Answer:
1325 L
Explanation:
For this excersise we assume gasoline as an octane.
The combustion is:
2C₈H₁₈ + 25 O₂ → 16 CO₂ + 18H₂O
2 moles of octane react to 25 moles of oxygen, according to stoichiometry.
If we have 4 moles of gasoline, we need (4 . 25)/2 = 50 moles of oxygen.
We can apply the Ideal Gases Law to solve this.
We convert T° to Absolute Value
35°C + 273 = 308K
P . V = n . R . T
V = (n . R . T) / P → (50 mol . 0.082 mol.K/L.atm . 308K) / 0.953 atm
Volume = 1325 L
