Questions: The combustion of propane produces carbon dioxide gas and water vapor according to the reaction C3H8(l) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) A sample of propane is added to a 260-L container containing excess oxygen to a final pressure of 10.8 atm and combusted at 650 K. Calculate the moles of carbon dioxide gas produced. 158 mol 52.6 mol 210 mol 263 mol

Solution Steps

We are given:

• Volume of the container, \( V = 260 \, \text{L} \)
• Final pressure, \( P = 10.8 \, \text{atm} \)
• Temperature, \( T = 650 \, \text{K} \)
• The reaction: \[ \mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{l}) + 5 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 3 \mathrm{CO}_{2}(\mathrm{~g}) + 4 \mathrm{H}_{2} \mathrm{O}(\mathrm{~g}) \]

The ideal gas law is given by: \[ PV = nRT \] where:

• \( P \) is the pressure,
• \( V \) is the volume,
• \( n \) is the number of moles,
• \( R \) is the ideal gas constant (\(0.0821 \, \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K}\)),
• \( T \) is the temperature.

Rearranging for \( n \): \[ n = \frac{PV}{RT} \]

Substituting the given values: \[ n = \frac{10.8 \, \text{atm} \times 260 \, \text{L}}{0.0821 \, \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K} \times 650 \, \text{K}} \]

\[ n = \frac{2808}{53.365} \approx 52.6255 \, \text{mol} \]

From the balanced chemical equation, 1 mole of propane (\(\mathrm{C}_{3}\mathrm{H}_{8}\)) produces 3 moles of carbon dioxide (\(\mathrm{CO}_{2}\)).

Thus, the moles of \(\mathrm{CO}_{2}\) produced: \[ n_{\mathrm{CO}_{2}} = 3 \times 52.6255 \approx 157.8765 \, \text{mol} \]

Final Answer