Question.
Calcium carbonate reacts with aqueous $\mathrm{HCl}$ to give $\mathrm{CaCl}_{2}$ and $\mathrm{CO}_{2}$ according to the
reaction, $\mathrm{CaCO}_{3}(s)+2 \mathrm{HCl}_{(a q)} \rightarrow \mathrm{CaCl}_{2(a q)}+\mathrm{CO}_{2(g)}+\mathrm{H}_{2} \mathrm{O}(l)$
What mass of $\mathrm{CaCO}_{3}$ is required to react completely with $25 \mathrm{~mL}$ of $0.75 \mathrm{M} \mathrm{HCl}$ ?
Calcium carbonate reacts with aqueous $\mathrm{HCl}$ to give $\mathrm{CaCl}_{2}$ and $\mathrm{CO}_{2}$ according to the
reaction, $\mathrm{CaCO}_{3}(s)+2 \mathrm{HCl}_{(a q)} \rightarrow \mathrm{CaCl}_{2(a q)}+\mathrm{CO}_{2(g)}+\mathrm{H}_{2} \mathrm{O}(l)$
What mass of $\mathrm{CaCO}_{3}$ is required to react completely with $25 \mathrm{~mL}$ of $0.75 \mathrm{M} \mathrm{HCl}$ ?
Solution:
0.75 M of HCl ≡ 0.75 mol of HCl are present in 1 L of water
$\equiv\left[(0.75 \mathrm{~mol}) \times\left(36.5 \mathrm{~g} \mathrm{~mol}^{-1}\right)\right] \mathrm{HCl}$ is present in $1 \mathrm{~L}$ of water
$\equiv 27.375 \mathrm{~g}$ of $\mathrm{HCl}$ is present in $1 \mathrm{~L}$ of water
Thus, $1000 \mathrm{~mL}$ of solution contains $27.375 \mathrm{~g}$ of $\mathrm{HCl}$.
$\therefore$ Amount of $\mathrm{HCl}$ present in $25 \mathrm{~mL}$ of solution
$=\frac{27.375 \mathrm{~g}}{1000 \mathrm{~mL}} \times 25 \mathrm{~mL}$
$=0.6844 \mathrm{~g}$
From the given chemical equation,
$\mathrm{CaCO}_{3(s)}+2 \mathrm{HCl}_{(a q)} \longrightarrow \mathrm{CaCl}_{2(\alpha))}+\mathrm{CO}_{2(g)}+\mathrm{H}_{2} \mathrm{O}_{(i)}$
$2 \mathrm{~mol}$ of $\mathrm{HCl}(2 \times 36.5=71 \mathrm{~g})$ react with $1 \mathrm{~mol}$ of $\mathrm{CaCO}_{3}(100 \mathrm{~g})$.
$\therefore$ Amount of $\mathrm{CaCO}_{3}$ that will react with $0.6844 \mathrm{~g}=\frac{100}{71} \times 0.6844 \mathrm{~g}$
$=0.9639 \mathrm{~g}$
0.75 M of HCl ≡ 0.75 mol of HCl are present in 1 L of water
$\equiv\left[(0.75 \mathrm{~mol}) \times\left(36.5 \mathrm{~g} \mathrm{~mol}^{-1}\right)\right] \mathrm{HCl}$ is present in $1 \mathrm{~L}$ of water
$\equiv 27.375 \mathrm{~g}$ of $\mathrm{HCl}$ is present in $1 \mathrm{~L}$ of water
Thus, $1000 \mathrm{~mL}$ of solution contains $27.375 \mathrm{~g}$ of $\mathrm{HCl}$.
$\therefore$ Amount of $\mathrm{HCl}$ present in $25 \mathrm{~mL}$ of solution
$=\frac{27.375 \mathrm{~g}}{1000 \mathrm{~mL}} \times 25 \mathrm{~mL}$
$=0.6844 \mathrm{~g}$
From the given chemical equation,
$\mathrm{CaCO}_{3(s)}+2 \mathrm{HCl}_{(a q)} \longrightarrow \mathrm{CaCl}_{2(\alpha))}+\mathrm{CO}_{2(g)}+\mathrm{H}_{2} \mathrm{O}_{(i)}$
$2 \mathrm{~mol}$ of $\mathrm{HCl}(2 \times 36.5=71 \mathrm{~g})$ react with $1 \mathrm{~mol}$ of $\mathrm{CaCO}_{3}(100 \mathrm{~g})$.
$\therefore$ Amount of $\mathrm{CaCO}_{3}$ that will react with $0.6844 \mathrm{~g}=\frac{100}{71} \times 0.6844 \mathrm{~g}$
$=0.9639 \mathrm{~g}$