Questions: The combustion of octane, C8H18, proceeds according to the reaction shown. 2 C8H18(l)+25 O2(g) -> 16 CO2(g)+18 H2O(l) If 506 mol of octane combusts, what volume of carbon dioxide is produced at 25.0°C and 0.995 atm? V= L

Solution Steps

The balanced chemical equation for the combustion of octane is:

\[ 2 \mathrm{C}_{8} \mathrm{H}_{18}(\mathrm{l}) + 25 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 16 \mathrm{CO}_{2}(\mathrm{~g}) + 18 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \]

From the equation, 2 moles of octane produce 16 moles of carbon dioxide. Therefore, the mole ratio of octane to carbon dioxide is 1:8.

Given that 506 moles of octane combust, we can calculate the moles of carbon dioxide produced using the stoichiometric ratio:

\[ \text{Moles of } \mathrm{CO}_2 = 506 \, \text{mol} \, \mathrm{C}_8\mathrm{H}_{18} \times \frac{16 \, \text{mol} \, \mathrm{CO}_2}{2 \, \text{mol} \, \mathrm{C}_8\mathrm{H}_{18}} = 4048 \, \text{mol} \, \mathrm{CO}_2 \]

The ideal gas law is given by:

\[ PV = nRT \]

Where:

• \( P = 0.995 \, \text{atm} \)
• \( V = ? \, \text{L} \)
• \( n = 4048 \, \text{mol} \)
• \( R = 0.0821 \, \text{L atm/mol K} \)
• \( T = 25.0^\circ \text{C} = 298.15 \, \text{K} \)

Rearranging the ideal gas law to solve for \( V \):

\[ V = \frac{nRT}{P} \]

Substituting the known values:

\[ V = \frac{4048 \, \text{mol} \times 0.0821 \, \text{L atm/mol K} \times 298.15 \, \text{K}}{0.995 \, \text{atm}} \]

\[ V = \frac{99112.8 \, \text{L atm}}{0.995 \, \text{atm}} \approx 99600.4 \, \text{L} \]

Final Answer