## How to balance complex chemical equations

While studying chemistry, the skill of balancing chemical equations is essential to master. Reaction analysis, lab work, and stoichiometry justify the importance of balancing chemical equations.

In previous articles we have discussed the methods for balancing general chemical equations and redox equations in basic solutions.

In this article, we'll discuss the easiest ways to balance complex chemical equations. Such an equation which comprises more than five elements.

## Why do general equation balancing methods not work well with complex equations?

There are two main methods to balance a general chemical equation. The first one is known as the trial and error method, while the second one is called atom counts.

Since the first method is being taught in high schools, most learners would be aware of it. This method of balancing chemical equations is, indeed, the easiest one.

While balancing complex chemical equations by the trial and error or atom counts method, the procedure gets prolonged.

According to a rough estimate, balancing a chemical equation that comprises 5-7 elements by one of these two general methods may take about thirty minutes! Meanwhile, the trial and error and atom count methods are suitable only for the equations containing only 2-3 atoms.

## Balancing complex chemical equations by algebra

The algebraic method is the precise and efficient way to balance complex equations.

The hybrid of chemical equations and algebra sounds terrifying at first, but it is not as difficult as it sounds.

This algebra part is more straightforward than the simultaneous equations of A-level maths. In other words, you can balance complex equations by learning algebraic methods.

You'll be able to solve unbalanced equations that are hard to equalize using inspection or the atom count method.

Related: Also find what happens during and after the reaction of Sodium (Na) and Chlorine (Cl).

### This algebraic method is further divided into two main types:

• General algebraic method
• Simplified algebraic method

## General algebraic method to solve complex chemical equations

The general strategy for balancing complex chemical equations by algebra includes the following steps

1. Add different suitable coefficient letters to each compound present in the equation
2. Apply algebraic rules or expressions for each element to equalize its atoms on both sides of the equation.
3. Substitute the coefficient to simplify the rules. It will help to digitized the balancing coefficient for each entity
4. To get the final balancing coefficient, replace the values with other rules.

These steps are like gobbledygook and won't make any sense without an example. So, let's balance a complex chemical equation by following the same strategy.

KMnO4 + HCl → MnCl2 + KCl + Cl2 + H2O

So first of all, we will add letter coefficient in front of each compound as follows

aKMnO4 + bHCl → cMnCl2 + dKCl + eCl2 + fH2O (2)

### Step# 2: Apply Algebraic Rules

After this, we will apply the algebraic rules to make the equations perfect according to the law of mass conservation.

We will start by writing rules for K since it is the first element from L.H.S in the equation. After this, we will use the same method to write different algebraic rules for each entity, such as

K: a = d
Mn: a = c
O: a = f
H: b = f
Cl: b = 2c + d + 2e

You could be confused on the last rule so let us explain this to you. According to the law of mass construction, the number of Cl atoms would be the same on both sides.

According to the formula [Cl: b = 2c + d + 2e], the reactant side contains b number of Cl atoms.

While cMnCl2 contains two chlorine atoms and its coefficient letter is C. We wrote the rule for cMnCl2 as 2c.

KCl would have a d number of chlorine atoms and so on. Thus, on adding up the coefficient values for c, d, and e at R.H.S, we will get the value of b at L H.S.

### Step# 3: Discard Unknown Coefficient

In the next step, we will discard the unknown coefficients and simplify the rules by adding common substitutes for each element.

Let's start with the chlorine atoms and substitute the rules for Mn and K to remove c and d from equation such as

b = 2c + d + 2e converted to b = 2a + a + 2e

On further simplifying the equation, it would become

b = 3a + 2e

Now, time for substituting the rules for H to eliminate b

2f = 3a + 2e

And lastly, substitute the rule for oxygen atom to add eliminate f

2(4a) = 3a + 2e

8a = 3a + 2e

5a = 2e : a = 2 and e=5

Now we have found the coefficient values for a and e. Use these values and the fact a = c = d = 2 into the rules for O and Cl to find the coefficient values for b and f

b = 2c + d + 2e

b = 3a + 2e

b = 3 . 2 + 2 . 5

b = 16

Now derive the value of f as follows

4a = f

4 . 2 = f = 8

### Step# 4: Add Coefficient Value

Now add all these coefficient values to the equation (2) to get your final balanced equation

2KMnO4 + 16HCl → 2MnCl4=2 + 2KCl + 5Cl2 + 8H2O

## Simplified algebraic method

If you have understood the general algebraic method thoroughly, you must be thinking that if a = c = d, then what is the need to add them in the first place?

Why not write the whole thing in a simplified manner from the very first point? That's precisely the purpose of using a simplified algebraic method.

To understand the exact procedure of solving unbalanced equations by the simplified algebraic method, let's consider the same complex chemical reaction example.

KMnO4 + HCl → MnCl2 + KCl + Cl2 + H2O

### Step#1: Identify Atoms

First of all, identify the atoms already balanced in the equation and use a similar coefficient letter. Such as one atom of Mn is present in KMnO4 and MnCl2 , which means it is balanced.

So we will add the coefficient such as

aKMnO4 + HCl → aMnCl2 + KCl + Cl2 + H2O

Now comes to balancing unbalanced atoms. There is one oxygen atom on R.H.S in the form of a water molecule. There are four atoms of oxygen on L.H.S, we will write the coefficient such as

aKMnO4 + HCl → MnCl2 + KCl + Cl2 + 4aH2O

According to the forced coefficient rules, the coefficient for KCl and MnCl2 would also be a. Since there are four water molecules at the product side, the coefficient letter for HCl would be 8a.

The final equation would be as follows

aKMnO4 + 8aHCl → aMnCl2 + aKCl + Cl2 + 4aH2O

For the Cl2, we'll write the coefficient b since no Cl2 molecule is present at L.H.S of the equation

aKMnO4 + 8aHCl → aMnCl2 + aKCl + bCl2 + 4aH2O

### Step#2: Write Algebraic Rules

Write the algebraic rules for each coefficient value, such as

8a = 2a + a + 2b

5a = 2b : a = 2 | b = 5

Thus, the final equation would be as follows

2KMnO4 + 16HCl → 2MnCl2 + 2KCl + 5Cl2 + 8H2O

#### Tips for balancing complex chemical equations by the simplified algebraic method

• Reduce the extra coefficient letter by using the same letters on both sides
• Always try not to use more than two-letter coefficients.
• To reduce the unknown numbers, use the charge conservation principle