First, we write the complete and balanced chemical equation describing complete combustion. Combustion involves rapid reaction with O2. The products of complete combustion of hydrocarbons are CO2 and H2O.The first step involves writing a reaction expression with single molecules or formula unit of the reactants and products. Then, we take the inventories of the number of atoms of each of the elements in the reaction mixture and we balance them starting with those that appear in only one reactant and product.
a. C5H10 l+ O2 g → CO2 g+ H2Og
Atoms: 5C+10H+2O →1C+3O+ 2H
There is only one C atom on the right and five on left hand side of reaction arrow. To balance C, we place a coefficient of 5 in front of CO2 and recalculate distribution of atoms on both sides.
C5H10 l+ O2 g → 5CO2 g+ H2Og
Atoms: 5C+10H+2O →5C+11O+ 2H
There are two H atoms on the right and ten on left hand side of reaction arrow. To balance H, we place a coefficient of 5 in front of H2O and recalculate distribution of atoms on both sides.
C5H10 l+ O2 g → 5CO2 g+ 5H2Og
Atoms: 5C+10H+2O →5C+15O+ 10H
This balances the number of C and H atoms but leaves an odd number of O atoms on the product side. The only source of O atoms is O2, so we need an even number of O atoms on the right. To balance O atoms, multiply all the coefficients in the expression by 2.
2C5H10 l+ 2O2 g → 10CO2 g+ 10H2Og
Atoms: 10C+20H+4O →10C+30O+ 20H
We can now balance the number of O atoms by replacing the coefficient 2 in front of O2 with 15, which gives us 30 O atoms and a balanced equation.
2C5H10 l+ 15O2 g → 10CO2 g+ 10H2Og
Atoms: 10C+20H+30O →10C+30O+ 20H
b. C6H14 l+ O2 g → CO2 g+ H2Og
Atoms: 6C+14H+2O →1C+3O+ 2H
There is only one C atom on the right and six on left hand side of reaction arrow. To balance C, we place a coefficient of 6 in front of CO2 and recalculate distribution of atoms on both sides.
C6H14 l+ O2 g → 6CO2 g+ H2Og
Atoms: 6C+14H+2O →6C+13O+ 2H
There are two H atoms on the right and 14 on left hand side of reaction arrow. To balance H, we place a coefficient of 7 in front of H2O and recalculate distribution of atoms on both sides.
C6H14 l+ O2 g → 6CO2 g+ 7H2Og
Atoms: 6C+14H+2O →6C+19O+ 14H
This balances the number of C and H atoms but leaves an odd number of O atoms on the product side. The only source of O atoms is O2, so we need an even number of O atoms on the right. To balance O atoms, multiply all the coefficients in the expression by 2.
2C6H14 l+ 2O2 g → 12CO2 g+ 14H2Og
Atoms: 12C+28H+4O →12C+38O+ 28H
We can now balance the number of O atoms by replacing the coefficient 2 in front of O2 with 19, which gives us 30 O atoms and a balanced equation.
2C6H14 l+ 19O2 g → 12CO2 g+ 14H2Og
Atoms: 12C+28H+38O →12C+38O+ 28H
c. C8H10 l+ O2 g → CO2 g+ H2Og
Atoms: 8C+10H+2O →1C+3O+ 2H
There is only one C atom on the right and eight on left hand side of reaction arrow. To balance C, we place a coefficient of 8 in front of CO2 and recalculate distribution of atoms on both sides.
C8H10 l+ O2 g → 8CO2 g+ H2Og
Atoms: 8C+10H+2O →8C+17O+ 2H
There are two H atoms on the right and ten on left hand side of reaction arrow. To balance H, we place a coefficient of 5 in front of H2O and recalculate distribution of atoms on both sides.
C8H10 l+ O2 g → 8CO2 g+ 5H2Og
Atoms: 8C+10H+2O →8C+21O+ 10H
This balances the number of C and H atoms but leaves an odd number of O atoms on the product side. The only source of O atoms is O2, so we need an even number of O atoms on the right. To balance O atoms, multiply all the coefficients in the expression by 2.
2C8H10 l+ 2O2 g → 16CO2 g+ 10H2Og
Atoms: 16C+20H+4O →16C+42O+ 20H
We can now balance the number of O atoms by replacing the coefficient 2 in front of O2 with 21, which gives us 40 O atoms and a balanced equation.
2C8H10 l+ 21O2 g → 16CO2 g+ 10H2Og
Atoms: 16C+20H+42O →16C+42O+ 20H
d. C9H12 g+ O2 g → CO2 g+ H2Og
Atoms: 9C+12H+2O →1C+3O+ 2H
There is only one C atom on the right and three on left hand side of reaction arrow. To balance C, we place a coefficient of 9 in front of CO2 and recalculate distribution of atoms on both sides.
C9H12 g+ O2 g → 9CO2 g+ H2Og
Atoms: 9C+12H+2O →9C+19O+ 2H
There are two H atoms on the right and eight on left hand side of reaction arrow. To balance H
, we place a coefficient of 6 in front of H2O and recalculate distribution of atoms on both sides.
C9H12 g+ O2 g → 9CO2 g+ 6H2Og
Atoms: 9C+12H+2O →9C+24O+ 12H
There are 24O atoms on the right and two on left hand side of reaction arrow. To balance O, we place a coefficient of 12 in front of O2 and recalculate distribution of atoms on both sides.
C9H12 g+ 12O2 g → 9CO2 g+ 6H2Og
Atoms: 9C+12H+24O →9C+24O+ 12H
Both sides of the reaction have the same number of each element. Therefore, this reaction is balanced.
Conclusion:
Balancing each atom on both sides of reaction arrow balances the entire reaction.