Example of Balancing Equations by Trial

The balancing of chemical equations consists of establishing the amount of substances that take part in a chemical reaction so that they correspond to the amount of substances produced, that is, the elements that react in the first member of the equation are the same that remain after the reaction in the second member of the equation. equation.

One of the methods for balancing an equation is the trial and error method. In this method we will try to balance the number of atoms in the chemical equation, modifying the values ​​of the substances present on one or both sides, so that there is equality between the number of atoms of the reactants and the substances produced. It is a method of trial and error.

When we study a complex chemical reaction, there is doubt as to whether the amount of reacting substances and the substances produced are the same on both sides of the equation. Applying trial and error balancing, we will follow the following steps:

Example of trial and error balancing of sodium sulfate with hydrochloric acid:

1. We take into account the radicals of the reacting substances, as well as those that are produced. Let's see the following neutralization reaction of sodium sulfate with hydrochloric acid:

Na₂SW₃ + HCl -> NaCl + H₂O + SO₂

As we can see, we have on the left side of the equation the reactants: sodium sulfate (Na₂SW₃) and hydrochloric acid (HCl). On the right side, we have the reaction products: Sodium chloride or common salt (NaCl), Water (H₂O) and sulfur oxide (SO₂).

We can see in this equation the substances that react and those that are produced, with their respective formulas. However, to know if this equation is balanced, we must count the number of atoms on either side; if the total is the same on both sides then we consider the equation to be balanced. Thus we have:

2 + 1 + 3 + 1+ 1 -- > 1 + 1 + 2 + 1 + 1 + 2

Na₂SW₃ + HCl -> NaCl + H₂O + SO₂

As we can see, the number of atoms in the first member of the equation is less than the second, so the equation is unbalanced.

2. We will start by identifying the number of atoms of each element on both sides of the equation:

Left side: Na = 2; S = 1; O = 3; H = 1; Cl = 1

Right side: Na = 1; S = 1; O = 3; H = 2; Cl = 1

So we have that on the right side of our equation we are missing one sodium atom, while we have one hydrogen atom left over.

3. To balance an equation by trial and error, we have to follow the following rules:

to. We will not add elements that do not belong to the equation.

b. We will not modify the radicals of the elements of the equation, that is, if on one side the hydrogen has a radical 2, it must continue with the radical 2.

c. Yes, we can express the increase in atoms by adding the number of atoms of any of the compounds in the mixture. Thus, if we want to express that there are 4 atoms of hydrochloric acid, we will write 4HCl.

d. It is convenient to start balancing with the elements that only appear once in each member, leaving those that appear more than once to the last, if necessary.

and. Hydrogen and oxygen are among the last elements to consider for balancing.

4. We do not have a definite place to start our balancing, so we can start with any of the members of the equation. We will start with the sodium atoms. As we can see, in the first member there are two sodium atoms to react in the sulfate molecule sodium, while on the right side, in the substance produced, sodium chloride, there is only one atom of sodium. This means that in order to balance the sodium and for there to be two atoms in the result, there must be two molecules of sodium chloride on the right side of the reaction. So we would have:

2 + 1 + 3 + 1+ 1 -- > 2 +2 + 2 + 1 + 1 + 2

Na₂SW₃ + HCl -> 2NaCl + H₂O + SO₂

5. As we can see, we already have the same number of sodium atoms. But our equation remains unbalanced. In effect, we now have:

Left side: Na = 2; S = 1; O = 3; H = 1; Cl = 1

Right side: Na = 2; S = 1; O = 3; H = 2; Cl = 2

6. Now we have two chlorine atoms in the result and only one in the reagents. If we consider that the result of the reaction produces two salt atoms, and there is only one chlorine atom in the reacting molecule, means that now we must consider that two molecules of the compound that contains chlorine act, that is, two molecules of acid hydrochloric. To check if our assumption is true, we add to our formula the indication that two HCl atoms are reacting and we count the atoms again:

2 + 1 + 3 + 2 + 2 -- > 2 +2 + 2 + 1 + 1 + 2

Na₂SW₃ + 2HCl -> 2NaCl + H₂O + SO₂

7. Now we already have the same number of atoms reacting from both sides of the equation. Finally we check that on both sides there is the same number of atoms of each element:

Left side: Na = 2; S = 1; O = 3; H = 2; Cl = 2

Right side: Na = 2; S = 1; O = 3; H = 2; Cl = 2

We have the same number of atoms of each element on both sides of the equation, which means that our formula is correctly balanced. We can also appreciate that when we begin to balance through the elements that only appear once, other atoms, in this case hydrogen, change their values. depending on the molecule in which it is combined and the number of molecules that act in the equation, also balancing together with the rest of elements.

Example of trial and error balancing of nitric acid with calcium hydroxide:

Now we are going to balance the equation for the reaction of nitric acid with calcium hydroxide, which produces calcium nitrate and water:

HNO₃ + Ca (OH)₂ -> Ca (NO₃)₂ + H₂OR

1. We start by counting the atoms in each side of the equation and the atoms in each element of the equation:

1 + 1 + 3 + 1 + 2 + 2 -- > 1 + 2 + 6 + 2 + 1

HNO₃ + Ca (OH)₂ -> Ca (NO₃)₂ + H₂OR

Left side: N = 1; Ca = 1; O = 5; H = 3

Right side: N = 2; Ca = 1; O = 7; H = 2

We will therefore begin our balance with nitrogen. On the side of reactions we have two atoms, while in the reactants, there is only one. We can balance this by considering that two molecules of nitric acid act, so our formula and our atom count would look like this:

2 + 2 + 6 + 1 + 2 + 2 -- > 1 + 2 + 6 + 2 + 1

2HNO₃ + Ca (OH)₂ -> Ca (NO₃)₂ + H₂OR

Left side: N = 2; Ca = 1; O = 8; H = 4

Right side: N = 2; Ca = 1; O = 7; H = 2

We already balanced the nitrogen, but the equation is still out of balance.

2. Looking at our equation, we see that we already have the same number of nitrogen and calcium atoms. This means that we already have the right amount of nitric acid and calcium hydroxide molecules to produce one molecule of calcium nitrate. Comparing the atoms of all the elements, we have that the equation on the right side lacks one molecule of oxygen and two of hydrogen to be balanced. What does this mean? Well, one molecule of oxygen and two of hydrogen produce water, and since there is already one water molecule present in the reaction, it means that not one but two water molecules are produced.

We add to our formula that two water molecules are produced, and we recount atoms and elements:

2 + 2 + 6 + 1 + 2 + 2 -- > 1 + 2 + 6 + 4 + 2

2HNO₃ + Ca (OH)₂ -> Ca (NO₃)₂ + 2H₂OR

Left side: N = 2; Ca = 1; O = 8; H = 4

Right side: N = 2; Ca = 1; O = 8; H = 4

Our equation is correctly balanced.