You have an equimolar mixture of the gases \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\), along with some He, in a container fitted with a piston. The density of this mixture at STP is \(1.924\) \(\mathrm{g/L}\). Assume ideal behavior and constant temperature and pressure. a. What is the mole fraction of He in the original mixture? b. The \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) react to completion to form \(\mathrm{SO}_{3}\). What is the density of the gas mixture after the reaction is complete?

We know that the density of the gas mixture at STP (0°C and 1 atm) is \(1.924 \ \mathrm{g/L}\). From the ideal gas law, we have: \[PV = nRT\] where P is the pressure in atm, V is the volume in L, n is the number of moles, R is the ideal gas constant \((0.08206 \ \mathrm{atm \cdot L \cdot mol^{-1} \cdot K^{-1})\), and T is the temperature in K. At STP, we have: \[1 \ \mathrm{atm} \cdot V = n \times 0.08206 \ \mathrm{atm \cdot L \cdot mol^{-1} \cdot K^{-1}} \times 273 \ \mathrm{K}\] Rearranging for n gives: \[n = \frac{1 \ \mathrm{atm} \cdot V}{0.08206 \ \mathrm{atm \cdot L \cdot mol^{-1} \cdot K^{-1}} \times 273 \ \mathrm{K}}\] From the density, we also know that the mass of the gas mixture per liter is \(1.924 \ \mathrm{g}\). Therefore, dividing mass by moles gives us the molar mass of the gas mixture: \[\text{Molar mass} = \frac{1.924 \ \mathrm{g} \cdot V}{n}\] Substituting n from the above equation and simplifying gives: \[\text{Molar mass} = 1.924 \ \mathrm{g} \cdot \frac{0.08206 \ \mathrm{L \cdot mol^{-1} \cdot K^{-1}} \cdot 273 \ \mathrm{K}}{1\ \mathrm{atm}}\] Now, calculate the molar mass of the total mixture:setuptools \[\text{Molar mass} = 1.924 \times 0.08206 \times 273 \approx 43.449 \ \mathrm{g/mol}\]

We know that the mixture contains an equimolar amount of \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\). The molar mass of \(\mathrm{SO}_{2}\) is \(32 + 16 \times 2 = 64 \ \mathrm{g/mol}\) and the molar mass of \(\mathrm{O}_{2}\) is \(16 \times 2 = 32 \ \mathrm{g/mol}\). Let mole fraction of \(\mathrm{SO}_{2}\) = mole fraction of \(\mathrm{O}_{2}\) = x. Mole fraction of He = y = 1 - 2x. Since the molar mass of the gas mixture is the weighted average of the molar masses of the individual gases, the equation that relates the molar mass of the mixture with the mole fractions and masses of the individual gases is: \[\text{Molar mass}_{\text{mixture}} = x \times \text{Molar mass}_{\mathrm{SO}_{2}} + x \times \text{Molar mass}_{\mathrm{O}_{2}} + y \times \text{Molar mass}_{\text{He}}\] We know that the molar mass of He is \(4 \ \mathrm{g/mol}\). Substituting the known values in the above equation gives: \[43.449 = x(64 + 32) + (1 - 2x)4\] Solving for x gives: \[x = (43.449 - 4 + 8x) / (64 + 32)\] \[8x = 43.449 - 4\] \[x = (39.449) / 8\] \[x = 4.931\] Now, we can find the mole fraction of He: \[y = 1 - 2x\] \[y = 1 - 2 \times 4.931\] \[y \approx 0.138\] So, the mole fraction of He in the original mixture is approximately 0.138.

Now, we know that all \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) react completely to form \(\mathrm{SO}_{3}\). The molar mass of \(\mathrm{SO}_{3}\) is \(32 + 16 \times 3 = 80 \ \mathrm{g/mol}\). In the post-reaction gas mixture, let the mole fraction of \(\mathrm{SO}_{3}\) = z and mole fraction of He = y'. Then, y' = 1 - z. The mass ratio of \(\mathrm{SO}_{3}\) formed to the initial mass of \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) remains the same since the entire mass is conserved. Therefore: \[2x = z\] We can now write the equation for the molar mass of the post-reaction mixture: \[\text{Molar mass}_{(\text{post-reaction mixture})} = z \times \text{Molar mass}_{\mathrm{SO}_{3}} + y' \times \text{Molar mass}_{\text{He}}\] We have the molar mass of He and \(\mathrm{SO}_{3}\) as \(4 \ \mathrm{g/mol}\) and \(80 \ \mathrm{g/mol}\), respectively. Substituting the values for z and y' along with the molar masses gives: \[\text{Molar mass}_{(\text{post-reaction mixture})} = (2x) \times 80 + (1 - 2x) \times 4\] Now, we can substitute the value of x found in step 2: \[\text{Molar mass}_{(\text{post-reaction mixture})} = (2 \times 4.931) \times 80 + (1 - 2 \times 4.931) \times 4\] \[\text{Molar mass}_{(\text{post-reaction mixture})} \approx 56.449 \ \mathrm{g/mol}\] Finally, we can find the density of the post-reaction gas mixture by multiplying the molar mass by the ratio of standard molar volume: \[\text{Density}_{(\text{post-reaction mixture})} = \frac{56.449 \ \mathrm{g/mol}}{0.08206 \ \mathrm{L \cdot mol^{-1} \cdot K^{-1}} \times 273 \ \mathrm{K}}\] Then calculate the density: \[\text{Density}_{(\text{post-reaction mixture})} \approx 2.385 \ \mathrm{g/L}\] So, the density of the gas mixture after the reaction is complete is approximately \(2.385 \ \mathrm{g/L}\).