1. Write the balanced reaction

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

2. Calculate moles of reactants

• For HCl:

$n(\mathrm{HCl}) = M \times V = 0.76\,\mathrm{mol\,L^{-1}} \times 0.250\,\mathrm{L} = 0.19\,\mathrm{mol}$

• For $\mathrm{CaCO_3}$:

Molar mass of $\mathrm{CaCO_3}$

$= 40 + 12 + 3(16) = 100\,\mathrm{g\,mol^{-1}}$

So,

$n(\mathrm{CaCO_3}) = \frac{1000}{100} = 10\,\mathrm{mol}$

3. Find the limiting reagent

From the balanced equation:

$1\,\text{mol } \mathrm{CaCO_3} : 2\,\text{mol } \mathrm{HCl}$

To react completely with $10$ mol $\mathrm{CaCO_3}$, required HCl would be:

$20\,\mathrm{mol}$

But available HCl is only $0.19$ mol, so HCl is the limiting reagent.

4. Calculate moles of $\mathrm{CaCl_2}$ formed

From stoichiometry:

$2\,\mathrm{mol\ HCl} \rightarrow 1\,\mathrm{mol\ CaCl_2}$

Therefore,

$n(\mathrm{CaCl_2}) = \frac{0.19}{2} = 0.095\,\mathrm{mol}$

5. Calculate mass of $\mathrm{CaCl_2}$ formed

Molar mass of $\mathrm{CaCl_2}$:

$40 + 2(35.5) = 111\,\mathrm{g\,mol^{-1}}$

Hence,

$m(\mathrm{CaCl_2}) = 0.095 \times 111 = 10.545\,\mathrm{g}$

6. Choose the correct option

$\boxed{\text{Option C: } 10.545\,\mathrm{g}}$

7. Comparison with stored answer

Stored correct answer is C, which matches the derived answer.