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?

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.