Balancing a Chemical Equation: We know that chemical compounds are broken up during a chemical reaction to form new compounds. A chemical reaction follows the law of conservation of mass in which atoms are neither created nor destroyed; they are merely reorganised into different arrangements. In a complete chemical equation, the atoms undergoing rearrangement are known as reactants and are present on the left-hand side of the chemical equation. The rearranged atoms are present on the right-hand side of the chemical equation and are known as products. Let’s understand how these atoms maintain a balance in their numbers while undergoing rearrangement.

Coefficients and Subscripts

There are two types of numerical values in a chemical equation. These are subscripts in the chemical formulas of the reactants and products and coefficients in front of the formulas to show how many molecules of that material are used or produced.

The subscripts are a part of the formulas, and once the formulas for the reactants and products are framed, they cannot change. 

The coefficients indicate the number of each substance involved in the reaction and may differ from equation to equation as per balancing requirements. 

In the above equation, one mole of solid copper reacts with two moles of aqueous silver nitrate to form one mole of aqueous copper (II) nitrate and two atoms of solid silver.

Balancing Chemical Equations

It involves the addition of stoichiometric coefficients to the chemical formulas of reactants and products in a chemical equation.

A chemical equation is a symbolic depiction of a chemical reaction in which the reactants transform to generate products, which are represented by chemical formulae.

The example above illustrates the fundamental aspects of any chemical equation:

1. The substances undergoing reaction are placed on the left side of the equation and are called reactants.
2. The substances generated by the reaction are placed on the right side of the equation are called products.
3. The individual reactant and product formulas are separated by a plus \((+)\) sign.
4. An arrow \((→)\) separates the reactant on the left and products on the right sides of the equation.
5. Coefficients represent the relative numbers of reactant and product species (numbers placed immediately to the left of each formula). In most cases, a coefficient of one is ignored.

Balancing Equations

A balanced chemical equation is an equation that has equal numbers of atoms for each element both on the reactant and product sides. This is a requirement based on the Law of conservation of mass. While balancing a chemical equation, certain rules are to be followed-

The number of atoms for a given element is calculated by multiplying its coefficient with the element’s subscript in its chemical formula. 

If an element appears in many formulas on one side of an equation, the number of atoms represented in each must be determined and then added.

Steps to follow while balancing equations:

Let the word-equation for the action of zinc on sulphuric acid be-

\({\rm{ Zinc }} + {\rm{ Sulphuric}}\,{\rm{acid}} \to {\rm{ Zinc}}\,{\rm{sulphate}} + {\rm{ Hydrogen }}\)

The above word-equation can be represented by the following chemical equation –

\({\rm{Zn}} + {{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4} \to {\rm{ZnS}}{{\rm{O}}_4} + {{\rm{H}}_2}\)

Examining the number of atoms of different elements on both sides of the arrow, we get-

Element	Number of Atoms in Reactants (LHS)	Number of Atoms in Products (RHS)
\({\rm{Zn}}\)	\(1\)	\(1\)
\({\rm{H}}\)	\(2\)	\(2\)
\({\rm{S}}\)	\(1\)	\(1\)
\({\rm{O}}\)	\(4\)	\(4\)

As the number of atoms of each element on the reactant and product side is the same, the equation can be considered a balanced chemical equation. 

Let us take another unbalanced chemical equation and try to balance it –

\({\rm{Fe}} + {{\rm{H}}_2}{\rm{O}} \to {\rm{F}}{{\rm{e}}_3}{{\rm{O}}_4} + {{\rm{H}}_2}\)

Step I: Listing the number of atoms of different elements present in the reactant as well as on the product side in the unbalanced chemical equation.

