The experiment is repeated with an eggshell sample, and the experimental data are recorded in the ta — Stoichiometry Chemistry Question

STEPS:

1. Calculate the pressure change ($\Delta P$) caused by the reaction:
The reaction produces carbon dioxide gas ($\text{CO}_2$), which increases the pressure inside the sealed, rigid vessel. The change in pressure is calculated by subtracting the initial pressure from the final pressure at the completion of the reaction:
$$\Delta P = P_{\text{completion}} - P_{\text{prior}} \quad$$
$$\Delta P = 0.870\text{ atm} - 0.800\text{ atm} = \mathbf{0.070\text{ atm}} \quad$$

2. Use the calibration line to find the mass of reacting $\text{CaCO}_3(s)$:
Locate $0.070\text{ atm}$ on the vertical axis ("Pressure of Gas Produced") of the calibration graph. Moving horizontally to the calibration line and then vertically down to the horizontal axis ("Mass of $\text{CaCO}_3(s)$") yields a corresponding mass of approximately $0.145\text{ g}$ of pure calcium carbonate.

3. Calculate the mass percent of $\text{CaCO}_3(s)$ in the eggshell:
To find the percentage of the eggshell sample that consists of calcium carbonate, divide the calculated mass of $\text{CaCO}_3$ by the total starting mass of the eggshell sample ($0.200\text{ g}$):
$$\text{Mass Percent of CaCO}_3 = \frac{\text{Mass of CaCO}_3}{\text{Mass of Eggshell Sample}} \times 100\% \quad$$
$$\text{Mass Percent of CaCO}_3 = \frac{0.145\text{ g}}{0.200\text{ g}} \times 100\% = \mathbf{72.5\% \approx 73\%} \quad$$

4. Identify the closest multiple-choice option:
The calculated value of $73\%$ is closest to 75% (Option D).

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WHY_OTHERS_WRONG:

• Option A is incorrect (30%): A student would arrive at this value by making a significant calculation or graph-reading error, such as estimating the reacting mass of $\text{CaCO}_3$ to be $0.060\text{ g}$ instead of $0.145\text{ g}$ ($\frac{0.060\text{ g}}{0.200\text{ g}} \times 100\% = 30\%$).
• Option B is incorrect (45%): This represents a value that is much too low. It corresponds to an incorrect reading of the calibration line where the mass of $\text{CaCO}_3$ is mistakenly read as $0.090\text{ g}$ ($\frac{0.090\text{ g}}{0.200\text{ g}} \times 100\% = 45\%$).
• Option C is incorrect (60%): This value is obtained if a student incorrectly identifies the mass of $\text{CaCO}_3$ as approximately $0.120\text{ g}$ ($\frac{0.120\text{ g}}{0.200\text{ g}} \times 100\% = 60\%$).