Calcium carbonate reacts with aqueous $HCl$ to give $CaCl_2$ and $CO_2$ according to the reaction given below: $CaCO_{3(s)} + 2HCl_{(aq)} \to CaCl_{2(aq)} + CO_{2(g)} + H_2O_{(l)}$
What mass of $CaCl_2$ will be formed when $250 \ mL$ of $0.76 \ M$ $HCl$ reacts with $1000 \ g$ of $CaCO_3$? Name the limiting reagent. Calculate the number of moles of $CaCl_2$ formed in the reaction.

(N/A) Molar mass of $CaCO_3 = 40 + 12 + 3 \times 16 = 100 \ g \ mol^{-1}$.
Moles of $CaCO_3$ in $1000 \ g$:
$n_{CaCO_3} = \frac{1000 \ g}{100 \ g \ mol^{-1}} = 10 \ mol$.
Moles of $HCl$ in $250 \ mL$ of $0.76 \ M$ solution:
$n_{HCl} = \text{Molarity} \times \text{Volume (L)} = 0.76 \ mol \ L^{-1} \times 0.250 \ L = 0.19 \ mol$.
According to the stoichiometry of the reaction $CaCO_{3(s)} + 2HCl_{(aq)} \to CaCl_{2(aq)} + CO_{2(g)} + H_2O_{(l)}$:
$1 \ mol$ of $CaCO_3$ requires $2 \ mol$ of $HCl$.
Since we have $10 \ mol$ of $CaCO_3$ and only $0.19 \ mol$ of $HCl$,$HCl$ is the limiting reagent.
$2 \ mol$ of $HCl$ produces $1 \ mol$ of $CaCl_2$.
Therefore,$0.19 \ mol$ of $HCl$ produces $\frac{0.19}{2} = 0.095 \ mol$ of $CaCl_2$.
Molar mass of $CaCl_2 = 40 + (2 \times 35.5) = 111 \ g \ mol^{-1}$.
Mass of $CaCl_2 = 0.095 \ mol \times 111 \ g \ mol^{-1} = 10.545 \ g$.