Introduction
To determine the volume of oxygen required to completely burn 2.2g of propane (C3H8), we need to follow these steps: calculate the moles of propane, use the combustion reaction to find the moles of oxygen needed, and then convert that to volume at standard conditions.
Step 1: Calculate Moles of Propane
- The molar mass of propane (C3H8) is approximately 44.1 g/mol.
- Moles of propane = mass (g) / molar mass (g/mol)
Moles of propane = 2.2 g / 44.1 g/mol ≈ 0.0500 mol
Step 2: Combustion Reaction
- The balanced chemical equation for the combustion of propane is:
C3H8 + 5 O2 → 3 CO2 + 4 H2O
- From the equation, 1 mole of propane requires 5 moles of oxygen.
Step 3: Calculate Moles of Oxygen Required
- Moles of oxygen needed = moles of propane × 5
Moles of oxygen = 0.0500 mol × 5 = 0.250 mol
Step 4: Convert Moles of Oxygen to Volume
- At standard conditions (1 atm and 273 K), 1 mole of gas occupies 22.4 liters.
- Volume of oxygen = moles of oxygen × volume per mole
Volume of oxygen = 0.250 mol × 22.4 L/mol = 5.6 liters
Conclusion
To completely burn 2.2g of propane, approximately 5.6 liters of oxygen at 1 atm and 273 K are required.