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 \(\chi_{\text {methane }}=\) \(0.915\) 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?

Short answer

The partial pressures of methane and ethane in the 15.00 L container of natural gas are approximately 1.318 atm and 0.122 atm, respectively. After complete combustion of both gases, the total mass of water formed is approximately 33.498 grams.

Step-by-step solution

Calculate the total moles of gases using the Ideal Gas Law equation

First, we need to calculate the total moles of the gases in the mixture. We can do this using the Ideal Gas Law equation, which is: PV = nRT Where: P = Pressure (1.44 atm), V = Volume (15.00 L), n = Moles of gas, R = Gas constant (0.0821 L.atm/mol.K), T = Temperature (20°C = 293 K) So, we can rewrite the Ideal Gas Law equation to find 'n': n = PV / RT Substitute the values: n = (1.44 atm)(15.00 L) / (0.0821 L.atm/mol.K)(293 K) n ≈ 0.8917 moles

Calculate the moles of each gas and their partial pressures

To find the moles of each gas, we need to use the mole fractions and the total moles calculated in step 1. For Methane: n_methane = mole_fraction_of_methane * total_moles n_methane = 0.915 * 0.8917 n_methane ≈ 0.8157 moles For Ethane: n_ethane = mole_fraction_of_ethane * total_moles n_ethane = 0.085 * 0.8917 n_ethane ≈ 0.0760 moles Now, we'll use Dalton's law of partial pressures to find the partial pressures of each gas: P_partial = mole_fraction * total_pressure (P) For Methane: P_methane = 0.915 * 1.44 atm P_methane ≈ 1.318 atm For Ethane: P_ethane = 0.085 * 1.44 atm P_ethane ≈ 0.122 atm

Balance the combustion equations and find the number of moles of water produced

Now, we need to balance the combustion equations for both methane and ethane gases to determine the number of moles of water produced. Methane combustion: CH4 + 2O2 → CO2 + 2H2O Ethane combustion: C2H6 + 3.5O2 → 2CO2 + 3H2O For Methane: moles of water produced = moles of methane * moles of water produced per mole of methane moles_water_methane = 0.8157 * 2 moles_water_methane ≈ 1.6314 moles For Ethane: moles of water produced = moles of ethane * moles of water produced per mole of ethane moles_water_ethane = 0.0760 * 3 moles_water_ethane ≈ 0.2280 moles

Calculate the total mass of water produced

Add the moles of water produced from both methane and ethane combustion and multiply by the molar mass of water to find the total mass. Total_moles_water = moles_water_methane + moles_water_ethane Total_moles_water = 1.6314 + 0.2280 Total_moles_water ≈ 1.8594 moles Now, calculate the mass of water produced: Mass_water = total_moles_water * molar_mass_water Mass_water = 1.8594 moles * 18.015 g/mole Mass_water ≈ 33.498 g So, the total mass of water formed after complete combustion of both methane and ethane gases is approximately 33.498 grams.

Key concepts

Partial Pressures

In a gas mixture, each individual gas contributes to the total pressure. This contribution is known as its partial pressure. The concept of partial pressures is governed by Dalton's Law, which states that the total pressure of a gas mixture is equal to the sum of the partial pressures of each individual gas. This becomes particularly relevant when dealing with a mixture of more than one type of gas, such as natural gas, which contains both methane and ethane.

To calculate the partial pressure of each gas, you would multiply the mole fraction of the gas by the total pressure of the gas mixture. The mole fraction is essentially the ratio of the number of moles of that gas to the total number of moles of all gases present. Using this method, we derived that the partial pressure for methane in the mixture is approximately 1.318 atm, and for ethane, it is around 0.122 atm. This means that out of the total pressure of 1.44 atm, methane and ethane contribute 1.318 atm and 0.122 atm respectively.

Combustion Reactions

A combustion reaction is a chemical process where a substance combines with oxygen to release energy, usually in the form of heat and light. This type of reaction commonly involves hydrocarbons, such as methane (CH₄) and ethane (C₂H₆), as these compounds are prevalent in natural fuels.

The complete combustion of methane can be represented as: \[\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}\]
This equation shows that one mole of methane reacts with two moles of oxygen to produce one mole of carbon dioxide and two moles of water.

Similarly, ethane undergoes complete combustion in the following way:\[\text{C}_2\text{H}_6 + 3.5\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O}\]
Here, one mole of ethane reacts with 3.5 moles of oxygen, resulting in the production of two moles of carbon dioxide and three moles of water.

By analyzing these equations, we can determine how much water is produced during the combustion of a given amount of fuel, essential for problems involving energy conversion and material balance.

Molar Mass Calculations

Molar mass is the mass of one mole of a given substance. It is expressed in grams per mole (g/mol). Calculating the molar mass of compounds like water is essential as it helps in converting between moles and grams, a common requirement in stoichiometric calculations like those found in combustion reactions.

For water, the molar mass can be calculated by considering the atomic masses of its constituent elements—hydrogen and oxygen. Water (H₂O) consists of two hydrogen atoms, each with an atomic mass of approximately 1.008 g/mol, and one oxygen atom with an atomic mass of about 16.00 g/mol. Thus, the molar mass of water is computed as:\[2(1.008) + 16.00 = 18.016\, \text{g/mol}\]
In the exercise provided, this molar mass is used to convert the number of moles of water produced from the combustion reaction into grams. The formula applied is:\[\text{Mass} = \text{Moles} \times \text{Molar Mass}\]
This results in approximately 33.498 grams of water formed from the complete combustion of the mix of methane and ethane, offering a practical application of molar mass calculations in real-world scenarios.