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

Consider the above reaction, what mass of $\mathrm{CaCl}_2$ will be formed if 250 mL of 0.76 M HCl reacts with 1000 g of $\mathrm{CaCO}_3$ ?

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

<p>The reaction is: </p> <p>$ \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}) $</p></p> <p><strong>Calculate Moles of $\mathrm{CaCO}_3$</strong>:</p> <p><p>The molar mass of $\mathrm{CaCO}_3$ is calculated as </p> <p>$ 40 + 12 + 3 \times 16 = 100 \ \mathrm{g/mol} $</p></p> <p><p>Given 1000 g of $\mathrm{CaCO}_3$, the number of moles is </p> <p>$ \frac{1000}{100} = 10 \ \mathrm{moles} $</p></p> <p><strong>Calculate Moles of $\mathrm{HCl}$</strong>:</p> <p><p>Using the given concentration and volume for $\mathrm{HCl}$, the moles are calculated as follows:</p> <p>$ 0.76 \times \frac{250}{1000} = 0.19 \ \mathrm{moles} $</p></p> <p><p>$\mathrm{HCl}$ is the limiting reagent (L.R.) because it has fewer moles than needed to completely react with the $\mathrm{CaCO}_3$.</p></p> <p><strong>Calculate Moles of $\mathrm{CaCl}_2$ Formed</strong>:</p> <p><p>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:</p> <p>$ \frac{0.19}{2} = 0.095 \ \mathrm{moles} $</p></p> <p><strong>Calculate Mass of $\mathrm{CaCl}_2$</strong>:</p> <p><p>The molar mass of $\mathrm{CaCl}_2$ is:</p> <p>$ 40 + 2 \times 35.5 = 111 \ \mathrm{g/mol} $</p></p> <p><p>Thus, the mass of $\mathrm{CaCl}_2$ is:</p> <p>$ 0.095 \times 111 = 10.545 \ \mathrm{g} $</p></p> <p>Hence, the mass of $\mathrm{CaCl}_2$ formed is 10.545 g.</p>