Balancing Chemical Equations

Written by tutor Nilsa S.

Why do we need to balance a chemical equation?

The law of conservation of energy: Energy cannot be created nor destroyed but only transformed from one form to another. This means that if you begin your reaction with hydrogen, the hydrogen will not magically transform to Oxygen. Elements do not mysteriously appear or disappear during a reaction; however, they rearrange to form new chemical substances. For instance, if you make a fruit salad and used an apple and a banana, although the fruit is chopped up, when you eat the fruit salad, you will still taste and see the apples and bananas. However, if you placed the cut apples and bananas in the blender, it would still be the same fruit ingredients (bananas and apples) but in a different state (liquid) and harder to visually distinguish from the physical change.

An accurate chemical equation has the correct ratio of reactants and products called Coefficients. These are the numbers in front of the reactant or product. For instance, in the fruit salad analogy if two apples were used with one banana one fruit salad would form. Let A represent the apples B for banana and A2B for fruit salad the balanced chemical equation would be: 2A + B -> A2B.

It is also important to denote the state of the reactants and products in an accurate balanced chemical equation. The three states are solid (s), liquid (l), and gas (g). Because water is abundant we only let water have the state (l). Everything that is dissolved in water has (aq) which means aqueous like aqua.

When we successfully balance a chemical equation you may calculate the amounts of each reactant and product using Stoichiometry.

Let’s practice balancing chemical equations
Example 1: Decomposition of water.

H2O(g) -> H2(g) + O2(g)

Step 1: Draw a line down from the arrow separating the reactants (left side) and products (right side) of the equation.

H2O(g) -> H2(g) + O2(g)

Balancing a chemical equation

Step 2: To the left of the arrow make a column of all the elements present

Balancing a chemical equation

Step 3: Now calculate the number of atoms of each element on the reactants and products side

      H2O(g) -> H2(g) + O2(g)

Balancing a chemical equation

Step 4: Evaluate the unequal amounts of elements to follow with the law of the conservation of energy.

We can see there are unequal amounts of oxygen in the water on the reactants side of the equation. We need 2 oxygen so we must put a 2 in front of the water molecule. This number represents the coefficient. The coefficient is the amount of water molecules needed for this reaction. Since the coefficient is two we have two water molecules.

      2H2O(g) -> H2(g) + O2(g)

Balancing a chemical equation

Step 5: Recalculating the numbers on the products and reactants side of the equation.

      2H2O(g) -> H2(g) + O2(g)

Balancing a chemical equation

Because we put the coefficient 2 in front of the water we now have unequal amounts of hydrogen in the products side, there is a 4 to 2 ratio so we must put another 2 coefficient in front of the hydrogen gas.

There are two factors involved in balancing: the first is the coefficient, which is how many molecules or compounds in each product or reactant. The second is the ratio of atoms in a compound or molecule that comprises the molecule or compound. For instance H2 has 2 hydrogen atoms to make one hydrogen molecule. The coefficient must be multiplied by the ratio of atoms in a molecule or compound to correctly add up all the elements present in a molecule or compound.

Step 6: Balancing the hydrogen by putting a 2 in front of the hydrogen gas molecule.

      2H2O(g) -> 2H2(g) + O2(g)

Balancing a chemical equation

All numbers are equal on both the reactants and products side so the chemical equation is now balanced.

Example 2: The combination/synthesis reaction of ammonium (NH4+) and amide ion (NH2-) produce ammonia. Follow steps 1-3.

NH4+(aq) + NH2-(aq) -> NH3(aq)            

Balancing a chemical equation

Inspecting the amount we can see if the atoms of the product side were doubled they would be equal to the reactant side. So a coefficient of 2 would be the right choice (Always use prime numbers and look for LCD).

Also notice the charges of each reactant. NH4+ is plus one over all and NH2- is minus one over all. Notice charges are on the top right corner of all atoms or compounds or molecules while the amount of atoms that make up a molecule or compound are always subscripts (at the bottom). The charges when added together equal zero so the reactants charges are neutral just like the charge of the product. The product NH3 is also neutral because it has no charge. One way to verify if you have a balanced chemical equation is to also inspect the charges and make sure they too are equal on the reactant and product side (this is helpful in oxidation reduction reactions).

Step 4: putting the 2and recalculating the amounts of products.

NH4+(aq) + NH2-(aq) -> 2NH3(aq)

Balancing a chemical equation


Example 3: Combustion of propane (C3H8).

C3H8(g) + O2 -> CO2(g) + H2O(l)

Balancing a chemical equation

Upon inspection we can see this is an ugly one!
Strategy: always balance the binary compounds (with two different atoms) before you balance the molecules that have the same type of atom in it.

