What volume of air at 1.0 atm and 25 degree Celsius must be taken to an engine to burn 1 gallon of gasoline?
Assuming that:
2 C8H18 + 25 O2 --> 16 CO2 + 18 H2O
provides reasonable model of the actual combustion process. ( The partial pressure of oxygen in air is 0.21 atm and the density of liquid octane is 0.70g/mL)
1 gallon = 3.785 L octane.
mass octane = volume x density = approx 0.7 x 3785 mL = approx 250 g
mols octane = grams/molar mass = approx 23
23 mols C8H18 x (25 mols O2/2 mol octane) = approx 300 mols oxygen needed. Check all of these numbers.
Then use PV = nRT to solve for volume oxygen needed and correct for the 21% O2 in air to find volume air needed.
thanks DrBob222
To determine the volume of air required to burn 1 gallon of gasoline, we need to consider the stoichiometry of the combustion reaction and the conditions given.
1. Convert 1 gallon of gasoline to the corresponding mass:
Since the density of liquid octane (the main component of gasoline) is given as 0.70 g/mL, we can use this information to convert 1 gallon to grams.
1 gallon ≈ 3.785 L (approximate volume conversion)
Mass = Volume × Density = 3.785 L × 0.70 g/mL = 2.6495 g
2. Calculate the molar mass of octane (C8H18):
C8H18 is octane, and its molar mass can be determined from the periodic table.
Molar mass of octane = (8 × molar mass of carbon) + (18 × molar mass of hydrogen)
The molar mass of carbon (C) is approximately 12.01 g/mol, and the molar mass of hydrogen (H) is approximately 1.01 g/mol.
Molar mass of octane = (8 × 12.01 g/mol) + (18 × 1.01 g/mol) = 114.23 g/mol
3. Determine the moles of octane in 2.6495 g:
Moles of octane = Mass of octane / Molar mass of octane = 2.6495 g / 114.23 g/mol ≈ 0.0232 mol
4. Use the stoichiometry of the combustion reaction to determine the moles of oxygen required:
From the balanced equation, we see that 2 moles of octane react with 25 moles of oxygen. Therefore, the moles of oxygen needed to burn 0.0232 mol of octane can be calculated using a proportion:
(25 mol O2) / (2 mol octane) = (x mol O2) / (0.0232 mol octane)
x = (25 mol O2) × (0.0232 mol octane) / (2 mol octane) ≈ 0.290 mol O2
5. Calculate the volume of oxygen (air) required at 1.0 atm and 25 °C:
The ideal gas law can be used to determine the volume of oxygen needed.
PV = nRT
where:
P = pressure = 1.0 atm
V = volume (unknown)
n = moles of oxygen = 0.290 mol
R = ideal gas constant = 0.0821 L·atm/(mol·K) (approximately)
T = temperature = 25 °C = 298 K (temperature in Kelvin)
Rearranging the ideal gas law equation to solve for V:
V = nRT / P = (0.290 mol) × (0.0821 L·atm/(mol·K)) × (298 K) / (1.0 atm) ≈ 7.2 L
Therefore, approximately 7.2 liters of air at 1.0 atm and 25 °C would be needed to burn 1 gallon of gasoline.