Of all the skills to know about in chemistry, balancing chemical equations is perhaps the most important to master. So many parts of chemistry depend on this vital skill, including stoichiometry, reaction analysis, and lab work. This comprehensive guide will show you the steps to balance even the most challenging reactions and will walk you through a series of examples, from simple to complex.

The Key to Balancing Chemical Equations

The ultimate goal for balancing chemical equations is to make both sides of the reaction, the reactants and the products, equal in the number of atoms per element. This stems from the universal law of the conservation of mass, which states that matter can neither be created nor destroyed. So, if we start with ten atoms of oxygen before a reaction, we need to end up with ten atoms of oxygen after a reaction. This means that chemical reactions do not change the actual building blocks of matter; rather, they just change the arrangement of the blocks. An easy way to understand this is to picture a house made of blocks. We can break the house apart and build an airplane, but the color and shape of the actual blocks do not change.

But how do we go about balancing these equations? We know that the number of atoms of each element needs to be the same on both sides of the equation, so it is just a matter of finding the correct coefficients (numbers in front of each molecule) to make that happen. It is best to start with the atom that shows up the least number of times on one side, and balance that first. Then, move on to the atom that shows up the second least number of times, and so on. In the end, make sure to count the number of atoms of each element on each side again, just to be sure.

Example of Balancing a Chemical Equation

Let’s illustrate this with an example by balancing this chemical equation:

P₄O₁₀ + H₂O → H₃PO₄

First, let’s look at the element that appears least often. Notice that oxygen occurs twice on the left-hand side, so that is not a good element to start out with. We could either start with phosphorus or hydrogen, so let’s start with phosphorus. There are four atoms of phosphorus on the left-hand side, but only one on the right-hand side. So, we can put the coefficient of 4 on the molecule that has phosphorous on the right-hand side to balance them out.

P₄O₁₀ + H₂O → 4 H₃PO₄

Now we can check hydrogen. We still want to avoid balancing oxygen, because it occurs in more than one molecule on the left-hand side. It is easiest to start with molecules that only appear once on each side. So, there are two molecules of hydrogen on the left-hand side and twelve on the right-hand side (notice that there are three per molecule of H₃PO₄, and we have four molecules). So, to balance those out, we have to put a six in front of H₂O on the left.

P₄O₁₀ + 6 H₂O → 4 H₃PO₄

At this point, we can check the oxygens to see if they balance. On the left, we have ten atoms of oxygen from P₄O₁₀ and six from H₂O for a total of 16. On the right, we have 16 as well (four per molecule, with four molecules). So, oxygen is already balanced. This gives us the final balanced equation of

P₄O₁₀ + 6 H₂O → 4 H₃PO₄

Balancing Chemical Equations Practice Problems

Try to balance these ten equations on your own, then check the answers below. They range in difficulty level, so don’t get discouraged if some of them seem too hard. Just remember to start with the element that shows up the least, and proceed from there. The best way to approach these problems is slowly and systematically. Looking at everything at once can easily get overwhelming. Good luck!

1. CO₂ + H₂O → C₆H₁₂O₆ + O₂
2. SiCl₄ + H₂O → H₄SiO₄ + HCl
3. Al + HCl → AlCl₃ + H₂
4. Na₂CO₃ + HCl → NaCl + H₂O + CO₂
5. C₇H₆O₂ + O₂ → CO₂ + H₂O
6. Fe₂(SO₄)₃ + KOH → K₂SO₄ + Fe(OH)₃
7. Ca₃(PO₄)₂ + SiO₂ → P₄O₁₀ + CaSiO₃
8. KClO₃ → KClO₄ + KCl
9. Al₂(SO₄)₃ + Ca(OH)₂ → Al(OH)₃ + CaSO₄
10. H₂SO₄ + HI → H₂S + I₂ + H₂O

Complete Solutions:

1. CO₂ + H₂O → C₆H₁₂O₆ + O₂

The first step to balancing chemical equations is to focus on elements that only appear once on each side of the equation. Here, both carbon and hydrogen fit this requirement. So, we will start with carbon. There is only one atom of carbon on the left-hand side, but six on the right-hand side. So, we add a coefficient of six to the carbon-containing molecule on the left.

6CO₂ + H₂O → C₆H₁₂O₆ + O₂

Next, let’s look at hydrogen. There are two hydrogen atoms on the left and twelve on the right. So, we will add a coefficient of six on the hydrogen-containing molecule on the left.

6CO₂ + 6H₂O → C₆H₁₂O₆ + O₂

Now, it is time to check the oxygen. There are a total of 18 oxygen molecules on the left (6×2 + 6×1). On the right, there are eight oxygen molecules. Now, we have two options to even out the right-hand side: We can either multiply C₆H₁₂O₆ or O₂ by a coefficient. However, if we change C₆H₁₂O₆, the coefficients for everything else on the left-hand side will also have to change, because we will be changing the number of carbon and hydrogen atoms. To prevent this, it usually helps to only change the molecule containing the fewest elements; in this case, the O₂. So, we can add a coefficient of six to the O₂ on the right. Our final answer will be:

6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂

2. SiCl₄ + H₂O → H₄SiO₄ + HCl

The only element that occurs more than once on the same side of the equation here is hydrogen, so we can start with any other element. Let’s start by looking at silicon. Notice that there is only one atom of silicon on either side, so we do not need to add any coefficients yet. Next, let’s look at chlorine. There are four chlorine atoms on the left side and only one on the right. So, we will add a coefficient of four on the right.

