## Overview

In a chemical reaction, the quantity of each element does not change. Thus, each side of the equation must represent the same quantity of any particular element. In case of net ionic reactions, the same charge must be present on both sides of the unbalanced equation. By changing the scalar number for each molecular formula, the equation may be balanced.

## Using Trial and Error/Inspection

### Example #1 (Simple)

Simple chemical equations can be balanced by inspection, that is, by trial and error. Generally, it is best to balance the most complicated molecule first. Hydrogen and oxygen are usually balanced last.

- Na + O
_{2}= Na_{2}O

In order for this equation to be balanced, there must be an equal amount of Na on the left hand side as on the right hand side. As it stands now, there is 1 Na on the left but 2 Na's on the right. This problem is solved by putting a 2 in front of the Na on the left hand side:

- 2Na + O
_{2}= Na_{2}O

In this there are 2 Na atoms on the left and 2 Na atoms on the right. In the next step the oxygen atoms are balanced as well. On the left hand side there are 2 O atoms and the right hand side only has one. This is still an unbalanced equation. To fix this a 2 is added in front of the Na_{2}O on the right hand side. Now the equation reads:

- 2Na + O
_{2}= 2Na_{2}O

Notice that the 2 on the right hand side is "distributed" to both the Na_{2} and the O. Currently the left hand side of the equation has 2 Na atoms and 2 O atoms. The right hand side has 4 Na's total and 2 O's. Again, this is a problem, there must be an equal amount of each chemical on both sides. To fix this 2 more Na's are added on the left side. The equation will now look like this:

- 4Na + O
_{2}= 2Na_{2}O

This equation is a balanced equation because there is an equal number of atoms of each element on the left and right hand sides of the equation.

### Example #2 (Complex)

- P
_{4}+ O_{2}= 2P_{2}O_{5}

This equation is not balanced because there is an unequal amount of O's on both sides of the equation. The left hand side has 4 P's and the right hand side has 4 P's. So the P atoms are balanced. The left hand side has 2 O's and the right hand side has 10 O's.

To fix this unbalanced equation a 5 in front of the O_{2} on the left hand side is added to make 10 O's on both sides resulting in

- P
_{4}+ 5O_{2}= 2P_{2}O_{5}

The equation is now balanced because there is an equal amount of substances on the left and the right hand side of the equation.

### Example #3 (Complex)

- C
_{2}H_{5}OH + O_{2}= CO_{2}+ H_{2}O

This equation is more complex than the previous examples and requires more steps. The most complicated molecule here is C_{2}H_{5}OH, so balancing begins by placing the coefficient 2 before the CO_{2} to balance the carbon atoms.

- C
_{2}H_{5}OH + O_{2}= 2CO_{2}+ H_{2}O

Since C_{2}H_{5}OH contains 6 hydrogen atoms, the hydrogen atoms can be balanced by placing 3 before the H_{2}O:

- C
_{2}H_{5}OH + O_{2}= 2CO_{2}+ 3H_{2}O

Finally the oxygen atoms must be balanced. Since there are 7 oxygen atoms on the right and only 3 on the left, a 3 is placed before O_{2}, to produce the balanced equation:

- C
_{2}H_{5}OH + 3O_{2}= 2CO_{2}+ 3H_{2}O

## Using Linear Systems

In reactions involving many compounds, equations may then be balanced using an algebraic method, based on solving set of linear equations.

1. Assign variables to each coefficient. (Coefficients represent both the basic unit and mole ratios in balanced equations.):

**a K**_{4}Fe(CN)_{6}+ b H_{2}SO_{4}+ c H_{2}O = d K_{2}SO_{4}+ e FeSO_{4}+ f (NH_{4})_{2}SO_{4}+ g CO

2. There must be the same quantities of each atom on each side of the equation. So, for each element, count its atoms and let both sides be equal.

**K: 4a = 2d****Fe: 1a = 1e****C: a = 6g****N: a = 3f****H: 2b+2c = 8f****S: b = d+e+f****O: 4b+c = 4d+4e+4f+g**

3. Solve the system (Direct substitution is usually the best way.)

**d=2a****e=a****g=6a****f=3a****b=6a****c=6a**

which means that all coefficients depend on a parameter a, just choose a=1 (a number that will make all of them small whole numbers), which gives:

**a=1 b=6 c=6 d=2 e=1 f=3 g=6**

4. And the balanced equation at last:

**K**_{4}Fe(CN)_{6}+ 6 H_{2}SO_{4}+ 6 H_{2}O = 2 K_{2}SO_{4}+ FeSO_{4}+ 3 (NH_{4})_{2}SO_{4}+ 6 CO

To speed up the process, one can combine both methods to get a more practical algorithm:

1. Identify elements which occur in one compound in each member. (This is very usual.)

2. Start with the one among those which has a big index (this will help to keep working with integers), and assign a variable, such as a.

**a K**_{4}Fe(CN)_{6}+ H_{2}SO_{4}+ H_{2}O = K_{2}SO_{4}+ FeSO_{4}+ (NH_{4})_{2}SO_{4}+ CO

3. K_{2}SO_{4} has to be 2a (because of K), and also, FeSO_{4} has to be 1a (because of Fe), CO has to be 6a (because of C) and (NH_{4})_{2}SO_{4} has to be 3a (because of N). This removes the first four equations of the system. It is already known that whatever the coefficients are, those proportions must hold:

**a K**_{4}Fe(CN)_{6}+ H_{2}SO_{4}+ H_{2}O = 2a K_{2}SO_{4}+ a FeSO_{4}+ 3a (NH_{4})_{2}SO_{4}+ 6a CO

4. One can continue by writing the equations now (and having simpler problem to solve) or, in this particular case (although not so particular) one could continue by noticing that adding the Sulfurs yields 6a for H_{2}SO_{4} and finally by adding the hydrogens (or the oxygens) one can find the lasting 6a for H_{2}SO_{4}.

5. Again, having a convenient value for a (in this case 1 will do, but if a results in fractional values in the other coefficients, one would like to cancel the denominators) The result is

**K**_{4}Fe(CN)_{6}+ 6 H_{2}SO_{4}+ 6 H_{2}O = 2 K_{2}SO_{4}+ FeSO_{4}+ 3 (NH_{4})_{2}SO_{4}+ 6 CO

## Using a Tool

You can also use our chemical equation balancer to balance the equations for you.