Question

A mixture of methane \(\left(\mathrm{CH}_{4}\right)\) and ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) is stored in a container at \(294 \mathrm{mmHg}\). The gases are burned in air to form \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\). If the pressure of \(\mathrm{CO}_{2}\) is 356 \(\mathrm{mmHg}\), measured at the same temperature and volume as the original mixture, calculate the mole fractions of the gases.

Short answer

The mole fractions are \(X_{CH₄} = 0.185\) and \(X_{C₂H₆} = 0.815\).

Step-by-step solution

Write the Balanced Chemical Equations

The burning reactions of methane (CH₄) and ethane (C₂H₆) are as follows: 1. CH₄ + 2O₂ → CO₂ + 2H₂O 2. 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O These equations will be used to determine the moles of CO₂ produced from each gas.

Use Partial Pressure and Mole Concept

Let the moles of CH₄ be \(n \) and moles of C₂H₆ be \( m \). The total initial pressure is 294 mmHg. The reactions indicate that burning 1 mole of CH₄ produces 1 mole of CO₂, and burning 1 mole of C₂H₆ produces 2 moles of CO₂.

Set Up the Pressure Equations

Using the ideal gas law and Dalton's law of partial pressures, we set the equations:\[ P_{CH₄} + P_{C₂H₆} = 294 \, \text{mmHg} \] \[ P_{CO₂} = n \cdot X_{CH₄} + 2m \cdot X_{C₂H₆} = 356 \, \text{mmHg} \]where \( X_{CH₄} \) and \( X_{C₂H₆} \) are mole fractions of CH₄ and C₂H₆ respectively.

Solve for Mole Fractions

Using the above equations, and knowing the CO₂ pressure equation:\[ X_{CH₄} + X_{C₂H₆} = 1 \]\[ X_{CH₄} \times 294 + 2X_{C₂H₆} \times 294 = 356 \]From here, substitute \( X_{C₂H₆} = 1 - X_{CH₄} \), and solve the equations:1. \( 294X_{CH₄} + 588(1 - X_{CH₄}) = 356 \ -294X_{CH₄} + 588 - 588X_{CH₄} = 356 \ 882 = 742X_{CH₄} + 356 \ 742X_{CH₄} = 882 - 356 \ X_{CH₄} = 0.185 \ X_{C₂H₆} = 0.815 \)

Confirm the Solution

Verify that the sum of the mole fractions is 1, confirming the correct proportions:\( X_{CH₄} + X_{C₂H₆} = 0.185 + 0.815 = 1 \). Thus, the solution is consistent.

Key concepts

Mole Fractions

In a chemical mixture, the mole fraction represents the ratio of the number of moles of a specific component to the total number of moles of all components in the mixture. To compute mole fractions, follow these steps:

• First, determine the total moles present in the mixture.
• Next, calculate the moles of each individual gas.
• Finally, divide the moles of each gas by the total moles to find its respective mole fraction.

Mole fractions are often represented by the symbol \(X\). For example, the mole fraction of methane (\(X_{CH_4}\)) is calculated as \(\frac{n_{CH_4}}{n_{total}}\). These fractions are crucial in gas law calculations, influencing properties like pressure and volume.

Chemical Reactions

Chemical reactions help us understand how substances interact and transform into new products. In this case, both methane (\(CH_4\)) and ethane (\(C_2H_6\)) are burned in air:

• The burning or combustion process converts chemical energy into heat energy.
• For methane: \(CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O\)
• For ethane: \(2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O\)

These balanced reactions indicate the stoichiometry, which is the ratio of reactants to products. This stoichiometry is key in predicting the amount of each product formed from a given amount of reactant.

Partial Pressure

Partial pressure is the pressure exerted by a single component of a gas mixture. Dalton's Law states that the total pressure of a gas mixture is equal to the sum of the partial pressures of each individual gas:\[ P_{total} = P_{CH_4} + P_{C_2H_6} \]In this problem, the partial pressures are calculated using mole fractions and the initial total pressure.

• If methane's mole fraction is \(X_{CH_4}\), its partial pressure is \(P_{CH_4} = X_{CH_4} \times P_{total}\).
• Similarly, ethane's partial pressure is \(P_{C_2H_6} = X_{C_2H_6} \times P_{total}\).

Recognizing each gas's contribution to the total pressure helps in understanding and managing gaseous reactions.

Ideal Gas Law

The Ideal Gas Law is an equation that relates the pressure, volume, temperature, and number of moles of a gas:\[ PV = nRT \]where:

• \(P\) is pressure.
• \(V\) is volume.
• \(n\) is the number of moles.
• \(R\) is the gas constant.
• \(T\) is temperature in Kelvin.

In controlled conditions, such as the same temperature and volume, the relationship between moles and pressure becomes evident. The ideal gas law demonstrates how a change in one state variable affects another, aiding in predicting outcomes like gas behavior during chemical reactions or when determining gas compositions from known pressures.