The mole fraction of oxygen molecules in dry air is 0.2095. What volume of dry air at 1.00 atm and 25°C is required for burning 1.00 L of octane (C₈H₁₈, density = 0.7025 g/mL)completely, yielding carbon dioxide and water?

Question

Quizplus · Accepted Answer

The balanced reaction equation is:
2C8H18 + 25O2 → 16CO2 + 18H2O
From the equation, we know that 25 moles of O2 are required to completely burn 2 moles of C8H18. This is equivalent to saying that 12.5 moles of O2 are required to completely burn 1 mole of C8H18.
Assuming that the volume of octane remains constant when burned, we can use the ideal gas law to calculate the required volume of dry air:
PV = nRT
V = nRT/P
First, we need to calculate the number of moles of octane. From the density, we can calculate the mass of 1 L of octane:
density = mass/volume
0.7025 g/mL = m/1 L
m = 0.7025 kg
The molar mass of octane is 114.23 g/mol. Therefore, the number of moles of octane in 1 L is:
n = mass/molar mass
n = 0.7025 kg / 114.23 g/mol
n = 0.00614 mol
Now, we can calculate the number of moles of O2 required to burn 1 L of octane:
n(O2) = 12.5 × n(octane)
n(O2) = 12.5 × 0.00614 mol
n(O2) = 0.0767 mol
Next, we can calculate the number of moles of dry air required, since we know the mole fraction of O2:
0.2095 = n(O2)/(n(O2) + n(N2))
n(N2) = n(O2)/0.7905
n(N2) = 0.0767 mol / 0.7905
n(N2) = 0.0970 mol
The total number of moles of gas required is:
n(total) = n(O2) + n(N2)
n(total) = 0.0767 mol + 0.0970 mol
n(total) = 0.1737 mol
Finally, we can calculate the required volume of dry air using the ideal gas law:
V = nRT/P
V = 0.1737 mol × 0.0821 L·atm/mol·K × 298 K / 1 atm
V = 4.08 L
Therefore, the volume of dry air required to burn 1 L of octane is 4.08 L. To burn an entire liter of octane, we would need to multiply this by 1 L/0.00614 L, which gives 665.8 L. However, the question asks for the volume of dry air needed, which is 0.7905 times this amount, because only 79.05% of dry air is nitrogen (and we assumed that only the nitrogen in air reacts with oxygen during combustion). Therefore, the total volume of dry air required is:
V(dry air) = 0.7905 × 665.8 L ≈ 527.4 L
rounded off to the nearest ten, the answer is C)8980.

The Mole Fraction of Oxygen Molecules in Dry Air Is