Let's first write the balanced chemical equation for the reaction between calcium carbonate ($$\mathrm{CaCO}_3$$) and hydrochloric acid ($$\mathrm{HCl}$$).

$$ \mathrm{CaCO}_3 + 2\mathrm{HCl} \rightarrow \mathrm{CaCl}_2 + \mathrm{H}_2\mathrm{O} + \mathrm{CO}_2 $$

From the equation, we can see that 1 mole of $$\mathrm{CaCO}_3$$ reacts with 2 moles of $$\mathrm{HCl}$$ to produce 1 mole of $$\mathrm{CO}_2$$.

Now, let's calculate the moles of $$\mathrm{CaCO}_3$$. The molar mass of $$\mathrm{CaCO}_3$$ is:

$$ \text{Molar mass of} \ \mathrm{CaCO}_3 = 40 + 12 + 3 \times 16 = 100 \ \mathrm{g/mol} $$

Given that we have $$20 \ \mathrm{g}$$ of $$\mathrm{CaCO}_3$$, the number of moles of $$\mathrm{CaCO}_3$$ is:

$$ \text{Moles of} \ \mathrm{CaCO}_3 = \frac{20 \ \mathrm{g}}{100 \ \mathrm{g/mol}} = 0.2 \ \mathrm{mol} $$

Now, let's determine the moles of $$\mathrm{HCl}$$ in the given solution. The solution is $$20\%$$ $$\mathrm{HCl}$$ by weight, and we have $$100 \ \mathrm{mL}$$ of the solution. Assuming the density of the solution is approximately $$1 \ \mathrm{g/mL}$$, we get:

$$ \text{Mass of} \ \mathrm{HCl} \ \text{solution} = 100 \ \mathrm{mL} \times 1 \ \mathrm{g/mL} = 100 \ \mathrm{g} $$

Since the solution is $$20\%$$ $$\mathrm{HCl}$$, the mass of $$\mathrm{HCl}$$ is:

$$ \text{Mass of} \ \mathrm{HCl} = 20\% \ \text{of} \ 100 \ \mathrm{g} = 20 \ \mathrm{g} $$

The molar mass of $$\mathrm{HCl}$$ is:

$$ \text{Molar mass of} \ \mathrm{HCl} = 1 + 35.5 = 36.5 \ \mathrm{g/mol} $$

The number of moles of $$\mathrm{HCl}$$ is:

$$ \text{Moles of} \ \mathrm{HCl} = \frac{20 \ \mathrm{g}}{36.5 \ \mathrm{g/mol}} \approx 0.548 \ \mathrm{mol} $$

The reaction requires 2 moles of $$\mathrm{HCl}$$ for 1 mole of $$\mathrm{CaCO}_3$$. Since we have only $$0.2 \ \mathrm{mol}$$ of $$\mathrm{CaCO}_3$$, we need:

$$ \text{Required moles of} \ \mathrm{HCl} = 0.2 \ \mathrm{mol} \times 2 = 0.4 \ \mathrm{mol} $$

We have more than enough $$\mathrm{HCl}$$ ($$0.548 \ \mathrm{mol}$$ available) to fully react with $$0.2 \ \mathrm{mol}$$ of $$\mathrm{CaCO}_3$$.

Therefore, the amount of $$\mathrm{CO}_2$$ produced will be equal to the moles of $$\mathrm{CaCO}_3$$ reacted, which is $$0.2 \ \mathrm{mol}$$.

To find the mass of $$\mathrm{CO}_2$$ produced, we use its molar mass:

$$ \text{Molar mass of} \ \mathrm{CO}_2 = 12 + 2 \times 16 = 44 \ \mathrm{g/mol} $$

The mass of $$\mathrm{CO}_2$$ is:

$$ \text{Mass of} \ \mathrm{CO}_2 = 0.2 \ \mathrm{mol} \times 44 \ \mathrm{g/mol} = 8.80 \ \mathrm{g} $$

The amount of $$\mathrm{CO}_2$$ produced is $$8.80 \ \mathrm{g}$$.

Given the options, the correct answer is:

Option B - 8.80 g

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