Complete combustion: conserve atoms when hydrocarbons burn
| English | Français |
|---|---|
| complete combustion/kəmˈpliːt kəmˈbʌstʃn/ | combustion complète |
| oxygen supply | apport d'oxygène |
What would explain this observation?
- The water formed when propane burns was not stored as liquid water inside the fuel. Its hydrogen atoms combine with oxygen during a reaction that also forms carbon dioxide.
- Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.
Build the model
- Combustion of hydrocarbon fuels releases energy. During complete combustion 完全燃烧 with sufficient oxygen, carbon is oxidised to carbon dioxide and hydrogen to water. Begin with hydrocarbon + oxygen → carbon dioxide + water. Use the given hydrocarbon formula to count carbon and hydrogen atoms, then balance oxygen. This chemical reaction makes new substances, unlike the physical separation in the fractionating column.
- complete combustion: Burning a hydrocarbon in sufficient oxygen to form carbon dioxide and water; oxygen supply 氧气供应: The oxygen available for a combustion reaction.
What are the products of complete combustion of a hydrocarbon?
One carbon atom requires one carbon dioxide molecule, while each pair of hydrogen atoms forms one water molecule. Oxygen atoms on the product side come from both products, so count both before finding O₂. If an intermediate oxygen coefficient is a half number, multiply every coefficient by two for smallest whole-number coefficients. Never alter the given hydrocarbon’s subscripts to force balance.
Match each technical term to its precise meaning.
Use the definitions to distinguish related quantities and processes.
Choose evidence that can test it
- One carbon atom requires one carbon dioxide molecule, while each pair of hydrogen atoms forms one water molecule. Oxygen atoms on the product side come from both products, so count both before finding O₂. If an intermediate oxygen coefficient is a half number, multiply every coefficient by two for smallest whole-number coefficients. Never alter the given hydrocarbon’s subscripts to force balance.
- Write correct formulae first, balance carbon and hydrogen, then oxygen, and check all three element totals. State the oxygen condition in the explanation. Supplied combustion observations can support energy release; they do not alone prove every carbon atom became carbon dioxide. Incomplete combustion products and atmospheric effects belong to their other course sections. Use teacher-approved demonstrations or written evidence rather than independent fuel-burning trials.
Which two habits make the investigation or model in this case more defensible?
Write correct formulae first, balance carbon and hydrogen, then oxygen, and check all three element totals. State the oxygen condition in the explanation. Supplied combustion observations can support energy release; they do not alone prove every carbon atom became carbon dioxide. Incomplete combustion products and atmospheric effects belong to their other course sections. Use teacher-approved demonstrations or written evidence rather than independent fuel-burning trials.
Work from known quantities
- State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
- Known: propane is C₃H₈. Three carbon atoms give 3CO₂ and eight hydrogen atoms give 4H₂O. Product oxygen count=3×2+4=10, requiring 5O₂. The balanced equation is C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. Check C:3→3, H:8→8 and O:10→10. For ethane, 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O avoids a fractional coefficient.
In C₃H₈ + 5O₂ → 3CO₂ + 4H₂O, find the O₂ molecules needed for three propane molecules. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.
In C₃H₈ + 5O₂ → 3CO₂ + 4H₂O, find the O₂ molecules needed for three propane molecules.
The result is 15 molecules. Known: propane is C₃H₈. Three carbon atoms give 3CO₂ and eight hydrogen atoms give 4H₂O. Product oxygen count=3×2+4=10, requiring 5O₂. The balanced equation is C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. Check C:3→3, H:8→8 and O:10→10. For ethane, 2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O avoids a fractional coefficient.
Check the conclusion and its limits
- Equal atom totals do not imply equal numbers of molecules or equal masses of carbon dioxide and water. A smoky flame is not proof of complete combustion. This task needs the supplied formula and atom conservation, not memorised combustion equations for every named fuel or an unrequired enthalpy calculation.
- Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.
Balancing combustion allows the fuel formula to change. This claim is false: Equal atom totals do not imply equal numbers of molecules or equal masses of carbon dioxide and water. A smoky flame is not proof of complete combustion. This task needs the supplied formula and atom conservation, not memorised combustion equations for every named fuel or an unrequired enthalpy calculation.
Complete combustion: conserve atoms when hydrocarbons burn: One carbon atom requires one carbon dioxide molecule, while each pair of hydrogen atoms forms one water molecule. Oxygen atoms on the product side come from both products, so count both before finding O₂. If an intermediate oxygen coefficient is a half number, multiply every coefficient by two for smallest whole-number coefficients. Never alter the given hydrocarbon’s subscripts to force balance.
Balancing combustion allows the fuel formula to change.
Equal atom totals do not imply equal numbers of molecules or equal masses of carbon dioxide and water. A smoky flame is not proof of complete combustion. This task needs the supplied formula and atom conservation, not memorised combustion equations for every named fuel or an unrequired enthalpy calculation.
Burning a hydrocarbon in sufficient oxygen to form carbon dioxide and water: write the technical term.
complete combustion means Burning a hydrocarbon in sufficient oxygen to form carbon dioxide and water.