$$CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + CO_2(g) + H_2O(l)$$

Consider the above reaction, what mass of $$CaCl_2$$ will be formed if 250 mL of 0.76 M HCl reacts with 1000 g of $$CaCO_3$$?

(Given: Molar mass of Ca, C, O, H and Cl are 40, 12, 16, 1 and 35.5 g mol$$^{-1}$$, respectively)

The balanced chemical equation is $$CaCO_3(s)+2HCl(aq)\rightarrow CaCl_2(aq)+CO_2(g)+H_2O(l)$$.

Step 1: Calculate moles of each reactant.

Molarity formula: $$M = \frac{\text{moles}}{\text{volume in L}} \; \Rightarrow \; \text{moles} = M \times V$$.

For $$HCl$$:
Volume $$V = 250\ \text{mL} = 0.250\ \text{L}$$,
Molarity $$M = 0.76\ \text{mol L}^{-1}$$.
Therefore, moles of $$HCl = 0.76 \times 0.250 = 0.19\ \text{mol}$$.

Molar mass of $$CaCO_3$$:
$$40 + 12 + 3(16) = 40 + 12 + 48 = 100\ \text{g mol}^{-1}$$.

Given mass = $$1000\ \text{g}$$, so moles of $$CaCO_3$$ are
$$\frac{1000}{100} = 10\ \text{mol}$$.

Step 2: Identify the limiting reagent.

From the equation, $$2$$ moles of $$HCl$$ react with $$1$$ mole of $$CaCO_3$$.
Required moles of $$HCl$$ for $$10$$ moles $$CaCO_3$$:
$$2 \times 10 = 20\ \text{mol}$$.

Available moles of $$HCl$$ = $$0.19\ \text{mol} \lt 20\ \text{mol}$$, so $$HCl$$ is the limiting reagent.

Step 3: Moles of $$CaCl_2$$ formed.

Stoichiometry: $$2\ \text{mol HCl} \rightarrow 1\ \text{mol CaCl}_2$$.
Thus, moles of $$CaCl_2 = \frac{0.19}{2} = 0.095$$.

Step 4: Mass of $$CaCl_2$$ produced.

Molar mass of $$CaCl_2$$:
$$40 + 2(35.5) = 40 + 71 = 111\ \text{g mol}^{-1}$$.

Mass = moles $$\times$$ molar mass = $$0.095 \times 111 = 10.545\ \text{g}$$.

Final Answer: $$10.545\ \text{g}$$ of $$CaCl_2$$ will be formed. This matches Option C.