Problem 168 - Chemistry An Atoms First Approach

Question

Natural gas is a mixture of hydrocarbons, primarily methane \(\left(\mathrm{CH}_{4}\right)\) and ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right) .\) A typical mixture might have \(\chi_{\text {mathane }}=\) 0.915 and \(\chi_{\text {ethane }}=0.085 .\) What are the partial pressures of the two gases in a \(15.00\) -\(\mathrm{L}\) container of natural gas at \(20 .^{\circ} \mathrm{C}\) and \(1.44\) atm? Assuming complete combustion of both gases in the natural gas sample, what is the total mass of water formed?

Step-by-Step Solution

Answer

The partial pressures of methane and ethane are obtained using their mole fractions (0.915 and 0.085) and total pressure (1.44 atm). They are respectively, \( P_{\text{methane}} = 0.915 \times 1.44 \, \text{atm} \) and \( P_{\text{ethane}} = 0.085 \times 1.44 \, \text{atm} \). The balanced equations for methane (\( CH_{4}(g) + 2O_{2}(g) \to CO_{2}(g) + 2H_{2}O(g) \)) and ethane (\( 2C_{2}H_{6}(g) + 7O_{2}(g) \to 4CO_{2}(g) + 6H_{2}O(g) \)) combustion are used to find the moles of gas in the container using the Ideal Gas Law \(PV = nRT\). The moles of methane and ethane are used to calculate the moles of water formed from their combustion using stoichiometry. From here, \( n_{H2O(\text{methane})} = 2 \times n_{\text{methane}} \) and \( n_{H2O(\text{ethane})} = 3 \times n_{\text{ethane}} \) allow us to calculate the total moles of water formed (\( n_{H2O(\text{total})} = n_{H2O(\text{methane})} + n_{H2O(\text{ethane})} \)). Finally, the mass of water formed is given by \( \text{mass}_{H2O} = n_{H2O(\text{total})} \times 18.015 \, g \cdot mol^{-1} \) using the molar mass of water.

Step 1: Determine the partial pressures of methane and ethane

We are given the mole fractions of methane (χ_methane = 0.915) and ethane (χ_ethane = 0.085), volume (15.00 L), temperature (20°C = 293 K), and total pressure (1.44 atm) in the container. To find the partial pressures, we will use Dalton's law of partial pressures: Partial Pressure = (mole fraction) × (total pressure) First, let's find the partial pressure of methane: \( P_{methane} = χ_{methane} × P_{total} \) Now, find the partial pressure of ethane: \( P_{ethane} = χ_{ethane} × P_{total} \)

Step 2: Write the balanced equation for the combustion of methane and ethane

In order to find the moles of water formed during combustion, we need the balanced chemical equations for the combustion of methane and ethane: Methane: \( CH_4(g) + 2O_2(g) → CO_2(g) + 2H_2O(g) \) Ethane: \( 2C_2H_6(g) + 7O_2(g) → 4CO_2(g) + 6H_2O(g) \)

Step 3: Calculate the moles of methane and ethane using the Ideal Gas Law

We will use the Ideal Gas Law equation to find the moles of methane and ethane in the container: \(PV = nRT\) For methane, we have: \( n_{methane} = \frac{P_{methane} \times V}{RT} \) For ethane, we have: \( n_{ethane} = \frac{P_{ethane} \times V}{RT} \)

Step 4: Calculate the moles of water formed in the combustion

Using stoichiometry, calculate the moles of water formed from the combustion of methane and ethane: For methane, from the balanced equation, 1 mole of methane forms 2 moles of water: \( n_{H2O(methane)} = 2 \times n_{methane} \) Similarly, for ethane, from the balanced equation, 1 mole of ethane forms 3 moles of water: \( n_{H2O(ethane)} = 3 \times n_{ethane} \) The total moles of water formed is the sum of water moles formed due to the combustion of methane and ethane: \( n_{H2O(total)} = n_{H2O(methane)} + n_{H2O(ethane)} \)

Step 5: Calculate the mass of water formed

Convert the moles of water formed to mass using the molar mass of water (18.015 g/mol): Mass of water formed = moles of water × molar mass of water \( mass_{H2O} = n_{H2O(total)} \times 18.015 \ g \cdot mol^{-1} \)

Key Concepts

Partial Pressure Calculation

Understanding partial pressure is key in dealing with gas mixtures like natural gas. Dalton's Law of Partial Pressures is our tool here. It tells us that the pressure of each gas component in a mixture is equal to its mole fraction multiplied by the total pressure of the mixture.

In our example, we have methane and ethane in natural gas. With the total pressure of the container given as 1.44 atm:

• The partial pressure of methane is calculated as: \( P_{methane} = \chi_{methane} \times P_{total} = 0.915 \times 1.44 \)


• The partial pressure of ethane is: \( P_{ethane} = \chi_{ethane} \times P_{total} = 0.085 \times 1.44 \)

These calculations tell us how much each gas contributes to the total pressure in the container.

Ideal Gas Law

The Ideal Gas Law is a convenience when dealing with gases, providing a relationship between pressure, volume, temperature, and moles of a gas. The formula is given by:\[ PV = nRT \]Where:

• \(P\) is the gas pressure
• \(V\) is the volume
• \(n\) is the number of moles of gas
• \(R\) is the ideal gas constant (0.0821 L atm K^{-1} mol^{-1})
• \(T\) is the temperature in Kelvin

In the example, you'll first want to find the moles of each component gas using this formula, given their respective partial pressures and the conditions in the container:

• For methane: \( n_{methane} = \frac{P_{methane} \times V}{RT} \)


• For ethane: \( n_{ethane} = \frac{P_{ethane} \times V}{RT} \)

This allows for determining the exact quantity of gas present, which is essential for further calculations like combustion analysis.

Stoichiometry

Understanding stoichiometry is vital when dealing with chemical reactions, such as the combustion of fuels. It gives the relationship between reactants and products in a chemical equation.

In the combustion reactions provided:

• Methane combusts to produce carbon dioxide and water: \( CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O \)


• Ethane combusts similarly: \( 2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O \)

With these balanced equations, we can identify the mole ratios for reactants and products. For example, each mole of methane results in two moles of water. This ratio helps in calculating the exact amount of water formed when using the gas quantities determined earlier.

By applying these principles, you ensure accurate predictions for the outcomes of gas reactions.

Mole Fraction

The mole fraction is an important concept in chemistry, particularly when working with mixtures. It represents the ratio of the moles of a component to the total moles in the mixture. For gas mixtures like natural gas, the mole fraction helps determine each gas component's role in overall volume and pressure.

Given in our problem:

• Mole fraction of methane, \( \chi_{methane} = 0.915 \)


• Mole fraction of ethane, \( \chi_{ethane} = 0.085 \)

These fractions show how much of the total gas mixture can be attributed to each gas. They aid in calculating partial pressures (as seen in solution calculations) and can inform us about the composition of the mixture.

Using the mole fraction helps simplify complex gas mixtures into understandable and computable terms.