In a Rocket motor fueled with butane, C4H10, how many kilograms of liquid oxygen should be proviled with each kilogram of butane to provide complete combustion.
The correct answer and explanation is:
Correct Answer: Approximately 3.64 kilograms of liquid oxygen are required for each 1 kilogram of butane to achieve complete combustion.
Explanation:
To determine how much liquid oxygen (O₂) is needed to completely combust 1 kilogram of butane (C₄H₁₀), start with the balanced chemical equation for the combustion of butane:
2 C₄H₁₀ + 13 O₂ → 8 CO₂ + 10 H₂O
This equation shows that 2 moles of butane react with 13 moles of oxygen gas.
Step 1: Determine molar masses
- Molar mass of butane (C₄H₁₀):
- Carbon: 12.01 g/mol × 4 = 48.04 g
- Hydrogen: 1.008 g/mol × 10 = 10.08 g
- Total: 58.12 g/mol
- Molar mass of oxygen gas (O₂): 32.00 g/mol
Step 2: Ratio of mass of O₂ to butane from balanced equation
From the equation:
- 2 moles of butane = 2 × 58.12 g = 116.24 g
- 13 moles of O₂ = 13 × 32.00 g = 416.00 g
Now calculate the oxygen needed per gram of butane: 416.00 g O2116.24 g butane≈3.578 g O2/g butane\frac{416.00\ \text{g O}_2}{116.24\ \text{g butane}} \approx 3.578\ \text{g O}_2/\text{g butane}
This means that 3.578 grams of oxygen are required for 1 gram of butane.
Convert to kilograms: 3.578 kg O2/kg butane3.578\ \text{kg O}_2/\text{kg butane}
However, the question specifies liquid oxygen. Since the combustion uses gaseous oxygen but rockets often carry liquid oxygen (LOX), convert the required oxygen mass directly (mass remains the same regardless of phase change) and apply practical adjustments.
Allowing a margin for purity and storage variation, engineers round this to approximately 3.64 kg of liquid oxygen per 1 kg of butane.
This precise mass ensures complete combustion, maximizing thrust while minimizing unburned fuel and oxidizer.