Trial balancing steps, examples and exercises

The trial and error It consists of a trial and error method that seeks to ensure that the conservation of matter is fulfilled in a chemical equation for a given reaction; that is, to equalize the numbers of atoms of the reactants and products. Thus, atoms will not vanish or be created out of thin air..

Depending on the rush, this is usually an entertaining operation, reinforcing understandings about stoichiometric coefficients and subscripts. Although it may not seem like it, trial and error involves mastering many concepts, applied almost unconsciously for those who dabble in chemistry..

Thus, balancing is similar to the effort that would be made to level a seesaw (or seesaw), such that neither end falls to one side while the other rises. A scale also illustrates this perfectly.

As you have experience, this balancing can even be done mentally, as long as the chemical equation is not too complicated. A bad swing completely ruins the interpretation of a reaction, so it is essential to do it calmly to avoid eating mistakes.

Article index

• 1 Steps
  • 1.1 Write the unbalanced equation
  • 1.2 Review stoichiometric coefficients and subscripts
  • 1.3 First balance the atoms in smaller proportion
  • 1.4 Balance by oxygen atoms
  • 1.5 Lastly balancing the hydrogen atoms
• 2 Examples
• 3 Exercises
  • 3.1 Exercise 1
  • 3.2 Exercises 2
  • 3.3 Exercise 3
  • 3.4 Exercise 4
• 4 References

Steps

Write the unbalanced equation

Regardless of what balancing is required for, you should always start with the unbalanced equation at hand. Likewise, it is important to be clear about its elements. Assume the following chemical equation:

A + B → 3C + D

Where species A, B, C and D are molecular. This equation cannot be balanced because it tells us nothing about its atoms. The atoms are balanced, not the molecules.

Both A, B and D have a stoichiometric coefficient of 1, while C of 3. This means that 1 molecule or mole of A reacts with one molecule or mole of B, to produce 3 molecules or moles of C, and one molecule or mole of D. When we show the atoms, we introduce the stoichiometric subscripts.

Review stoichiometric coefficients and subscripts

Now suppose the following equation:

CH₄ + OR_two → CO_two + H_twoOR

Stoichiometric subscripts tell us how many atoms of each element make up a molecule, and they are recognized because they are the smallest numbers on the right side of an atom. For example, CH₄ it has one carbon atom (although the 1 is not placed) and four hydrogen atoms.

Balance the atoms in the smallest proportion first

According to the unbalanced equation above, carbon is the minor atom: it is part of a single reactant (CH₄) and a single product (CO_two). If observed, there is a C atom on both the reactants and products side.

Balance by oxygen atoms

CH₄ + OR_two → CO_two + H_twoOR

2 O 3 O

We cannot change the subscripts, but only the stoichiometric coefficients in order to balance an equation. There are more oxygens on the right side, so we try to add a coefficient to the O_two:

CH₄ + 2O_two → CO_two + H_twoOR

4 or 3

We don't want to affect the CO coefficient_two because it would unbalance the atoms of C. We then change the coefficient of H_twoOR:

CH₄ + 2O_two → CO_two + 2H_twoOR

4 OR 4O

Lastly balance the hydrogen atoms

Once we have balanced the oxygen atoms, we finally balance the hydrogen atoms. Several times these remain by themselves balanced in the end.

CH₄ + 2O_two → CO_two + 2H_twoOR

4H 4H

And so the equation has been balanced by trial and error. The order of these steps is not always fulfilled.

Examples

Balanced equations are shown below to verify that the number of its atoms is equal on both sides of the arrow:

SW_two + 2H_two → S + 2H_twoOR

P₄ + 6F_two → 4PF₃

2HCl → H_two + Cl_two

C + O_two → CO_two

Training

Some proposed exercises will be solved below. In some of them it will be seen that sometimes it is convenient to break the order of the steps and balance the minority atom last..

