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 \(X_{\text {ethane }}=0.085 .\) What are the partial pressures of the two gases in a \(15.00\) - 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?

In the given exercise, we have the mole fraction of methane \((\chi_{\text{methane}}) = 0.915\) and ethane \((\chi_{\text{ethane}}) = 0.085\).

The partial pressure of a gas in a mixture can be determined using Dalton's Law: Partial Pressure of Gas = (Mole Fraction of Gas) × (Total Pressure) The total pressure is given as \(1.44\) atm. For methane: \(P_{\text{methane}} = 0.915 × 1.44\,\text{atm} ≈ 1.317\,\text{atm}\) For ethane: \(P_{\text{ethane}} = 0.085 × 1.44\,\text{atm} ≈ 0.122\,\text{atm}\)

Using the ideal gas law, we can determine the number of moles of each gas: \(PV = nRT\) \(n = \frac{PV}{RT}\) where \(P\) is the partial pressure, \(V\) is the volume (15.00 L), \(R\) is the ideal gas constant (0.0821 L atm/mol K), and \(T\) is the temperature in Kelvin (293 K). For methane: \(n_{\text{methane}} = \frac{(1.317\,\text{atm})(15.00\,\text{L})}{(0.0821\,\text{L}\,\text{atm}\,\text{mol}^{-1}\,\text{K}^{-1})(293\,\text{K})} ≈ 0.906\,\text{mol}\) For ethane: \(n_{\text{ethane}} = \frac{(0.122\,\text{atm})(15.00\,\text{L})}{(0.0821\,\text{L}\,\text{atm}\,\text{mol}^{-1}\,\text{K}^{-1})(293\,\text{K})} ≈ 0.0837\,\text{mol}\)

The combustion reactions for methane and ethane are as follows: Methane: \(\mathrm{CH}_{4} + 2\,\mathrm{O}_{2} → \mathrm{CO}_{2} + 2\,\mathrm{H}_{2}\mathrm{O}\) Ethane: \(\mathrm{C}_{2}\mathrm{H}_{6} + 3.5\,\mathrm{O}_{2} → 2\,\mathrm{CO}_{2} + 3\,\mathrm{H}_{2}\mathrm{O}\) Stoichiometry of methane reaction: 1 mole methane = 2 moles water Stoichiometry of ethane reaction: 1 mole ethane = 3 moles water Total moles of water = (moles of methane × 2) + (moles of ethane × 3): = (0.906 mol × 2) + (0.0837 mol × 3) ≈ 2.941 moles Now, we calculate the mass of water, knowing that the molar mass of water is about 18.015 g/mol: Mass of water = (2.941 moles) × (18.015 g/mol) ≈ 52.95 g So, the total mass of water formed is approximately 52.95 g.