## What is Chemical Equation?

A chemical equation is written for the symbolic representation of the chemical reaction. In chemical equation, the chemical formulas of reactants and products are expressed.

A chemical equation help us to monitor the direction of a chemical reaction and the physical state of the entities involved in the chemical reaction.

We can write a complete chemical reaction on paper with the help of chemical equations. Follow this chemical equations examples:

H_{2} + O → H_{2}O

Hydrogen reacts with Oxygen to produce water. In above chemical equation, the reactants and written on the left side and the product (H_{2}O) is written on the right side.

## Why must Chemical Equations be balanced?

A chemical reaction undergos a lot of changes. A chemical reaction is initiated so that we could get something. In order to get the desired output from the chemical reaction, chemical equations must be balanced.

Chemistry equations are very crucial for us to obtain the right product from the chemical reaction. Thats why complex equations must be balanced.

## How to balance Complex Chemical Equations?

In chemistry, the concept of how to balance 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 balancing redox reactions in basic solution.

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

**Related:** Find what you need to know about the periodic table and the synthesis reaction example with solution.

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

There are two main methods of balancing difficult chemical equations. The first one is the trial and error method, while the second one is the atom counts.

The first method of balancing chemical equations is taught in school. This method of balancing chemical equations is simple and easy to understand.

Balancing complex chemical equations gets prolonged if we use other method. Trial and error or atom counts method make the procedure quite lengthy.

The trial and error and atom count methods are suitable for the equations having 2-3 atoms. Since both are general methods, it will take us some time to complete the process. On average, it takes around 30 minuets for balancing a chemical equation having 5-7 element.

**Related:**What is the general chemical equation for an endothermic reaction?

## Balancing complex chemical equations by algebra

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

Combination of chemical equations and algebra sounds terrifying, but it is not as difficult as it sounds.

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

In the process of equation balancer, 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

**Related:** How to balance complex chemical equations?

## How to balance Complex Equations using General Algebraic Method

The general strategy for the process of how to balance chemical equations by algebra includes the following steps

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

**Related:** Also find about metal displacement reactions and how to determine the chemical factors.

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

_{4}+ HCl → MnCl

_{2}+ KCl + Cl

_{2}+ H

_{2}O

**Related:** Get step by step guide on how to calculate the oxidation states of elements in a chemical compound.

### Step# 1: Add Coefficient

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

_{4}+ bHCl → cMnCl

_{2}+ dKCl + eCl

_{2}+ fH

_{2}O (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 process of equation balancer. 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.

**Related:** How to Determine the Heat of Combustion in Organic Chemistry?

### 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

**Related:** How to find the percent yield of a chemical reaction step by step?

### Step# 4: Add Coefficient Value

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

_{4}+ 16HCl → 2MnCl

_{4=2}+ 2KCl + 5Cl

_{2}+ 8H

_{2}O

**Related:** Find useful online calculators on this website like we offer redox equation balancer and atomic mass calculator which you can use for free.

## How to balance Complex Equations using 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, There are a lot of complex chemical reaction examples. One of complex chemical reaction balancer example is given below:

_{4}+ HCl → MnCl

_{2}+ KCl + Cl

_{2}+ H

_{2}O

**Related:** What are Gas Laws and How are Gas Laws Used in Real Life?

### 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

_{4}+ HCl → aMnCl

_{2}+ KCl + Cl

_{2}+ H

_{2}O

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

_{4}+ HCl → MnCl

_{2}+ KCl + Cl

_{2}+ 4aH

_{2}O

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

_{4}+ 8aHCl → aMnCl

_{2}+ aKCl + Cl

_{2}+ 4aH

_{2}O

For the Cl_{2}, we'll write the coefficient b since no Cl_{2} molecule is present at L.H.S of the equation

_{4}+ 8aHCl → aMnCl

_{2}+ aKCl + bCl

_{2}+ 4aH

_{2}O

### 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

As per this balancing equations examples, the final equation would be as follows:

_{4}+ 16HCl → 2MnCl

_{2}+ 2KCl + 5Cl

_{2}+ 8H

_{2}O

#### Tips for complex balancing 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