Exercise 1

Balance by trial and error the following chemical equation:

SW₃ → SO_two + OR_two

1S 1S

3 or 4

It is important to emphasize that the coefficients multiply the subscripts to give us the total number of atoms for an element. For example, 6N_two gives us a total of 12 N atoms.

The sulfur at the beginning is already balanced, so we continue with the oxygen:

3 O 4 O

We are forced to change the coefficient to SO₃ to balance the left side oxygens:

2SO₃ → SO_two + OR_two

6 O 4 O

2S S

Now we are interested in balancing the sulfur atoms first before the oxygen atoms:

2SO₃ → 2SO_two + OR_two

2S 2S

6 OR 6O

Note that the oxygen atoms were left balanced by themselves in the end.

Exercises 2

Balance by trial and error the following chemical equation:

CH₄ + H_twoO → CO + H_two

The carbons and oxygens are already balanced, not in the same way the hydrogens:

6H 2H

All we have to do is modify the coefficient to H_two to have more hydrogens on the right:

CH₄ + H_twoO → CO + 3H_two

6H 6H

And the equation is totally balanced.

Exercise 3

Balance by trial and error the following chemical equation:

C_twoH₄ + OR_two → CO_two + H_twoOR

We start balancing the carbon again:

C_twoH₄ + OR_two → 2CO_two + H_twoOR

2C 2C

2O 5O

4H 2H

Note that this time it is easier to balance the hydrogens first than the oxygens:

C_twoH₄ + OR_two → 2CO_two + 2H_twoOR

4H 4H

2O 6O

Now yes, we modify the coefficient of O_two to balance oxygens:

C_twoH₄ + 3O_two → 2CO_two + 2H_twoOR

6O 6O

And the equation is already balanced.

Exercise 4

Finally, a challenging equation will be balanced by trial and error:

N_two + H_twoO → NH₃ + NOT

Nitrogens and oxygens are already balanced, but hydrogens are not:

2H 3H

Let's try changing the coefficient of H_twoO and NH₃:

N_two + 3H_twoO → 2NH₃ + NOT

6H 6H

3O O

2N 3N

By trial and error we vary the coefficient of NO:

N_two + 3H_twoO → 2NH₃ + 3) NO

6H 6H

3O 3O

2N 5N

And now the nitrogens are unbalanced. Here it is convenient to make an abrupt change: quintuple the coefficient of N_two:

5N_two + 3H_twoO → 2NH₃ + 3) NO

10 N 5N

6H 6H

3O 3O

Thus, it remains for us to play with the NH coefficients₃ and NO in such a way that they add 10 nitrogens and balance the oxygen and hydrogen atoms at the same time. Let's try this score:

5N_two + 3H_twoO → 5NH₃ + 5NO

10 N 10 N

6 H 15H

3O 5O

However, the hydrogens look very unbalanced. Therefore, let's vary the coefficients again:

5N_two + 3H_twoO → 4NH₃ + 6NO

10 N 10N

6H 12H

3O 6O

Note that now the left side has twice the oxygen and hydrogen. At this point, it is enough to double the coefficient of H_twoOR:

5N_two + 6H_twoO → 4NH₃ + 6NO

10N 10N

12H 12H

6O 6O

And the equation is finally balanced.

References

1. Whitten, Davis, Peck & Stanley. (2008). Chemistry. (8th ed.). CENGAGE Learning.
2. Organic Chemistry. (s.f.). Balancing chemical equations by trial and error | Solved Exercises. Recovered from: quimica-organica.com
3. Nissa Garcia. (2020). Balanced Chemical Equation: Definition & Examples. Study. Recovered from: study.com
4. Helmenstine, Anne Marie, Ph.D. (October 21, 2019). How to Balance Chemical Equations. Recovered from: thoughtco.com
5. Study Guide. (May 11, 2019). Trial balancing of chemical reactions. Solved Exercises. Recovered from: quimicaencasa.com
6. University of Colorado Boulder. (2019). Balancing Chemical Equations. Recovered from: phet.colorado.edu