Limestone granules react with excess HCl. Curve X shows the reaction of 0.2g. Which curve represents the reaction of 0.4g under identical conditions?

From the graph, Curve X shows that the reaction with $0.2\text{g}$ of limestone produces approximately $22\text{cm}^3$ of carbon dioxide. The curve flattens out, indicating the reaction has completed. Since the acid is in excess, the amount of carbon dioxide produced is limited by the amount of limestone.

The second experiment uses $0.4\text{g}$ of limestone, which is double the amount used in the first experiment ($0.4\text{g} / 0.2\text{g} = 2$). All other conditions (temperature, acid concentration, volume of acid, granule size) are identical.

Since the amount of limestone is doubled, and limestone is the limiting reactant, the total volume of carbon dioxide produced will also double.
Volume of $\text{CO}_2$ for $0.2\text{g}$ limestone $\approx 22\text{cm}^3$.
Volume of $\text{CO}_2$ for $0.4\text{g}$ limestone $\approx 2 \times 22\text{cm}^3 = 44\text{cm}^3$.

Using $0.4\text{g}$ of the same limestone granules means there is a larger total surface area available for the reaction compared to $0.2\text{g}$. A larger surface area leads to a faster rate of reaction because there are more sites for collisions between reactant particles. Therefore, the initial slope of the curve for the second experiment should be steeper than Curve X.

• Curve (i): Reaches a final volume of approximately $44\text{cm}^3$ and has a steeper initial slope than Curve X. This matches our predictions.
• Curve (ii): Reaches a final volume of approximately $33\text{cm}^3$. This is not double the volume of X.
• Curve (iii): Reaches a final volume of approximately $22\text{cm}^3$ (same as X). This is incorrect.
• Curve (iv): Reaches a final volume of approximately $15\text{cm}^3$ and has a less steep initial slope. This is incorrect.

Therefore, Curve (i) correctly represents the results of the second experiment.

The final answer is $\text{A}$.