The reaction is:

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

Calculate Moles of $\mathrm{CaCO}_3$:

The molar mass of $\mathrm{CaCO}_3$ is calculated as

$ 40 + 12 + 3 \times 16 = 100 \ \mathrm{g/mol} $

Given 1000 g of $\mathrm{CaCO}_3$, the number of moles is

$ \frac{1000}{100} = 10 \ \mathrm{moles} $

Calculate Moles of $\mathrm{HCl}$:

Using the given concentration and volume for $\mathrm{HCl}$, the moles are calculated as follows:

$ 0.76 \times \frac{250}{1000} = 0.19 \ \mathrm{moles} $

$\mathrm{HCl}$ is the limiting reagent (L.R.) because it has fewer moles than needed to completely react with the $\mathrm{CaCO}_3$.

Calculate Moles of $\mathrm{CaCl}_2$ Formed:

According to the stoichiometry of the reaction, 2 moles of $\mathrm{HCl}$ produce 1 mole of $\mathrm{CaCl}_2$. Therefore, the moles of $\mathrm{CaCl}_2$ formed are:

$ \frac{0.19}{2} = 0.095 \ \mathrm{moles} $

Calculate Mass of $\mathrm{CaCl}_2$:

The molar mass of $\mathrm{CaCl}_2$ is:

$ 40 + 2 \times 35.5 = 111 \ \mathrm{g/mol} $

Thus, the mass of $\mathrm{CaCl}_2$ is:

$ 0.095 \times 111 = 10.545 \ \mathrm{g} $

Hence, the mass of $\mathrm{CaCl}_2$ formed is 10.545 g.