Another sample of eggshell reacts completely with 4.0 mL of an HCl(aq) solution of unknown concentra — Stoichiometry Chemistry Question

### STEPS:

1. Relate the gas pressure produced to the mass of reactant on the calibration graph:
Find $0.095 \text{ atm}$ 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 reacting mass of approximately $0.20 \text{ g}$ of pure calcium carbonate ($\text{CaCO}_3$).

2. Calculate the moles of reacted calcium carbonate:
The molar mass of $\text{CaCO}_3$ is approximately $100. \text{ g/mol}$. Convert the mass from step 1 into moles ($n$) using the formula $n = \frac{m}{\mathcal{M}}$:
$$\text{Moles of } \text{CaCO}_3 = \frac{0.20 \text{ g}}{100. \text{ g/mol}} = \mathbf{0.0020 \text{ mol of } \text{CaCO}_3} \text{}$$

3. Apply reaction stoichiometry to find the minimum moles of hydrochloric acid ($\text{HCl}$) needed:
From the balanced net ionic equation representing the reaction of solid calcium carbonate with hydrochloric acid:
$$2\text{ H}^+(aq) + \text{CaCO}_3(s) \rightarrow \text{Ca}^{2+}(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g) \text{}$$
This chemical equation shows a $2:1$ stoichiometric ratio between $\text{H}^+$ ions (from the strong acid $\text{HCl}$) and $\text{CaCO}_3(s)$. Therefore, reacting $0.0020 \text{ mol}$ of calcium carbonate completely requires exactly twice as many moles of acid:
$$\text{Moles of } \text{HCl} \text{ needed} = 0.0020 \text{ mol } \text{CaCO}_3 \times \frac{2 \text{ mol } \text{HCl}}{1 \text{ mol } \text{CaCO}_3} = \mathbf{0.0040 \text{ mol of } \text{HCl}} \text{}$$

4. Calculate the minimum molar concentration (Molarity) of the $\text{HCl}$ solution:
The sample of eggshell reacts completely with exactly $4.0 \text{ mL}$ of the acid solution. Convert this volume to liters ($4.0 \text{ mL} = 0.0040 \text{ L}$) and solve for molarity ($M = \frac{\text{moles}}{\text{Liters}}$):
$$\text{Molarity of } \text{HCl} = \frac{0.0040 \text{ mol of } \text{HCl}}{0.0040 \text{ L of solution}} = \mathbf{1.0 \text{ M}} \text{}$$
This confirms that the concentration of the hydrochloric acid solution was at least $1.0 \text{ M}$, identifying Option D as the correct answer.

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### WHY_OTHERS_WRONG:

• Option A is incorrect ($0.0020 \text{ M}$): This option represents a severe decimal place or unit conversion error. A student might select this value if they mistakenly equate the calculated moles of $\text{CaCO}_3$ ($0.0020 \text{ mol}$) directly with the molarity, or fail to convert milliliters to liters when dividing the moles of acid by the volume.
• Option B is incorrect ($0.050 \text{ M}$): This represents a calculation error where the $2:1$ reaction stoichiometry is neglected (treating it as $1:1$, yielding $0.0020 \text{ mol}$ of acid), and then dividing this incorrect mole value by $4.0 \text{ mL}$ while introducing an additional factor-of-ten decimal placement error ($0.0020 \text{ mol} \div 0.0040 \text{ L} = 0.50 \text{ M}$, miscalculated as $0.050 \text{ M}$).
• Option C is incorrect ($0.50 \text{ M}$): This is a classic stoichiometric distractor. It is the result of failing to apply the $2:1$ ratio of $\text{HCl}$ to $\text{CaCO}_3$ from the balanced equation. If a student assumes a $1:1$ reaction ratio, they would calculate that only $0.0020 \text{ mol}$ of $\text{HCl}$ is needed. Dividing $0.0020 \text{ mol}$ of $\text{HCl}$ by $0.0040 \text{ L}$ yields exactly $0.50 \text{ M}$.