SiCl₄ + H₂O → H₄SiO₄ + 4HCl

Next, let’s look at oxygen. Remember that we first want to analyze all the elements that only occur once on one side of the equation. There is only one oxygen atom on the left, but four on the right. So, we will add a coefficient of four on the left-hand side of the equation.

SiCl₄ + 4H₂O → H₄SiO₄ + 4HCl

We are almost done! Now, we just have to check the number of hydrogen atoms on each side. The left has eight and the right also has eight, so we are done. Our final answer is

SiCl₄ + 4H₂O → H₄SiO₄ + 4HCl

As always, make sure to double-check that the number of atoms of each element balances on each side before continuing.

3. Al + HCl → AlCl₃ + H₂

This problem is a bit tricky, so be careful. Whenever a single atom is alone on either side of the equation, it is easiest to start with that element. So, we will start by counting the aluminum atoms on both sides. There is one on the left and one on the right, so we do not need to add any coefficients yet. Next, let’s look at hydrogen. There is also one on the left, but two on the right. So, we will add a coefficient of two on the left.

Al + 2HCl → AlCl₃ + H₂

Next, we will look at chlorine. There are now two on the left, but three on the right. Now, this is not as straightforward as just adding a coefficient to one side. We need the number of chlorine atoms to be equal on both sides, so we need to get two and three to be equal. We can accomplish this by finding the lowest common multiple. In this case, we can multiply two by three and three by two to get the lowest common multiple of six. So, we will multiply 2HCl by three and AlCl₃ by two:

Al + 6HCl → 2AlCl₃ + H₂

We have looked at all the elements, so it is easy to say that we are done. However, always make sure to double-check. In this case, because we added a coefficient to the aluminum-containing molecule on the right-hand side, aluminum is no longer balanced. There is one on the left but two on the right. So, we will add one more coefficient.

2Al + 6HCl → 2AlCl₃ + H₂

We are not quite done yet. Looking over the equation one final time, we see that hydrogen has also been unbalanced. There are six on the left but two on the right. So, with one final adjustment, we get our final answer:

2Al + 6HCl → 2AlCl₃ + 3H₂

4. Na₂CO₃ + HCl → NaCl + H₂O + CO₂

Hopefully, by this point, balancing equations is becoming easier and you are getting the hang of it. Looking at sodium, we see that it occurs twice on the left, but once on the right. So, we can add our first coefficient to the NaCl on the right.

Na₂CO₃ + HCl → 2NaCl + H₂O + CO₂

Next, let’s look at carbon. There is one on the left and one on the right, so there are no coefficients to add. Since oxygen occurs in more than one place on the left, we will save it for last. Instead, look at hydrogen. There is one on the left and two on the right, so we will add a coefficient to the left.

Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂

Then, looking at chlorine, we see that it is already balanced with two on each side. Now we can go back to look at oxygen. There are three on the left and three on the right, so our final answer is

Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂

5. C₇H₆O₂ + O₂ → CO₂ + H₂O

We can start balancing this equation by looking at either carbon or hydrogen. Looking at carbon, we see that there are seven atoms on the left and only one on the right. So, we can add a coefficient of seven on the right.

C₇H₆O₂ + O₂ → 7CO₂ + H₂O

Then, for hydrogen, there are six atoms on the left and two on the right. So, we will add a coefficient of three on the right.

C₇H₆O₂ + O₂→ 7CO₂ + 3H₂O

Now, for oxygen, things will get a little tricky. Oxygen occurs in every molecule in the equation, so we have to be very careful when balancing it. There are four atoms of oxygen on the left and 17 on the right. There is no obvious way to balance these numbers, so we must use a little trick: fractions. Now, when balancing chemical equations, we cannot include fractions as it is not proper form, but it sometimes helps to use them to solve the problem. Also, try to avoid over-manipulating organic molecules. You can easily identify organic molecules, otherwise known as CHO molecules, because they are made up of only carbon, hydrogen, and oxygen. We don’t like to work with these molecules, because they are rather complex. Also, larger molecules tend to be more stable than smaller molecules, and less likely to react in large quantities.

So, to balance out the four and seventeen, we can multiply the O₂ on the left by 7.5. That will give us

C₇H₆O₂ + 7.5O₂ → 7CO₂ + 3H₂O

Remember, fractions (and decimals) are not allowed in formal balanced equations, so multiply everything by two to get integer values. Our final answer is now

2C₇H₆O₂ + 15O₂ → 14CO₂ + 6H₂O

6. Fe₂(SO₄)₃ + KOH → K₂SO₄ + Fe(OH)_3­-

We can start by balancing the iron on both sides. The left has two while the right only has one. So, we will add a coefficient of two to the right.

