91Ó°ÊÓ

When discussing the composition of a gas mixture, the term "mole fraction" is commonly used. The mole fraction indicates the proportion of a component in the total mixture. It is a way to express concentrations and is useful in calculations involving gases. To calculate the mole fraction, you divide the number of moles of the substance of interest by the total number of moles in the mixture.

For example, in the problem given, we need to calculate the mole fraction of helium in a gas mixture consisting of \ \(\mathrm{SO}_{2}\), \ \(\mathrm{O}_{2}\), and helium (He). By knowing the molar masses of these gases, we use the molar mass of the entire gas mixture to find the fraction for helium:

• \(M_{SO_{2}} = 64 \, \text{g/mol} \)
• \(M_{O_{2}} = 32 \, \text{g/mol} \)
• \(M_{He} = 4 \, \text{g/mol} \)

Using the calculated total molar mass of the mixture \(M = 43.3 \, \text{g/mol}\), we found that the mole fraction of helium \(x_{He}\) is given by \(x_{He} = \frac{4}{43.3} \approx 0.0924\). This tells us that 9.24% of the gas mixture is helium.

A crucial part of understanding this exercise is recognizing the chemical reactions that occur between the components of the gas mixture. The problem statement involves the reaction between sulfur dioxide (\(\mathrm{SO}_{2}\)) and oxygen (\(\mathrm{O}_{2}\)) to form sulfur trioxide (\(\mathrm{SO}_{3}\)):

• The balanced chemical equation is \(\mathrm{SO}_{2} + \frac{1}{2} \mathrm{O}_{2} \rightarrow \mathrm{SO}_{3}\).
• This reaction is exothermic, meaning it releases energy.
• Equimolar conditions imply that every mole of \(\mathrm{SO}_{2}\) reacts with half a mole of \(\mathrm{O}_{2}\) to produce one mole of \(\mathrm{SO}_{3}\).

The importance of this reaction being complete means that no \(\mathrm{SO}_{2}\) or \(\mathrm{O}_{2}\) remains post-reaction. This reaction changes the gas mixture composition, requiring a new density calculation based on the presence of \(\mathrm{SO}_{3}\) and the non-reactive \(\mathrm{He}\).

Density calculations are essential when dealing with gases, especially under the assumptions of ideal behavior. Density (\(\rho\)) is defined as mass divided by volume. For gases, it can be expressed using the relation:

• \(\rho = \frac{M \cdot P}{R \cdot T}\)

where:

• \(M\) is the molar mass of the gas,
• \(P\) is the pressure,
• \(R\) is the ideal gas constant,
• \(T\) is the temperature.

For our gas mixture after the chemical reaction, we recalculated the molar mass \(M_{\text{new}}\) considering the produced \(\mathrm{SO}_{3}\) (80 g/mol) and helium's mole fraction (0.0924). Using these values, the density of the new gas mixture was determined:

• \(M_{\text{new}} = (1 - 0.0924) \times 80 + 0.0924 \times 4 = 75.04 \, \text{g/mol}\)

Applying the density formula:

• \(\rho = \frac{75.04 \times 1}{0.0821 \times 273} = 3.14 \, \text{g/L}\)

This calculation reveals the increased density of the new gas mixture compared to the initial one.