How to Balance ANY Chemical Equation

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How to Balance ANY Chemical Equation

Step 1:  List out the elements present on BOTH the PRODUCT and REACTANT sides of the equation.  Then, count how many of each atom are present initially.

EXAMPLE:  Cu(s) + HNO3(aq)  LaTeX: \longrightarrow⟶ Cu(NO3)2(aq) + NO2(g) + H2O(l)

Reactants

 

Products

Cu

1

 

Cu

1

H

1

 

H

2

N

1

 

N

3

O

3

 

O

9

 

 

Step 2:  Determine which element to balance first.  You want to balance the easiest elements first, then the harder elements later.  The easiest elements to balance will be those which appear in the fewest substances in the equation.  Therefore, to determine balancing order, count how many substances each element appears in.

EXAMPLE:   Cu(s) + HNO3(aq)  LaTeX: \longrightarrow⟶ Cu(NO3)2(aq) + NO2(g) + H2O(l)

 

The element Cu appears in 2 substances [Cu(s) and Cu(NO3)2(aq)]

The element H appears in 2 substances [HNO3(aq) and H2O(l)]

The element N appears in 3 substances [HNO3(aq) and Cu(NO3)2(aq) and NO2(g)]

The element O appears in 4 substances [H2O(l) and HNO3(aq) and Cu(NO3)2(aq) and NO2(g)]

 

Therefore, I will balance Cu and H first, then N, and lastly O.

REMEMBER!  If at any point one of the easier elements to balance becomes unbalanced, rebalance that element before continuing to the harder elements.

 

 

Step 3:  When balancing an element, you may only change coefficients (the numbers in front of the substance), NEVER CHANGE SUBSCRIPTS.  Change the coefficient first, then record how that changes the numbers of the elements in your list.  Be sure to apply the coefficient to ALL of the elements within the compound.

EXAMPLE:   Cu(s) + 2HNO3(aq)  LaTeX: \longrightarrow⟶ Cu(NO3)2(aq) + NO2(g) + H2O(l)

Reactants

 

Products

Cu

1

 

Cu

1

H

2

 

H

2

N

2

 

N

3

O

6

 

O

9

 

 

Step 4:  In many equations, you will come across a situation where you are trying to balance an odd number of atoms on one side with an even number of atoms on the other.  In the example equation, this is the case with N.  When this happens, DOUBLE the coefficient of the compound that is causing the number of atoms to be ODD.  Then, rebalance.

EXAMPLE:   Cu(s) + 2HNO3(aq)  LaTeX: \longrightarrow⟶ Cu(NO3)2(aq) + 2NO2(g) + H2O(l)

Reactants

 

Products

Cu

1

 

Cu

1

H

2

 

H

2

N

2

 

N

4

O

6

 

O

11

NO2 was causing the number of N’s to be odd on the product side, therefore I doubled the coefficient of NO2 from 1 to 2.

 

 

Step 5: Continue rebalancing until all elements have the same number of atoms on both the product and reactant sides.

EXAMPLE:   Cu(s) + 4HNO3(aq)  LaTeX: \longrightarrow⟶ Cu(NO3)2(aq) + 2NO2(g) + H2O(l)

Reactants

 

Products

Cu

1

 

Cu

1

H

4

 

H

2

N

4

 

N

4

O

12

 

O

11

 I rebalanced N by changing the coefficient of HNO3 to 4.  However, now I have unbalanced H again, therefore I must rebalance H before trying to balance O.

 

 

Step 5 continued... :

EXAMPLE:   Cu(s) + 4HNO3(aq)  LaTeX: \longrightarrow⟶ Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)

Reactants

 

Products

Cu

1

 

Cu

1

H

4

 

H

4

N

4

 

N

4

O

12

 

O

12

Notice that by balancing the easiest elements first, the hardest element to balance ended up balancing itself!     :o)

 

And now we have a balanced equation!

EXAMPLE:  Cu(s) + 4HNO3(aq)  LaTeX: \longrightarrow⟶ Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)