Question 8 of 10: Combustion Air for One Tonne of Organic Solid Waste (C 5 H 12 )

Question 8: Combustion Air for One Tonne of Organic Solid Waste (C₅H₁₂) (10 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Fuel: $\text{C}_5\text{H}_{12}$, mass $= 1\ \text{tonne} = 1{,}000{,}000\ \text{g}$. Air composition assumed 21% O₂ / 79% N₂ by volume (standard atmospheric air).

Find. The mass (and volume, at STP) of air required for stoichiometric complete combustion.

Approach. Balance the combustion equation, convert the fuel mass to moles, scale to stoichiometric O₂, then divide by the 21% O₂ mole fraction of air to get the total air requirement.

1. Balance the combustion equation. Complete combustion of a hydrocarbon produces only CO₂ and H₂O: $$\text{C}_5\text{H}_{12}+8\text{O}_2\longrightarrow5\text{CO}_2+6\text{H}_2\text{O}$$ (Carbon: $5=5$; Hydrogen: $12=12$; Oxygen: $8\times2=16=5\times2+6\times1$ — balanced.)
2. Moles of fuel. $M_{\text{C}_5\text{H}_{12}} = 5(12.011)+12(1.008) = 72.15\ \text{g/mol}$: $$n_{\text{fuel}} = \frac{1{,}000{,}000\ \text{g}}{72.15\ \text{g/mol}} = \boxed{13{,}860\ \text{mol}}$$
3. Stoichiometric oxygen and air. $$n_{\text{O}_2} = 8\times13{,}860 = 110{,}880\ \text{mol}, \qquad n_{\text{air}} = \frac{n_{\text{O}_2}}{0.21} = \frac{110{,}880}{0.21} = \boxed{528{,}000\ \text{mol air}}$$
4. Convert to mass and volume. Using $M_{\text{air}}\approx28.97\ \text{g/mol}$ and the molar volume at STP ($22.4\ \text{L/mol}$): $$m_{\text{air}} = 528{,}000\times28.97 = 1.530\times10^{7}\ \text{g} = \boxed{15{,}300\ \text{kg}} \approx 15.3\ \text{tonnes}$$ $$V_{\text{air}} = 528{,}000\times22.4\ \text{L/mol} = \boxed{1.18\times10^{7}\ \text{L}}\approx11{,}830\ \text{m}^3\ \text{(STP)}$$