1. Balanced Chemical Equation:
The combustion of benzene (C₆H₆) produces carbon dioxide (CO₂) and water (H₂O):
2 C₆H₆ + 15 O₂ → 12 CO₂ + 6 H₂O
2. Calculate Moles of Benzene:
* Molar mass of benzene (C₆H₆) = 78.11 g/mol
* Moles of benzene = (39 g) / (78.11 g/mol) = 0.5 mol
3. Determine Moles of Oxygen Required:
* From the balanced equation, 2 moles of benzene require 15 moles of oxygen.
* Moles of oxygen needed = (0.5 mol benzene) * (15 mol O₂ / 2 mol benzene) = 3.75 mol O₂
4. Convert Moles of Oxygen to Liters at STP:
* At standard temperature and pressure (STP), 1 mole of any ideal gas occupies 22.4 liters.
* Liters of oxygen = (3.75 mol O₂) * (22.4 L/mol) = 84 liters
Therefore, approximately 84 liters of oxygen are required for the combustion of 39 grams of liquid benzene at STP.
