Question

Two 750 -mL (three significant figures) tanks contain identical amounts of a gaseous mixture containing \(\mathrm{O}_{2}, \mathrm{~N}_{2}\), and \(\mathrm{CO}_{2} .\) Each tank has a pressure of 1.38 atm and is kept at \(25^{\circ} \mathrm{C}\). The gas in the first tank is passed through a \(\mathrm{CO}_{2}\) absorber. After all the \(\mathrm{CO}_{2}\) is absorbed, the mass of the absorber increases by \(0.114 \mathrm{~g}\). All the nitrogen in the second tank is removed. The pressure in the tank without the nitrogen at \(25^{\circ} \mathrm{C}\) is 1.11 atm. What is the composition (in mass percent) of the gaseous mixture in the tanks?

Short answer

The mass percent composition of the gaseous mixture is O₂: 76.18%, N₂: 15.28%, and CO₂: 8.55%.

Step-by-step solution

Determine the moles of total gas in a tank

To find the moles of gas in a tank, we can use the ideal gas law equation \(PV = nRT\), where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the ideal gas constant (\(0.0821 \frac{\mathrm{L} \cdot \mathrm{atm}}{\mathrm{mol} \cdot \mathrm{K}}\)), and \(T\) is the temperature in Kelvin. We are given pressure \(P = 1.38 \, \mathrm{atm}\), volume \(V = 750 \, \mathrm{mL} = 0.750 \, \mathrm{L}\), and temperature \(T = 25^{\circ} \mathrm{C} = 298 \, \mathrm{K}\). Rearrange the equation to solve for \(n\): \(n = \frac{PV}{RT} = \frac{1.38 \, \mathrm{atm} \cdot 0.750 \, \mathrm{L}}{0.0821 \frac{\mathrm{L} \cdot \mathrm{atm}}{\mathrm{mol} \cdot \mathrm{K}} \cdot 298 \, \mathrm{K}} = 0.04159 \, \mathrm{mol}\)

Determine the moles of \(\mathrm{CO}_2\)

Given the increase in the mass of the absorber after absorbing all the \(\mathrm{CO}_2\), which is 0.114 g, we can determine the moles of \(\mathrm{CO}_2\) using its molar mass (\(44.01 \, \frac{\mathrm{g}}{\mathrm{mol}}\)): Moles of \(\mathrm{CO}_2 = \frac{0.114 \, \mathrm{g}}{44.01 \, \frac{\mathrm{g}}{\mathrm{mol}}} = 0.002590 \, \mathrm{mol}\)

Determine the moles of \(\mathrm{O}_2\) and \(\mathrm{N}_2\)

In the second tank, after removing all the nitrogen, the pressure is 1.11 atm. First, calculate the moles of gas remaining in the second tank using the same equation as in Step 1: \(n_{\mathrm{O}_2\,+\, \mathrm{CO}_2} = \frac{PV}{RT} = \frac{1.11 \, \mathrm{atm} \cdot 0.750 \, \mathrm{L}}{0.0821 \frac{\mathrm{L} \cdot \mathrm{atm}}{\mathrm{mol} \cdot \mathrm{K}} \cdot 298 \, \mathrm{K}} = 0.03432 \, \mathrm{mol}\) Now, subtract the moles of \(\mathrm{CO}_2\) to get the moles of \(\mathrm{O}_2\): Moles of \(\mathrm{O}_2 = n_{\mathrm{O}_2\,+\, \mathrm{CO}_2} - \text{Moles of }\mathrm{CO}_2 = 0.03432 \, \mathrm{mol} - 0.002590 \, \mathrm{mol} = 0.03173 \, \mathrm{mol}\) Finally, subtract the moles of \(\mathrm{O}_2\) and \(\mathrm{CO}_2\) from the moles of total gas to get the moles of \(\mathrm{N}_2\): Moles of \(\mathrm{N}_2 = n - (\text{Moles of }\mathrm{O}_2 + \text{Moles of }\mathrm{CO}_2) = 0.04159 \, \mathrm{mol} - (0.03173 \, \mathrm{mol} + 0.002590 \, \mathrm{mol}) = 0.007270 \, \mathrm{mol}\)

Calculate the mass of \(\mathrm{O}_2\), \(\mathrm{N}_2\), and \(\mathrm{CO}_2\)

Now, convert the moles of each gas to mass using their molar masses: \(\mathrm{O}_2\) = \(32.00 \frac{\mathrm{g}}{\mathrm{mol}}\), \(\mathrm{N}_2\) = \(28.02 \frac{\mathrm{g}}{\mathrm{mol}}\), and \(\mathrm{CO}_2\) = \(44.01 \frac{\mathrm{g}}{\mathrm{mol}}\): Mass of \(\mathrm{O}_2 = 0.03173 \, \mathrm{mol} \cdot 32.00 \, \frac{\mathrm{g}}{\mathrm{mol}} = 1.015 \, \mathrm{g}\) Mass of \(\mathrm{N}_2 = 0.007270 \, \mathrm{mol} \cdot 28.02 \, \frac{\mathrm{g}}{\mathrm{mol}} = 0.2036 \, \mathrm{g}\) Mass of \(\mathrm{CO}_2 = 0.114 \, \mathrm{g}\) (given)

Calculate the mass percent composition of each gas

To find the mass percent of each gas, divide each mass by the total mass of the gas mixture and multiply by 100: Total mass = Mass of \(\mathrm{O}_2 +\) Mass of \(\mathrm{N}_2 +\) Mass of \(\mathrm{CO}_2 = 1.015 \, \mathrm{g} + 0.2036 \, \mathrm{g} + 0.114 \, \mathrm{g} = 1.3326 \, \mathrm{g}\) Mass percent of \(\mathrm{O}_2 = \frac{\text{Mass of }\mathrm{O}_2}{\text{Total mass}} \times 100 = \frac{1.015 \, \mathrm{g}}{1.3326 \, \mathrm{g}} \times 100 = 76.18\%\) Mass percent of \(\mathrm{N}_2 = \frac{\text{Mass of }\mathrm{N}_2}{\text{Total mass}} \times 100 = \frac{0.2036 \, \mathrm{g}}{1.3326 \, \mathrm{g}} \times 100 = 15.28\%\) Mass percent of \(\mathrm{CO}_2 = \frac{\text{Mass of }\mathrm{CO}_2}{\text{Total mass}} \times 100 = \frac{0.114 \, \mathrm{g}}{1.3326 \, \mathrm{g}} \times 100 = 8.55\%\) The mass percent composition of the gaseous mixture in the tanks is \(\mathrm{O}_2: 76.18\%\), \(\mathrm{N}_2: 15.28\%\), and \(\mathrm{CO}_2: 8.55\%\).