCO}_{3}} = 40.078\,(\mathrm{Ca}) + 12.011\,(\mathrm{C}) + 3 \times 15.999\,(\mathrm{O}) \approx 100.086\,\text{g/mol}
$$

Thus, the mass of pure $$\mathrm{CaCO}_{3}$$ is:

$$
m_{\mathrm{pure\,CaCO}_{3}} = n_{\mathrm{CaCO}_{3}} \times M_{\mathrm{CaCO}_{3}} \approx 0.004288\,\text{mol} \times 100.086\,\text{g/mol} \approx 0.4292\,\text{g}
$$

6. Determine the percent purity of the sample. Given that the total sample mass is $$0.4510\,\text{g}$$:

$$
\%\text{Purity} = \left(\frac{m_{\mathrm{pure\,CaCO}_{3}}}{m_{\mathrm{sample}}}\right) \times 100\% = \left(\frac{0.4292\,\text{g}}{0.4510\,\text{g}}\right) \times 100\% \approx 95.18\%
$$
The sample is approximately 95.18% pure $$\mathrm{CaCO}_{3}$$.

A mass of 0.4510 g of sample was transferred into a 250 mL . Erlenmeyer
fask.
Then 50.00 mL of 0.2275 M HCl was pipetted into the sample and the solution was
boiled for two minutes.
The excess acid was titrated with 0.1158 M NaOH . The average volume of NaOH
to reach the end-point was 24.19 mL .

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A mass of 0.4510 g of sample was transferred into a 250 mL . Erlenmeyer fask. Then 50.00 mL of 0.2275 M HCl was pipetted into the sample and the solution was boiled for two minutes. The excess acid was titrated with 0.1158 M NaOH . The average volume of NaOH to reach the end-point was 24.19 mL .