Topic 4 · page 12 of 17

Combustion of alkanes

Alkanes burn in plenty of oxygen to give carbon dioxide and water. With limited oxygen they give carbon monoxide or carbon as well. You must balance both kinds of equation, and use them in calculations.

13 min read4.174.17(i)

Read first: Alkanes and cycloalkanes

Complete combustion 4.174.17(i)

When an alkane burns in a plentiful supply of air or oxygen, every carbon atom ends up in carbon dioxide and every hydrogen atom in water. This is complete combustion. It is also an oxidation, because the alkane gains oxygen, and it is strongly exothermic, which is why alkanes make good fuels.

CHX4(g)+2 OX2(g)→COX2(g)+2 HX2O(l) \ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)}
CX3HX8(g)+5 OX2(g)→3 COX2(g)+4 HX2O(l) \ce{C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(l)}

In an engine or a flame the water is formed as steam, so the state symbol is (g) rather than (l). Check which one the question uses.

Balancing a combustion equation

MethodBalancing combustion

  1. Write the alkane, and O₂ on the left; CO₂ and H₂O on the right.
  2. Balance carbon first: one CO₂ for each carbon.
  3. Balance hydrogen next: one H₂O for every two hydrogens.
  4. Balance oxygen last, by counting the oxygen atoms on the right and halving.
  5. A fraction in front of O₂ is allowed. Alternatively, multiply every coefficient by 2 to clear the fraction.

Worked exampleOctane, C₈H₁₈

  1. Carbon: 8 carbons, so 8CO₂.
  2. Hydrogen: 18 hydrogens, so 9H₂O.
  3. Oxygen: 8 × 2 + 9 = 25 oxygen atoms on the right, so 25/2 O₂ on the left.
CX8HX18+252 OX2→8 COX2+9 HX2O \ce{C8H18 + 25/2 O2 -> 8CO2 + 9H2O}

Doubling every coefficient avoids the fraction:

2 CX8HX18+25 OX2→16 COX2+18 HX2O \ce{2C8H18 + 25O2 -> 16CO2 + 18H2O}

For any alkane CₙH₂ₙ₊₂ the amount of oxygen is (3n + 1)/2. Check it: n = 8 gives 25/2, and n = 3 gives 5.

Incomplete combustion 4.174.17(i)

When the supply of oxygen is limited, some carbon is not fully oxidised. The products are carbon monoxide, CO, or carbon (soot), C, as well as water. A mixture of all of these is common.

CX8HX18+172 OX2→8 CO+9 HX2O \ce{C8H18 + 17/2 O2 -> 8CO + 9H2O}
CHX4+OX2→C+2 HX2O \ce{CH4 + O2 -> C + 2H2O}
CHX4+32 OX2→CO+2 HX2O \ce{CH4 + 3/2 O2 -> CO + 2H2O}
Products When Oxygen needed by CH₄
CO₂ and H₂O plenty of oxygen 2 mol per mol
CO and H₂O limited oxygen 1.5 mol per mol
C and H₂O very little oxygen 1 mol per mol

Using the equations 4.17

Worked exampleGas volumes

Propane burns completely: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. What volume of carbon dioxide is made when 10.0 cm³ of propane burns, and what volume of oxygen is needed? All gases are measured at the same temperature and pressure.

  1. Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules, so the volume ratio equals the mole ratio.
  2. 1 propane : 3 CO₂ : 5 O₂.
  3. CO₂ = 3 × 10.0 = 30.0 cm³. O₂ = 5 × 10.0 = 50.0 cm³.

Worked exampleMass of product

Calculate the mass of carbon dioxide made when 2.20 g of propane burns completely.

  1. MrM_\text{r} (C₃H₈) = 3 × 12.0 + 8 × 1.0 = 44.0.
  2. n(C₃H₈) = 2.20 ÷ 44.0 = 0.0500 mol.
  3. From the equation, 1 mol propane gives 3 mol CO₂: n(CO₂) = 3 × 0.0500 = 0.150 mol.
  4. MrM_\text{r} (CO₂) = 12.0 + 2 × 16.0 = 44.0. Mass = 0.150 × 44.0 = 6.60 g.

Check yourself

  1. Write a balanced equation for the complete combustion of butane, C₄H₁₀, with no fractions.
  2. Write a balanced equation for the incomplete combustion of hexane, C₆H₁₄, to carbon monoxide and water.
  3. Write an equation for the incomplete combustion of ethane, C₂H₆, to carbon and water.
  4. 50.0 cm³ of butane is burned completely. Calculate the volume of oxygen needed and the volume of carbon dioxide formed, at constant temperature and pressure.
  5. Calculate the mass of water formed when 1.60 g of methane burns completely.
  6. Explain why incomplete combustion is a problem in a gas boiler or a car engine.

Answers cover these until you have tried the questions

  1. Carbon: 4CO₂. Hydrogen: 5H₂O. Oxygen: 4 × 2 + 5 = 13 atoms, which is 13/2 O₂. Double everything: 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O.
  2. Carbon: 6CO. Hydrogen: 7H₂O. Oxygen: 6 + 7 = 13 atoms, so 13/2 O₂: C₆H₁₄ + 13/2 O₂ → 6CO + 7H₂O (or 2C₆H₁₄ + 13O₂ → 12CO + 14H₂O).
  3. Carbon: 2C. Hydrogen: 3H₂O. Oxygen: 3 atoms, which is 3/2 O₂: C₂H₆ + 3/2 O₂ → 2C + 3H₂O (or 2C₂H₆ + 3O₂ → 4C + 6H₂O).
  4. From 2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O the ratio is butane : O₂ : CO₂ = 2 : 13 : 8. O₂ = 50.0 × 13/2 = 325 cm³. CO₂ = 50.0 × 8/2 = 200 cm³.
  5. CH₄ + 2O₂ → CO₂ + 2H₂O. MrM_\text{r} (CH₄) = 12.0 + 4 × 1.0 = 16.0. n(CH₄) = 1.60 ÷ 16.0 = 0.100 mol. n(H₂O) = 2 × 0.100 = 0.200 mol. MrM_\text{r} (H₂O) = 2 × 1.0 + 16.0 = 18.0. Mass = 0.200 × 18.0 = 3.60 g.
  6. Incomplete combustion makes carbon monoxide, which is toxic, and carbon (soot), which blocks the equipment. It also releases less energy than complete combustion for the same amount of fuel.