Being able to balance chemical equations is a vital skill for chemistry. Here's a look at the steps involved in balancing equations, plus a worked example of how to balance an equation.

### Steps of Balancing a Chemical Equation

- Identify each element found in the equation. The number of atoms of each type of atom must be the same on each side of the equation once it has been balanced.

- What is the net charge on each side of the equation? The net charge must be the same on each side of the equation once it has been balanced.

- If possible, start with an element found in one compound on each side of the equation. Change the coefficients (the numbers in front of the compound or molecule) so that the number of atoms of the element is the same on each side of the equation. Remember, to balance an equation, you change the coefficients, not the subscripts in the formulas.

- Once you have balanced one element, do the same thing with another element. Proceed until all elements have been balanced. It's easiest to leave elements found in pure form for last.

- Check your work to make certain the charge on both sides of the equation is also balanced.

### Example of Balancing a Chemical Equation

? CH_{4} + ? O_{2} → ? CO_{2} + ? H_{2}O

Identify the elements in the equation: C, H, O

Identify the net charge: no net charge, which makes this one easy!

- H is found in CH
_{4}and H_{2}O, so it's a good starting element. - You have 4 H in CH
_{4}yet only 2 H in H_{2}O, so you need to double the coefficient of H_{2}O to balance H.1 CH

_{4}+ ? O_{2}→ ? CO_{2}+ 2 H_{2}O

- Looking at carbon, you can see that CH
_{4}and CO_{2}must have the same coefficient.1 CH

_{4}+ ? O_{2}→ 1 CO_{2}+ 2 H_{2}O - Finally, determine the O coefficient. You can see you need to double the O
_{2}coefficient in order to get 4 O seen on the product side of the reaction.1 CH

_{4}+ 2 O_{2}→ 1 CO_{2}+ 2 H_{2}O - Check your work. It's standard to drop a coefficient of 1, so the final balanced equation would be written:
CH

_{4}+ 2 O_{2}→ CO_{2}+ 2 H_{2}O

Take a quiz to see if you understand how to balance simple chemical equations.

### How to Balance a Chemical Equation for a Redox Reaction

Once you understand how to balance an equation in terms of mass, you're ready to learn how to balance an equation for both mass and charge. Reduction/oxidation or redox reactions and acid-base reactions often involve charged species. Balancing for charge means you have the same net charge on both the reactant and product side of the equation. This isn't always zero!

Here's an example of how to balance the reaction between potassium permanganate and iodide ion in aqueous sulfuric acid to form potassium iodide and manganese(II) sulfate. This is a typical acid reaction.

- First, write the unbalanced chemical equation:

KMnO_{4 }+ KI + H2SO_{4 }→ I_{2 }+ MnSO_{4} - Write down the oxidation numbers for each type of atom on both sides of the equation:

Left hand side: K = +1; Mn = +7; O = -2; I = 0; H = +1; S = +6

Right hand side: I = 0; Mn = +2, S = +6; O = -2 - Find the atoms that experience a change in oxidation number:

Mn: +7 → +2; I: +1 → 0 - Write a skeleton ionic equation that only covers the atoms that change oxidation number:

MnO_{4}^{-}→ Mn^{2+}

I^{-}→ I_{2} - Balance all of the atoms besides the oxygen (O) and hydrogen (H) in the half-reactions:

MnO4^{-}→ Mn^{2+}

2I^{-}→ I_{2}

- Now add O and H
_{2}O as needed to balance oxygen:

MnO_{4}^{-}→ Mn^{2+}+ 4H_{2}O

2I^{-}→ I_{2} - Balance the hydrogen by adding H
^{+}as needed:

MnO_{4}^{-}+ 8H^{+}→ Mn^{2+}+ 4H_{2}O

2I^{-}→ I_{2} - Now, balance charge by adding electrons as needed. In this example, the first half-reaction has a charge of 7+ on the left and 2+ on the right. Add 5 electrons to the left to balance the charge. The second half-reaction has 2- on the left and 0 on the right. Add 2 electrons to the right.

MnO_{4}^{-}+ 8H^{+}+ 5e^{-}→ Mn^{2+}+ 4H_{2}O

2I^{-}→ I_{2}+ 2e^{-} - Multiply the two half-reactions by the number that yields the lowest common number of electrons in each half-reaction. For this example, the lowest multiple of 2 and 5 is 10, so multiply the first equation by 2 and the second equation by 5:

2 x [MnO_{4}^{-}+ 8H^{+}+ 5e^{-}→ Mn^{2+}+ 4H_{2}O]

5 x [2I^{-}→ I_{2}+ 2e^{-}] - Add together the two half-reactions and cancel out species that appear on each side of the equation:

2MnO_{4}^{-}+ 10I^{-}+ 16H^{+}→ 2Mn^{2+}+ 5I_{2}+ 8H_{2}O

Now, it's a good idea to check your work by making sure the atoms and charge are balanced:

Left hand side: 2 Mn; 8 O; 10 I; 16 H

Right hand side: 2 Mn; 10 I; 16 H; 8 O

Left hand side: −2 – 10 +16 = +4

Right hand side: +4