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Chapter 5: Problem 164

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

Expert verified
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

01

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
02

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
03

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
04

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.

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Most popular questions from this chapter

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