Element	Number of Atoms in Reactants (LHS)	Number of Atoms in Products (RHS)
\({\rm{Fe}}\)	\(1\)	\(3\)
\({\rm{H}}\)	\(2\)	\(2\)
\({\rm{O}}\)	\(1\)	\(4\)

Step II: We must always start balancing with the compound that contains the maximum number of atoms either on the reactant or product sides. Having chosen the compound, the element with the maximum number of atoms is chosen for balancing.
Using these criteria, we select \({\rm{F}}{{\rm{e}}_3}{{\rm{O}}_4}\) and the element oxygen in it. In the given equation, there are four oxygen atoms on the product and only one atom on the reactant side. To balance the oxygen atoms –

Atoms of Oxygen	In Reactants	In Products
(i) Initial	\(1\left( {{\rm{ in }}\,{{\rm{H}}_2}{\rm{O}}} \right)\)	\(4\left( {{\rm{ in}}\,{\rm{F}}{{\rm{e}}_3}{{\rm{O}}_4}} \right)\)
(ii) To Balance	\(1 \times 4\)	\(4\)

To equalise the number of atoms, we must remember that we cannot alter the formulae of the compounds or elements involved in the reactions. This means the numbers in subscripts cannot be altered. However, to balance the number of atoms, stoichiometric coefficients are used. The numerical values are placed before the compound in a chemical reaction and can vary from equation to equation for the same chemical compound. 

In the above equation, to balance oxygen atoms, we can put coefficient ‘\(4\)’ as \(4{{\rm{H}}_2}{\rm{O}}\) and not \({{\rm{H}}_2}{{\rm{O}}_4}\) or \(\left( {{{\rm{H}}_2}{{\rm{O}}}} \right)_4\). Now the partly balanced equation becomes–

\({\rm{Fe}} + 4{{\rm{H}}_2}{\rm{O}} \to {\rm{F}}{{\rm{e}}_3}{{\rm{O}}_4} + {{\rm{H}}_2}\)

Step III: Though oxygen atoms are balanced, Fe and H atoms are still unbalanced. Proceeding with hydrogen atoms, the number of molecules of hydrogen is made as four on the product side.

Atoms of Hydrogen	In Reactants	In Products
(i) Initial	\(8\left({\rm{in}}\,4{\rm{H}}_2 {\rm{O}} \right)\)	\(2\left( {{\rm{ in }}\,{{\rm{H}}_2}} \right)\)
(ii) To Balance	\(8\)	\(2 \times 4\)

The equation would be:

\({\rm{Fe}} + 4{{\rm{H}}_2}{\rm{O}} \to {\rm{F}}{{\rm{e}}_3}{{\rm{O}}_4} + 4{{\rm{H}}_2}\)

Step IV: Once hydrogen atoms are balanced, the only element left to be balanced is, iron. To equalise Fe atoms, we make it three as on the reactant side.

Atoms of Iron	In Reactants	In Products
(i) Initial	\(1{\rm{ (in}}\,{\rm{Fe) }}\)	\(3\left( {{\rm{ in}}\,{\rm{F}}{{\rm{e}}_3}{{\rm{O}}_4}} \right)\)
(ii) To Balance	\(1 \times 3\)	\(3\)

\(3{\rm{Fe}} + 4{{\rm{H}}_2}{\rm{O}} \to {\rm{F}}{{\rm{e}}_3}{{\rm{O}}_4} + 4{{\rm{H}}_2}\)

Step V: To check the correctness of the balanced equation, we count atoms of each element on both sides of the equation.

\(3{\rm{Fe}} + 4{{\rm{H}}_2}{\rm{O}} \to {\rm{F}}{{\rm{e}}_3}{{\rm{O}}_4} + 4{{\rm{H}}_2}\)

The number of atoms in each element on both sides of the equation is now equal, and the equation is balanced. This approach of balancing chemical equations is known as the hit-and-trial method because we use the smallest whole number coefficient to balance the equation.

Step VI: Writing Symbols of Physical States for the compounds

The physical states of the reactants and products are mentioned together with their chemical formulae to make a chemical equation more meaningful. The notations \(({\rm{s}}),\,({\rm{l}}),\,({\rm{aq}})\) and \(({\rm{g}})\) is used to designate the solid, liquid, aqueous, and gaseous states, respectively.