Fe₂(SO₄)₃ + KOH → K₂SO₄ + 2Fe(OH)_3­-

Then, we can look at sulfur. There are three on the left, but only one on the right. So, we will add a coefficient of three to the right-hand side.

Fe₂(SO₄)₃ + KOH → 3K₂SO₄ + 2Fe(OH)_3­-

We are almost done. All that is left is to balance the potassium. There is one atom on the left and six on the right, so we can balance these by adding a coefficient of six. Our final answer, then, is

Fe₂(SO₄)₃ + 6KOH → 3K₂SO₄ + 2Fe(OH)_3­-

7. Ca₃(PO₄)₂ + SiO₂ → P₄O₁₀ + CaSiO₃

Looking at calcium, we see that there are three on the left and one on the right, so we can add a coefficient of three on the right to balance them out.

Ca₃(PO₄)₂ + SiO₂ → P₄O₁₀ + 3CaSiO₃

Then, for phosphorus, we see that there are two on the left and four on the right. To balance these, add a coefficient of two on the left.

2Ca₃(PO₄)₂ + SiO₂ → P₄O₁₀ + 3CaSiO₃

Notice that by doing so, we changed the number of calcium atoms on the left. Every time you add a coefficient, double check to see if the step affects any elements you have already balanced. In this case, the number of calcium atoms on the left has increased to six while it is still three on the right, so we can change the coefficient on the right to reflect this change.

2Ca₃(PO₄)₂ + SiO₂ → P₄O₁₀ + 6CaSiO₃

Since oxygen occurs in every molecule in the equation, we will skip it for now. Focusing on silicon, we see that there is one on the left, but six on the right, so we can add a coefficient to the left.

2Ca₃(PO₄)₂ + 6SiO₂ → P₄O₁₀ + 6CaSiO₃

Now, we will check the number of oxygen atoms on each side. The left has 28 atoms and the right also has 28. So, after checking that all the other atoms are the same on both sides as well, we get a final answer of

2Ca₃(PO₄)₂ + 6SiO₂ → P₄O₁₀ + 6CaSiO₃

8. KClO₃ → KClO₄ + KCl

This problem is particularly tricky because every atom, except oxygen, occurs in every molecule in the equation. So, since oxygen appears the least number of times, we will start there. There are three on the left and four on the right. To balance these, we find the lowest common multiple; in this case, 12. By adding a coefficient of four on the left and three on the right, we can balance the oxygens.

4KClO₃ → 3KClO₄ + KCl

Now, we can check potassium and chlorine. There are four potassium molecules on the left and four on the right, so they are balanced. Chlorine is also balanced, with four on each side, so we are finished, with a final answer of

4KClO₃ → 3KClO₄ + KCl

9. Al₂(SO₄)₃ + Ca(OH)₂ → Al(OH)₃ + CaSO₄

We can start here by balancing the aluminum atoms on both sides. The left has two molecules while the right only has one, so we will add a coefficient of two on the right.

Al₂(SO₄)₃ + Ca(OH)₂ → 2Al(OH)₃ + CaSO₄

Now, we can check sulfur. There are three on the left and only one on the right, so adding a coefficient of three will balance these.

Al₂(SO₄)₃ + Ca(OH)₂ → 2Al(OH)₃ + 3CaSO₄

Moving right along to calcium, there is only one on the left but three on the right, so we should add a coefficient of three.

Al₂(SO₄)₃ + 3Ca(OH)₂ → 2Al(OH)₃ + 3CaSO₄

Double-checking all the atoms, we see that all the elements are balanced, so our final equation is

Al₂(SO₄)₃ + 3Ca(OH)₂ → 2Al(OH)₃ + 3CaSO₄

10. H₂SO₄ + HI → H₂S + I₂ + H₂O

Since hydrogen occurs more than once on the left, we will temporarily skip it and move to sulfur. There is one atom on the left and one on the right, so there is nothing to balance yet. Looking at oxygen, there are four on the left and one on the right, so we can add a coefficient of four to balance them.

H₂SO₄ + HI → H₂S + I₂ + 4H₂O

There is only one iodine on the left and two on the right, so a simple coefficient change can balance those.

H₂SO₄ + 2HI → H₂S + I₂ + 4H₂O

Now, we can look at the most challenging element: hydrogen. On the left, there are four and on the right, there are ten. So, we know we have to change the coefficient of either H₂SO₄ or HI. We want to change something that will require the least amount of tweaking afterwards, so we will change the coefficient of HI. To get the left-hand side to have ten atoms of hydrogen, we need HI to have eight atoms of hydrogen, since H₂SO₄ already has two. So, we will change the coefficient from 2 to 8.

H₂SO₄ + 8HI → H₂S + I₂ + 4H₂O

However, this also changes the balance for iodine. There are now eight on the left, but only two on the right. To fix this, we will add a coefficient of 4 on the right. After checking that everything else balances out as well, we get a final answer of

H₂SO₄ + 8HI → H₂S + 4I₂ + 4H₂O

As with most skills, practice makes perfect when balancing chemical equations. Keep working hard and try to do as many problems as you can to help you hone your balancing skills.