Step 4: Target one atom, I chose carbon. Since we need 3 on the product side to match the reactant side let us use 3as our coefficient then recalculate the amounts.

C3H8(g) + O2 -> 3CO2(g) + H2O(l)

Balancing a chemical equation

Step 5: Recalculate the amounts of atoms and balance Hydrogen because it is in two binary compounds. We want to leave O2 last because it is alone and we can touch it up lastly with out affecting any other atoms. Since there are 8 hydrogen atoms on the reactant side and there are 2 hydrogen atoms in a water molecule we need the coefficient 4in front of hydrogen to make the number of hydrogen 8. This will affect the number of oxygen so we will have to recalculate the numbers.

C3H8(g) + O2 -> 3CO2(g) + 4H2O(l)

Balancing a chemical equation

Step 6: We left the O2 last because it is by itself. There are 2 oxygen atoms on the reactant side and 10 on the product side. The LCD would be 5and that is the correct coefficient to use for the reactant side. The final step will be to recalculate all of the atoms and they will all be equal.

C3H8(g) + 5O2 -> 3CO2(g) + 4H2O(l)

Balancing a chemical equation

*Notice: The coefficients represent the amount of molecules of each reactant and product needed for a chemical reaction.
*The coefficient affects the amounts of atoms, which we have to calculate.
* If you have found your own way of finding the correct coefficient by all means do what comes natural. The point of this tutorial is to give suggestions and help beginners at balancing.
*It is helpful to make a table dividing the left and right side of the equation to tally the amount of atoms on the reactant side and product side.
*In order to balance you have to find the amount of each atom and always balance the compounds with more that one type of atom in the equation before you try the ones by themselves.

Remember once you practice you may find your own way through discovery. There is always more than one approach. With lots of practice you will eventually not need so many steps.

Balancing Equations Practice Quiz

Choose the answer with the correct coefficients. If the coefficient is 1, there will be a 1 present in the answer choice.

_I2(g) + _F2(g) -> _IF5(g)

A. 5, 2, 1
B. 1, 5, 2
C. 1, 2, 5
D. 1, 1, 2
The correct answer here would be B.

Since the product has a binary compound we will start with the product side: there are 5 Fluorine (F) atoms in IF5 and 2 on the reactants (F2). Their LCD is 10 so cross multiply 2 and 5 so that the amount of F is 10. We can leave the I2 alone in the reactant side because conveniently we have 2 atoms of each product and reactant.

I2(g) + 5 F2(g) -> 2 IF5(g)

_I2(g) + _XeF2(g) -> _IF3(g) + _Xe(g)

A. 1, 1, 3, 2
B. 1, 2, 3, 2
C. 1, 3, 2, 3
D. 1, 3, 3, 2
The correct answer here would be C.

For this question reserve I2 on the reactant side for the very last because it is solo and easy to fix at the end. For the same reason Xe is as well. Lets just investigate F: There are 2 on the reactant side and 3 on the product side. The LCD between them is 6 so cross multiply with the 2 and 3 and the F will be equal in amount. After wards we can see we will have 3 Xe which we can easily place a 3 in front of Xe in the product side because it is alone and we can leave the I2 alone on the reactant side because we conveniently have enough.

I2(g) + 3XeF2(g) -> 2IF3(g) + 3Xe(g)

_NaOH(aq) + _H3PO4(aq) -> _Na3PO4(aq) + _H2O(l)

A. 1, 1, 3, 3
B. 3, 1, 1, 3
C. 1, 3, 3, 1
D. 1, 3, 1, 3
The correct answer here would be C.

Lastly, this is a neutralization reaction we have a phosphate polyatomic anion which if you are experienced with balancing I recommend you treat as an entity. For instance one phosphate ion on the reactant and one phosphate on the product equals balanced. It is not necessary to tally the polyatomic ions up because they remain in their polyatomic (grouped) form for the most part. With that said treat them as a unit. For Na there are 3 Na in the product side so I will put a 3 coefficient on NaOH. Now using my trick of treating the PO4 as a unit I will only look at the number of oxygen in OH. There are 3 oxygen atoms in OH and 1 in water. I will then put a coefficient of 3 in front of the water. The oxygen atoms are now balanced but the hydrogen atoms are not. There are 3 hydrogen atoms in 3 NaOH compounds and 3 in the H3PO4 that totals to 6. Conveniently there 6 hydrogen in the 3 water compounds so we are finished.

3NaOH(aq) + H3PO4(aq) -> Na3PO4(aq) + 3H2O(l)

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