Balancing chemical equations

Balancing a chemical equation can sometimes prove very difficult. The method shown below uses fractions and may prove to be useful for some equations.
Firstly, lets see how to use the information in a chemical equation.

3Br2 + 8NH3 =>6NH4Br + N2

The number of each atom can be calculated by first multiplying the prefix in front of the compound and the subscript on the particular atom we are interested in finding.

Example, to calculate the number of hydrogen atoms on the left of the equation we multiply the prefix (8) by the subscript (3) to get 8 X 3 = 24 hydrogen atoms.
Example, to calculate the number of hydrogen atoms on the right of the equation we multiply the prefix (6) by the subscript(4) to get 6 X 4 = 24 hydrogen atoms.
Example, to calculate the number of nitrogen atoms on the left of the equation we multiply the prefix (8) by the subscript (1) to get 8 X 1 = 8 nitrogen atoms.
Example, to calculate the number of bromine atoms on the left of the equation we multiply the prefix (3) by the subscript (2) to get 3X 2 = 6 hydrogen atoms.

Balance the equation below.

Br2 + NH3 =>NH4Br + N2

Step 1) Start with Br. We have 2 bromine atoms on the left and 1 on the right. So place a 1/2 in front of the Br2 on the left.
1/2Br2 + NH3 =>NH4Br + N2
We now have one Br atom on both sides.


Step 2) Continue with H. We have 3 hydrogen atoms on the left and 4 on the right. So place a 4/3 in front of the NH3 on the left.
1/2Br2 + 4/3NH3 =>NH4Br + N2
We now have 4 hydrogen atoms on both sides.

Step 3) Continue with N. We have 4/3 atoms on the left and 4 on the right. So place a 1/6 in front of the N2 on the right.
1/2Br2 + 4/3NH3 =>NH4Br + 1/6N2

Step 4) Multiply the entire equation by the lowest number necessary to change all the fractions to a whole number.
(
1/2Br2 + 4/3NH3 =>NH4Br + 1/6N2 ) X 6
=> 3Br2 + 8NH3 =>6NH4Br + N2 This is a balanced equation.
We have 6 bromine atoms, 8 nitrogen atoms and 24 hydrogen atoms on both sides of the equation.

Using this method we sometimes refer to an equation as having a fraction of an atom. This is not the case and we multiply through by a number to remove fractions.

Example 2 - Balance the equation below.

C2H6+ O2 => CO2+ H2O

Step 1) Start with C. We have 2 carbon atoms on the left and 1 on the right. So place a 2 in front of the CO2 on the right.
C2H6+ O2 => 2CO2+ H2O
We now have two carbon atoms on both sides.


Step 2) Continue with H. We have 6 hydrogen atoms on the left and 2 on the right. So place a 3 in front of the H2O on the rightt.
C2H6+ O2 => 2CO2+ 3H2O
We now have 6 hydrogen atoms on both sides.

Step 3) Continue with O. We have 2 atoms of oxygen on the left and 7 on the right. So place a 7/2 in front of the O2 on the left.
C2H6+ 7/2O2 => 2CO2+ 3H2O

Step 4) Multiply the entire equation by the lowest number necessary to change all the fractions to a whole number.
(C2H6+ 7/2O2 => 2CO2+ 3H2O
) X 2
=> 2C2H6+ 7O2 => 4CO2+ 6H2O This is a balanced equation.
We have 4 carbon atoms, 14 oxygen atoms and 12 hydrogen atoms on both sides of the equation.

Try these

C3H4 + O2 → H2O + CO2 Solution
HCl + O2 → Cl2 + H2O Solution
H2S + O2 → SO2 + H2O Solution
H2O + Al2S3 → H2S + Al(OH)3  
Worksheet Solutions