Question

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

(b) What is the mass of $4.5$ moles of calcium carbonate? [Relative molecular mass of calcium carbonate is $100$]

Answer 1

Given information

• The given reaction is between Calcium carbonate$\left({\mathrm{CaCO}}_3\right)$ and dilute hydrochloric acid $\left(\mathrm{HCl}\right)$.
• The given moles of calcium carbonate$=4.5\mathrm{moles}$
• Relative molecular mass of calcium carbonate$=100$

Balanced chemical equation

• When Calcium carbonate$\left({\mathrm{CaCO}}_3\right)$ reacts with dilute Hydrochloric acid$\left(\mathrm{HCl}\right)$, Calcium chloride$\left({\mathrm{CaCl}}_2\right)$, Water$\left({\mathrm{H}}_2\mathrm{O}\right)$ and Carbon dioxide$\left({\mathrm{CO}}_2\right)$ are formed.
• 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 mass of $4.5$ moles of calcium carbonate

• The mass of calcium carbonate is calculated as follows:

$\begin{array}{rcl}\mathrm{Number}\mathrm{of}\mathrm{moles}\mathrm{of}{\mathrm{CaCO}}_3& =& \frac{\mathrm{Mass}\mathrm{of}{\mathrm{CaCO}}_3\mathrm{in}\mathrm{grams}}{\mathrm{Relative}\mathrm{molecular}\mathrm{mass}\mathrm{of}{\mathrm{CaCO}}_3}\\ \mathrm{Mass}\mathrm{of}{\mathrm{CaCO}}_3\mathrm{in}\mathrm{grams}& =& \mathrm{Number}\mathrm{of}\mathrm{moles}\mathrm{of}{\mathrm{CaCO}}_3\times \mathrm{Relative}\mathrm{molecular}\mathrm{mass}\mathrm{of}{\mathrm{CaCO}}_3\\ & =& \left(4.5\times 100\right)\mathrm{g}\\ & =& 450\mathrm{g}\end{array}$

Therefore, the mass of $4.5$ moles of calcium carbonate is $450\mathrm{g}$.