Question

Carbon dioxide is liberated by the reaction of aqueous hydrochloric acid with calcium carbonate:

${\mathbf{CaCO}}_{\mathbf{3}}\left(s\right)\mathbf{+}\mathbf{2}{\mathbf{H}}^{\mathbf{+}}\left(\mathrm{aq}\right)\mathbf{\to }{\mathbf{Ca}}^{\mathbf{+}\mathbf{2}}\left(\mathrm{aq}\right)\mathbf{+}{\mathbf{CO}}_{\mathbf{2}}\left(g\right)\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(l\right)$

A volume of 722 ml${\mathbf{CO}}_{\mathbf{2}}\left(\mathbf{g}\right)$ is collected over water at ${\mathbf{20}}^{\mathbf{o}}\mathbf{C}$ and a total pressure of 0.9963 atm . At this temperature, water has a vapor pressure of 0.0231 atm. Calculate the mass of calcium carbonate that has reacted, assuming no losses of carbon dioxide.

Short answer

The mass of calcium carbonate is 2.9 g.

Step-by-step solution

Given data

Total pressure of water and carbon dioxide = 0.09963 atm

$\text{V}=722\text{ml}=722\times {10}^{-6}{\text{m}}^3$

$\text{T}=20°\text{C}+273=293\text{K}$

Concept of Ideal gas equation

The ideal gas law shows the relation between various parameters of the gas, pressure, volume, moles and temperature. The mathematical expression of ideal gas: PV =nRT , Where

$$\begin{gathered} P=\mathrm{pressure} \\ V=\mathrm{volume} \\ n=\mathrm{number} \mathrm{of} \mathrm{moles} \\ R=\left(8.314\frac{\mathrm{J}}{\mathrm{K}\times \mathrm{mol}}\right) \\ T=\mathrm{temperature} \end{gathered}$$

Calculation of  number of moles of carbon dioxide

Pressure of ${\text{CO}}_2$= total pressure - water vapor pressure.

$$\begin{gathered} \text{p}\left({\text{CO}}_2\right)=0.9963\text{atm}-0.0231\text{atm}=0.9732\text{atm} \\ \text{p}\left({\text{CO}}_2\right)=0.9963\text{atm}\times \frac{101325\text{Pa}}{1\text{atm}}=98609.49\text{Pa} \end{gathered}$$

Using the ideal gas equation and putting the values as above we calculate number of moles of carbon dioxide

$\begin{array}{rcl}\text{n}& =& \frac{\text{P}·\text{V}}{\text{R}·\text{T}}\\ & =& \frac{98609.49\text{Pa}\times 722\times {10}^{-6}{\text{m}}^3}{8.314\frac{\text{J}}{\text{K}·\text{mol}}\times 293\text{K}}\\ & =& 0.029\times {10}^{-10}{\text{mol CO}}_2\end{array}$

So, of carbon dioxide$=0.029\times {10}^{-10}\mathrm{mol}$

Calculation of molar mass  of calcium carbonate

Balanced chemical equation calcium carbonate with hydrochloric acid

${\text{CaCO}}_{3(\text{s})}+2{\text{H}}_{\left(\text{aq}\right)}^{+}\to {\text{Ca}}_{\left(\text{aq}\right)}^{2+}+{\text{CO}}_{2(\text{g})}+{\text{H}}_2{\text{O}}_{\left(\text{l}\right)}$

Mol ratio is $\frac{1{\text{molCaCO}}_3}{1{\text{molCO}}_2}$

$\begin{array}{rcl}\text{n}\left({\text{CaCO}}_3\right)& =& 0.029 {\text{molCO}}_2\times \frac{1 {\text{molCaCO}}_3}{1 {\text{molCO}}_2}\\ & =& 0.029 \text{molCaCO}3\end{array}$

$\text{M}\left({\text{CaCO}}_3\right)=40+12+16·3=100\text{g/mol}$

$\begin{array}{rcl}m\left({\text{CaCO}}_3\right)& =& 0.029{\text{molCaCO}}_3\times 100\frac{\text{g}}{\text{mol}}\\ & =& 2.9 {\text{g CaCO}}_3\end{array}$

Therefore, mass of calcium carbonate is 2.9 g .