7. If 129.0 g of oxygen gas and 300 g of propane are mixed and allowed to react as shown below, determine the volume of water vapour formed at 116 kPa and \( 120^{\circ} \mathrm{C} \). \[ \mathrm{C}_{3} \mathrm{H}_{8(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CO}_{2(\mathrm{~g})}+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \]

To determine the volume of water vapor formed in the reaction between propane (\( \mathrm{C}_3\mathrm{H}_8 \)) and oxygen (\( \mathrm{O}_2 \)), we need to begin with the balanced chemical equation for the combustion of propane: \[ \mathrm{C}_3\mathrm{H}_8 + 5 \mathrm{O}_2 \rightarrow 3 \mathrm{CO}_2 + 4 \mathrm{H}_2\mathrm{O} \] From this balanced equation, one mole of propane reacts with five moles of oxygen to produce four moles of water vapor. 1. **Calculate moles of reactants:** - Molar mass of \( \mathrm{C}_3\mathrm{H}_8 \): \( 3(12.01) + 8(1.008) = 44.096 \, \text{g/mol} \) - Moles of propane: \( \frac{300 \, \text{g}}{44.096 \, \text{g/mol}} \approx 6.8 \, \text{moles} \) - Molar mass of \( \mathrm{O}_2 \): \( 2(16.00) = 32.00 \, \text{g/mol} \) - Moles of oxygen: \( \frac{129.0 \, \text{g}}{32.00 \, \text{g/mol}} \approx 4.0 \, \text{moles} \) 2. **Determine the limiting reagent:** - From the balanced equation, for 6.8 moles of \( \mathrm{C}_3\mathrm{H}_8 \), we would need \( 5 \times 6.8 = 34 \, \text{moles of } \mathrm{O}_2 \). However, we only have 4.0 moles, indicating that oxygen is the limiting reagent. 3. **Find the moles of water produced:** - Based on the stoichiometry, 5 moles of \( \mathrm{O}_2 \) produces 4 moles of \( \mathrm{H}_2\mathrm{O} \). - Hence, \( 4.0 \, \text{moles of } \mathrm{O}_2 \) will yield \( \frac{4.0 \, \text{moles of } \mathrm{O}_2 \times 4 \, \text{moles of } \mathrm{H}_2\mathrm{O}}{5 \, \text{moles of } \mathrm{O}_2} = 3.2 \, \text{moles of } \mathrm{H}_2\mathrm{O} \). 4. **Calculate the volume of water vapor at the given conditions:** - Use the ideal gas law \( PV = nRT \), - Convert temperature to Kelvin: \( T = 120 + 273.15 = 393.15 \, \text{K} \) - Use \( R = 8.314 \, \text{J/(mol K)} = 8.314 \, \text{L kPa/(mol K)} \) for pressures in kPa. - Rearranging the ideal gas law gives \( V = \frac{nRT}{P} \). Calculating the volume: \[ V = \frac{3.2 \, \text{moles} \times 8.314 \, \text{L kPa/(mol K)} \times 393.15 \, \text{K}}{116 \, \text{kPa}} \] \[ V \approx \frac{3.2 \times 8.314 \times 393.15}{116} \approx \frac{10,244.2 \, \text{L kPa}}{116} \approx 88.18 \, \text{L} \] Thus, the volume of water vapor formed is approximately **88.18 liters**.