Determine the moles of CaCO3 used given the volume of CO2 produced in Experiment A and the molar gas volume at 40°C.

From the graph, the reaction for Experiment A stops at a maximum volume of CO$_2$ gas.

$$\text{Volume of CO}_2 = 460 \text{ cm}^3$$

Since the molar gas volume is given in dm^3$$\cdotmol$^{-1}$, convert the volume of CO$_2$ from cm$^3$ to dm$^3$.

$$1 \text{ dm}^3 = 1000 \text{ cm}^3$$ $$\text{Volume of CO}_2 = \frac{460}{1000} \text{ dm}^3 = 0.460 \text{ dm}^3$$

Step 3: Calculate the moles of CO$_2$ produced.
Use the given molar gas volume at 40°C, which is 25.7 dm^3$$\cdotmol$^{-1}$.

$$\text{Moles of CO}_2 = \frac{\text{Volume of CO}_2}{\text{Molar gas volume}}$$ $$\text{Moles of CO}_2 = \frac{0.460 \text{ dm}^3}{25.7 \text{ dm}^3\cdot\text{mol}^{-1}}$$ $$\text{Moles of CO}_2 \approx 0.0178988 \text{ mol}$$

From the balanced chemical equation:

$$\text{CaCO}_3(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{CaCl}_2(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})$$

The mole ratio of CaCO$_3$ to CO$_2$ is 1:1.
Therefore, moles of CaCO$_3$ used = moles of CO$_2$ produced.

$$\text{Moles of CaCO}_3 = 0.0178988 \text{ mol}$$

Molar mass of Ca = 40.08 g/mol
Molar mass of C = 12.01 g/mol
Molar mass of O = 16.00 g/mol

$$\text{Molar mass of CaCO}_3 = 40.08 + 12.01 + (3 \times 16.00)$$ $$\text{Molar mass of CaCO}_3 = 40.08 + 12.01 + 48.00 = 100.09 \text{ g/mol}$$

$$\text{Mass of CaCO}_3 = \text{Moles of CaCO}_3 \times \text{Molar mass of CaCO}_3$$ $$\text{Mass of CaCO}_3 = 0.0178988 \text{ mol} \times 100.09 \text{ g/mol}$$ $$\text{Mass of CaCO}_3 \approx 1.7915 \text{ g}$$

Rounding to two decimal places:

$$\text{Mass of CaCO}_3 \approx 1.79 \text{ g}$$

The mass of CaCO$_3$(s) used in experiment A is $\boxed{\text{1.79 g}}$.