Calculate the moles of calcium carbonate remaining after reacting with 0.2 moles of HCl.

The atomic masses are: Ca = 40.08 g/mol, C = 12.01 g/mol, O = 16.00 g/mol.

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

The balanced chemical equation is:

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

From the stoichiometry, 1 mole of $\text{CaCO}_3$ reacts with 2 moles of $\text{HCl}$.
Therefore, the moles of $\text{CaCO}_3$ required to react with 0.2 moles of $\text{HCl}$ are:

$$\text{Moles of CaCO}_3 \text{ (reacted)} = 0.2 \text{ mol HCl} \times \frac{1 \text{ mol CaCO}_3}{2 \text{ mol HCl}} = 0.1 \text{ mol CaCO}_3$$

Initial moles of $\text{CaCO}_3 = 0.149865 \text{ mol}$
Moles of $\text{CaCO}_3$ that will react = $0.1 \text{ mol}$
Since the initial moles of $\text{CaCO}_3$ ($0.149865 \text{ mol}$) are greater than the moles of $\text{CaCO}_3$ that will react ($0.1 \text{ mol}$), $\text{HCl}$ is the limiting reactant, and some $\text{CaCO}_3$ will remain.

$$\text{Moles of CaCO}_3 \text{ (remaining)} = \text{Moles of CaCO}_3 \text{ (initial)} - \text{Moles of CaCO}_3 \text{ (reacted)}$$ $$\text{Moles of CaCO}_3 \text{ (remaining)} = 0.149865 \text{ mol} - 0.1 \text{ mol} = 0.049865 \text{ mol}$$

$$\text{Mass of CaCO}_3 \text{ (remaining)} = \text{Moles remaining} \times \text{Molar mass}$$ $$\text{Mass of CaCO}_3 \text{ (remaining)} = 0.049865 \text{ mol} \times 100.09 \text{ g/mol} \approx 4.991 \text{ g}$$

Rounding to three significant figures:

$$\text{Mass of CaCO}_3 \text{ (remaining)} \approx 4.99 \text{ g}$$

The amount of calcium carbonate that would remain is $\text{4.99 g}$.