Introduction
To determine the volume of air needed for the complete combustion of 2L of ethane (C2H6), we must first understand the stoichiometry of the combustion reaction.
Combustion Reaction of Ethane
The balanced chemical equation for the combustion of ethane is:
C2H6 + 7/2 O2 → 2 CO2 + 3 H2O
This equation indicates that one mole of ethane reacts with 3.5 moles of oxygen.
Calculating Required Oxygen
1. Volume of Ethane:
- Given volume of ethane = 2L
2. Moles of Ethane:
- At standard conditions, 1 mole of gas occupies 22.4L.
- Moles of ethane = Volume of ethane / Molar volume = 2L / 22.4L/mol ≈ 0.089 mol
3. Moles of Oxygen Required:
- According to the balanced equation, 1 mol of ethane requires 3.5 mol of O2.
- Therefore, moles of O2 needed = 0.089 mol x 3.5 ≈ 0.312 mol
Calculating Volume of Air
1. Volume of Oxygen Needed:
- Volume of O2 = Moles of O2 x Molar volume = 0.312 mol x 22.4L/mol ≈ 7L
2. Volume of Air:
- Air is approximately 21% oxygen by volume.
- Volume of air needed = Volume of O2 / 0.21 ≈ 7L / 0.21 ≈ 33.33L
Conclusion
For the complete combustion of 2L of ethane, approximately 33.33L of air is required.