Chapter 13, Problem 112AP

Interpretation Introduction

Interpretation:

The number of moles of ${\text{CO}}_2\left(\text{g}\right)$ should be calculated. The volume of ${\text{CO}}_{\text{2}}$ occupied should be calculated under both conditions.

Concept Introduction:

According to ideal gas equation:

$\text{P}\text{V}=\text{n}\text{R}\text{T}$

Here, P is pressure, V is volume, n is number of moles, R is Universal gas constant and T is temperature.

The value of Universal gas constant can be taken as $0.082{\text{ L atm K}}^{-1}{\text{ mol}}^{-1}$.

Mass of gas can be calculated from number of moles and molar mass as follows:

$\text{m}=\text{n}\times \text{M}$

Here, n is number of moles, m is mass and M is molar mass.

Answer to Problem 112AP

Volume of wet ${\text{CO}}_2\left(\text{g}\right)$ is 2.68 L and dry ${\text{CO}}_2\left(\text{g}\right)$ is 3.32 L.

Explanation of Solution

Calculation:

The balanced chemical reaction is as follows:

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

Calculate number of moles of ${\text{CaCO}}_{\text{3}}$ as follows:

$\text{n}=\frac{\text{m}}{\text{M}}$

Molar mass of ${\text{CaCO}}_{\text{3}}$ is 100.1 g/mol thus,

$\begin{array}{c}\text{n}=\frac{10\text{ g}}{100.1\text{ g/mol}}\\ =0.0999\text{ mol}\end{array}$

From the balanced chemical reaction, 1 mol of ${\text{CaCO}}_{\text{3}}$ produces 1 mol of ${\text{CO}}_2\left(\text{g}\right)$ thus, 0.0999 mol of ${\text{CaCO}}_{\text{3}}$ will produces 0.0999 mol of ${\text{CO}}_2\left(\text{g}\right)$

For the given temperature and pressure, volume of wet ${\text{CO}}_2\left(\text{g}\right)$ can be calculated using the ideal gas equation as follows:

$\text{V}=\frac{\text{n}\text{R}\text{T}}{\text{P}}$

Convert the given temperature from ${}^{\circ }\text{C}$ to K:

$0{}^{\circ }\text{C}=273.15\text{ K}$

Thus,

$\begin{array}{c}{\text{60 }}^{\circ }\text{C}=\left(60+273.15\right)\text{ K}\\ =333.15\text{ K}\end{array}$

Also, convert pressure from torr to atm as follows:

$1\text{ atm}=\text{760 torr}$

Thus,

$\begin{array}{c}774\text{ torr}=774\text{ torr}\left(\frac{1\text{ atm}}{760\text{ torr}}\right)\\ =1.018\text{ atm}\end{array}$

Putting the values,

$\begin{array}{c}\text{V}=\frac{\left(0.0999\text{ mol}\right)\left(0.082{\text{ L atm K}}^{-1}{\text{ mol}}^{-1}\right)\left(333.15\text{ K}\right)}{\left(1.018\text{ atm}\right)}\\ =2.68\text{ L}\end{array}$

The volume of wet ${\text{CO}}_2\left(\text{g}\right)$ is 2.68 L.

The vapour pressure of water at $60{}^{\circ }\text{C}$ is 149.4 mm Hg. Convert the given pressure from mm Hg to atm as follows:

$1\text{ atm}=\text{760 mm Hg}$

Thus,

$\begin{array}{c}149.4\text{ mm Hg}=149.4\text{ mm Hg}\left(\frac{1\text{ atm}}{760\text{ mm Hg}}\right)\\ =0.1966\text{ atm}\end{array}$

This pressure should be subtracted from the total pressure to calculate the volume of ${\text{CO}}_2\left(\text{g}\right)$ after the addition of drying agent.

Thus, pressure of carbon dioxide becomes $\left(1.018-0.1966\right)\text{ atm}=0.8214\text{ atm}$

Volume can be calculated using the ideal gas equation as follows:

$\text{V}=\frac{\text{n}\text{R}\text{T}}{\text{P}}$

Putting the values,

$\begin{array}{c}\text{V}=\frac{\left(0.0999\text{ mol}\right)\left(0.082{\text{ L atm K}}^{-1}{\text{ mol}}^{-1}\right)\left(333.15\text{ K}\right)}{\left(0.8214\text{ atm}\right)}\\ =3.32\text{ L}\end{array}$.

Conclusion

Therefore, volume of wet ${\text{CO}}_2\left(\text{g}\right)$ and dry ${\text{CO}}_2\left(\text{g}\right)$ is 2.68L and 3.32 L respectively.