Burning ethane (C₂H₆) via complete combustion follows this balanced equation:
2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O
This shows a mole ratio of 2 mol ethane to 4 mol CO₂, which simplifies to 1 mol ethane producing 2 mol CO₂.
To find the moles of CO₂ produced from 5.45 mol of ethane, multiply by this ratio:
5.45 mol C₂H₆ × (4 mol CO₂ / 2 mol C₂H₆) = 5.45 mol C₂H₆ × 2 = 10.9 mol CO₂
Why this ratio works: Each ethane molecule (C₂H₆) contains 2 carbon atoms. In complete combustion, every carbon atom ends up in a CO₂ molecule, so each mole of ethane yields exactly 2 moles of CO₂ — the carbon atoms are simply redistributed, not created or destroyed, following the law of conservation of mass.
Key assumptions:
- This calculation assumes complete combustion, meaning there's enough oxygen present for the reaction to fully convert all carbon to CO₂ and all hydrogen to H₂O.
- If oxygen is limited, incomplete combustion can occur, producing carbon monoxide (CO) or even elemental soot (C) instead of CO₂, which would lower the actual CO₂ yield below 10.9 mol.
- The result also assumes the reaction goes to completion (100% yield), which is typical for stoichiometry problems but may differ from real-world experimental results due to side reactions or practical inefficiencies.
So, under standard complete-combustion conditions, 5.45 moles of ethane produce 10.9 moles of carbon dioxide.