Question

$4.5$ moles of calcium carbonate are reacted with dilute hydrochloric acid.

(e) How many moles of $\mathrm{HCl}$ are used in this reaction?

Answer 1

Given information

• The given reactants are Calcium carbonate $\left({\mathrm{CaCO}}_3\right)$ and dilute hydrochloric acid $\left(\mathrm{HCl}\right)$.
• Moles of calcium carbonate$=4.5\mathrm{moles}$.

Balanced chemical reaction

• When Calcium carbonate reacts with dilute Hydrochloric acid, formation of Calcium chloride$\left({\mathrm{CaCl}}_2\right)$, Water$\left({\mathrm{H}}_2\mathrm{O}\right)$ and Carbon dioxide$\left({\mathrm{CO}}_2\right)$ occurs.
• The balanced chemical equation is as follows:

$\begin{array}{rcl}{\mathrm{CaCO}}_3\left(\mathrm{s}\right)+2\mathrm{HCl}\left(\mathrm{aq}\right)& \to & {\mathrm{CaCl}}_2\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+{\mathrm{CO}}_2\left(\mathrm{g}\right)\end{array}$

Calculation of moles of Hydrochloric acid

• As per the balanced reaction, it is clear that $1$ mole of calcium carbonate requires $2$ moles of hydrochloric acid.
• Thus, $4.5$ moles of calcium carbonate will require $\frac{2}{1}\times 4.5=9\mathrm{moles}$ of hydrochloric acid.

Therefore, $9$ moles of Hydrochloric acid are used in this reaction.