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.

(D) $1$. Calculate the moles of reactants:
Moles of $HCl = \text{Molarity} \times \text{Volume (L)} = 0.76 \ M \times 0.250 \ L = 0.19 \ mol$.
Moles of $CaCO_3 = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{1000 \ g}{100 \ g/mol} = 10 \ mol$.
$2$. Identify the limiting reagent:
According to the stoichiometry,$1 \ mol$ of $CaCO_3$ requires $2 \ mol$ of $HCl$.
For $10 \ mol$ of $CaCO_3$,we would need $20 \ mol$ of $HCl$.
Since we only have $0.19 \ mol$ of $HCl$,$HCl$ is the limiting reagent.
$3$. Calculate moles of $CaCl_2$ formed:
From the reaction,$2 \ mol$ of $HCl$ produces $1 \ mol$ of $CaCl_2$.
So,$0.19 \ mol$ of $HCl$ produces $\frac{0.19}{2} = 0.095 \ mol$ of $CaCl_2$.
$4$. Calculate mass of $CaCl_2$ formed:
Mass = $\text{Moles} \times \text{Molar Mass of } CaCl_2 = 0.095 \ mol \times 111 \ g/mol = 10.545 \ g$.