A balanced chemical equation has the same number of atoms of each element on the left (reactant) and right (product) sides.
We balance a chemical equation by adding coefficients in front of molecules.
Example1
Let's do this step by step with an example.
N2 + H2 → NH3
1. Write out the elements present on both sides of the equation.
N2 + H2 →NH3
N | N |
---|---|
H | H |
2. Put the number of atoms of each element on both sides. If there is no subscript next to an element, that means there is 1 atom of that element.
N2 + H2 → NH3
N=2 | N=1 |
H=2 | H=3 |
3. Put the coefficients in front of compounds so that the numbers of each element are equal on both side of the equation. The coefficient in front of a compounds gets distributed to all the elements in that compound. For example, 3H2O means 3*2 = 6 atoms of hydrogen and 3 atoms of oxygen.
N2 + H2 → 2NH3
N=2 | N=2 |
H=2 | H=6 |
Now, the number of nitrogen atoms is equal but not hydrogen atoms.
N2 + 3H2 → 2NH3
N=2 N=2
H=6 H=6
The numbers of atoms of each element are equal on both sides of the equation. Equation is balanced!
Example 2
Na3PO4 + KOH → NaOH + K3PO4
1. Write out the elements present on both sides of the equation. If we see the same polyatomic ion on both sides of the equation, we can put it instead of separating it into elements.
Na3PO4 + KOH → NaOH + K3PO4
Na | Na |
PO4 | PO4 |
K | K |
OH | OH |
2. Put the number of atoms of each element on both sides.
Na3PO4 + KOH→ NaOH + K3PO4
Na =3 | Na =1 |
PO4=1 | PO4=1 |
K=1 | K=3 |
OH=1 | OH=1 |
3. Put the coefficients in front of compounds so that the numbers of each element are equal on both side of the equation.
We first, add 3 in front of NaOH to make the number of Na atoms equal. Then, we put a 3 in front of KOH to make the number of K atoms equal.
Na3PO4 + 3KOH → 3NaOH + K3PO4
Na =3 | Na =3 |
PO4=1 | PO4=1 |
K=3 | K=3 |
OH=3 | OH=3 |
Example 3
1. Write out the elements present on both sides of the equation.
C3H6 + O2 → CO2 + H2O
C | C |
H | H |
O | O |
2. Put the number of atoms of each element on both sides.
C3H6 + O2→ CO2 + H2O
C=3 | C=1 |
H=6 | H=2 |
O=2 | O=3 |
3. Put the coefficients in front of compounds so that the numbers of each element are equal on both side of the equation.
First we balance the carbons.
C3H6 + O2 → 3CO2 + H2O
C=3 | C=3 |
H=6 | H=2 |
O=2 | O=7 |
Next, we balance the hydrogens.
C3H6 + O2 → 3CO2 + 3H2O
C=3 | C=3 |
H=6 | H=6 |
O=2 | O=9 |
Last, we need to balance the oxygens. There is no whole number that we can multiply by to get 9. Therefore, we will use a fraction: 9/2
C3H6 + 9/2O2→ 3CO2 + 3H2O
C=3 | C=3 |
H=6 | H=6 |
O=9 | O=9 |
Finally, we need to get rid of the fraction by multiplying the whole equation by 2.
2(C3H6 + 9/2O2 → 3CO2 + 3H2O)
2C3H6 + 9O2 → 6CO2 + 6H2O
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