Chapter 5, Problem 89QAP

Interpretation Introduction

Interpretation:

The mass percent of composition of the gaseous mixture needs to be calculated.

Concept introduction:

Mass percent for any component is known as the ratio of mass of the component to the mass of all the component.

$\text{Mass percent of A = }\frac{{\text{m}}_{\text{A}}}{{\text{m}}_{\text{tot}}}\times 100\%$

Here m_A is the mass of component A, m_tot is the mass of total components.

According to the ideal gas law volume i.e. V, pressure i.e. P, number of moles i.e. m, temperature i.e. t and universal gas constant i.e. R are interrelated as below:

$\text{PV = nRT}$

Moles are known as the ratio of mass and molar mass. Below is the formula:

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

Here, MM is molar mass and m is the mass.

Answer to Problem 89QAP

The composition of O₂ is $74.6\%$, N₂ is $16.8\%$ and CO₂ is $8.6\%$.

Explanation of Solution

The pressure, volume and temperature of both tanks A and B is 1.38 atm, 750 ml or 0.750 L, and 25 ⁰C respectively.

Converting the temperature from Celsius to Kelvin

$\begin{array}{l}{\text{T}}_{\text{K}}\text{=}\left({\text{t}}_{{}^{\text{0}}\text{C}}\text{+273}\right)\text{K}\\ \text{ = 25+273 = 298 K}\end{array}$

Therefore, temperature in kelvin is 298 K

Both the tanks have CO₂, O₂ and N₂ . Substituting the values in ideal gas equation and total number of moles of gases:

$\begin{array}{l}{\text{P}}_{\text{tot}}{\text{V}}_{\text{tot}}{\text{ = n}}_{\text{tot}}\text{RT}\\ \left(1.38\text{ atm}\right)\left(0.750\text{ L}\right)={\text{n}}_{\text{tot}}\left(0.08205\frac{\text{L}\times \text{atm }}{\text{K}\times \text{mol}}\right)\left(29\text{8 K}\right)\\ {\text{ n}}_{\text{tot}}=\text{ 0}\text{.042 mol}\end{array}$

The total no. of moles of gases in each tank are 0.042 mol.

The mass of absorber containing CO₂ from tank A is 0.114 g. The molar mass of CO₂ is 44 g/mol. The mol of CO₂ in each tank is:

$\begin{array}{l}{\text{n}}_{{\text{CO}}_{\text{2}}}\text{ = }\frac{{\text{m}}_{{\text{CO}}_{\text{2}}}}{{\left(\text{MM}\right)}_{{\text{CO}}_{\text{2}}}}\\ =\frac{0.144\text{ g }}{44\text{ g/mol}}\\ =0.0026\text{ mol}\end{array}$

After removal of N₂ from tank B, the pressure of tank B is 1.11 atm. Now, the mol of gases remaining are:

$\begin{array}{l}{\text{P}}_{\left({\text{CO}}_2+{\text{O}}_2\right)}{\text{V = n}}_{\left({\text{CO}}_2+{\text{O}}_2\right)}\text{RT}\\ \left(1.11\text{ atm}\right)\left(0.750\text{ L}\right)={\text{n}}_{\left({\text{CO}}_2+{\text{O}}_2\right)}\left(0.08205\frac{\text{L}\times \text{atm }}{\text{K}\times \text{mol}}\right)\left(29\text{8 K}\right)\\ {\text{ n}}_{{}_{\left({\text{CO}}_2+{\text{O}}_2\right)}}=\text{ 0}\text{.034 mol}\end{array}$

Therefore, the total mole of CO₂ and O₂ in each tank is 0.034 mol

The moles of O₂ and N₂ are:

$\begin{array}{l}{n}_{O_2}=n_{\left(C{O}_2+O_2\right)}-n_{C{O}_2}\\ =0.034\text{ mol }-\text{ 0}\text{.0026 mol}\\ =\text{ 0}\text{.031 mol}\end{array}$

And

$\begin{array}{l}{n}_{N_2}=n_{tot}-n_{\left(C{O}_2+O_2\right)}\\ =0.042\text{ mol }-\text{ 0}\text{.034 mol}\\ =\text{ 0}\text{.008 mol}\end{array}$

Therefore, the mole of O₂ and N₂ is 0.031 mol and 0.008 mol respectively.

The mass of CO₂ is 0.114. The molar mass of O₂ is 32 g/mol. The mol of N₂ is 28 g/mol. Now the mass of O₂ :

$\begin{array}{l}{\text{ n}}_{{\text{O}}_{\text{2}}}\text{ = }\frac{{\text{m}}_{{\text{O}}_{\text{2}}}}{{\left(\text{MM}\right)}_{{\text{O}}_{\text{2}}}}\\ \text{0}\text{.031 mol }=\frac{{\text{m}}_{{\text{O}}_{\text{2}}}}{32\text{ g/mol}}\\ {\text{ m}}_{{\text{O}}_{\text{2}}}=0.99\text{ g}\end{array}$

The mass of N₂ would be:

$\begin{array}{l}{\text{ n}}_{{\text{N}}_{\text{2}}}\text{ = }\frac{{\text{m}}_{{\text{N}}_{\text{2}}}}{{\left(\text{MM}\right)}_{{\text{N}}_{\text{2}}}}\\ \text{0}\text{.008 mol }=\frac{{\text{m}}_{{\text{N}}_{\text{2}}}}{28\text{ g/mol}}\\ {\text{ m}}_{{\text{N}}_{\text{2}}}=0.224\text{ g}\end{array}$

Therefore, the mass of O₂ is 0.99 g and mass of N₂ is 0.224 g

The total mass of gases in each tank:

$\begin{array}{l}{m}_{tot}=m_{C{O}_2}+m_{O_2}+m_{N_2}\\ \text{ = 0}\text{.114 g + 0}\text{.992 g + 0}\text{.224 g}\\ \text{ = }1.330\text{ g}\end{array}$

The total mass of gases is 1.330 g.

Now calculating the mass percent O₂, N₂ and CO₂.

$\begin{array}{l}{\text{Mass percent of O}}_2=\frac{{\text{m}}_{{\text{O}}_{\text{2}}}}{{\text{m}}_{\text{tot}}}\times 100\%\\ =\frac{0.992\text{ g}}{1.330\text{ g}}\times 100\%\\ =74.6\%\\ {\text{Mass percent of N}}_2=\frac{{\text{m}}_{{\text{N}}_{\text{2}}}}{{\text{m}}_{\text{tot}}}\times 100\%\\ =\frac{0.224\text{ g}}{1.330\text{ g}}\times 100\%\\ =16.8\%\\ {\text{Mass percent of CO}}_2=\frac{{\text{m}}_{{\text{CO}}_{\text{2}}}}{{\text{m}}_{\text{tot}}}\times 100\%\\ =\frac{0.114\text{ g}}{1.330\text{ g}}\times 100\%\\ =8.6\%\end{array}$

Thus, the composition of O₂ is $74.6\%$, N₂ is $16.8\%$ and CO₂ is $8.6\%$.

Conclusion

The composition of O₂ is $74.6\%$, N₂ is $16.8\%$ and CO₂ is $8.6\%$.