If a reactant or product exists as a solution in water, the word aqueous \(({\rm{aq}})\) is used. As a result, in the case above, the balanced equation is:

\(3{\rm{Fe}}({\rm{s}}) + 4{{\rm{H}}_2}{\rm{O}}({\rm{g}}) \to {\rm{F}}{{\rm{e}}_3}{{\rm{O}}_4}(\;{\rm{s}}) + 4{{\rm{H}}_2}(\;{\rm{g}})\)

The symbol (g) is used with \({{\rm{H}}_2}{\rm{O}}\) indicates that in this reaction, water is in the form of steam. The reaction conditions are sometimes provided above and/or below the arrow in the equation, such as temperature, pressure, catalyst, and so on. For instance —

\({\text{CO}}\left( {\text{g}} \right){\text{ + 2}}{{\text{H}}_{\text{2}}}\left( {\text{g}} \right)\xrightarrow{{{\text{340}}\,{\text{atm}}}}{\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}}\left( {\text{l}} \right)\)

\({\text{6C}}{{\text{O}}_{\text{2}}}\left( {{\text{aq}}} \right){\text{ + 12}}{{\text{H}}_{\text{2}}}{\text{O}}\left( {\text{l}} \right)\xrightarrow[{{\text{Chlorophyll}}}]{{{\text{Sunlight}}}}\mathop {{{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{12}}}}{{\text{O}}_{\text{6}}}\left( {{\text{aq}}} \right)}\limits_{\left( {{\text{Glucose}}} \right)} {\text{ + 6}}{{\text{O}}_{\text{2}}}\left( {{\text{aq}}} \right){\text{ + 6}}{{\text{H}}_{\text{2}}}{\text{O}}\left( {\text{l}} \right)\)

Solved Examples

Q.1. Balance the following equation-

\({\rm{Ca}}{({\rm{OH}})_2} + {{\rm{H}}_3}{\rm{P}}{{\rm{O}}_4} \to {\rm{C}}{{\rm{a}}_3}{\left( {{\rm{P}}{{\rm{O}}_4}} \right)_2} + {{\rm{H}}_2}{\rm{O}}\)

Solution:

Step I: List the number of atoms of different elements present in the reactant as well as on the product side in the unbalanced chemical equation.

Element	Number of Atoms in Reactants (LHS)	Number of Atoms in Products (RHS)
\({\rm{Ca}}\)	\(1\)	\(3\)
\({\rm{P}}\)	\(1\)	\(2\)
\({\rm{O}}\)	\(6\)	\(9\)
\({\rm{H}}\)	\(5\)	\(2\)

Step II: In the given equation, there are three calcium atoms on the product side and one atom on the reactant side. To balance the calcium atoms, we add a \(3\) in front of the \({\rm{Ca}}{({\rm{OH}})_2}\) on the left hand side.

\(3{\rm{Ca}}{({\rm{OH}})_2} + {{\rm{H}}_3}{\rm{P}}{{\rm{O}}_4} \to {\rm{C}}{{\rm{a}}_3}{\left( {{\rm{P}}{{\rm{O}}_4}} \right)_2} + {{\rm{H}}_2}{\rm{O}}\)

Step III: There is one Phosphorus atom on the reactant side and two atoms on the product side as \({\rm{P}}{{\rm{O}}_4}\) ions. Balancing the phosphorus atom, we add a \(2\) in front of the \({{\rm{H}}_3}{\rm{P}}{{\rm{O}}_4}\) on the left side.

\(3{\rm{Ca}}{({\rm{OH}})_2} + 2{{\rm{H}}_3}{\rm{P}}{{\rm{O}}_4} \to {\rm{C}}{{\rm{a}}_3}{\left( {{\rm{P}}{{\rm{O}}_4}} \right)_2} + {{\rm{H}}_2}{\rm{O}}\)

Element	Number of Atoms in Reactants (LHS)	Number of Atoms in Products (RHS)
\({\rm{Ca}}\)	\(3\)	\(3\)
\({\rm{P}}\)	\(2\)	\(2\)
\({\rm{O}}\)	\(14\)	\(9\)
\({\rm{H}}\)	\(12\)	\(2\)

Step IV: Finally, there are fourteen oxygen atoms on the left side and only nine atoms on the right side. And twelve and two hydrogen atoms on the left and right sides of the equation. So, adding a \(6\) in front of the \(\mathrm{H}_{2} \mathrm{O}\) on the right-hand side, we get-

\(3{\rm{Ca}}{({\rm{OH}})_2} + 2{{\rm{H}}_3}{\rm{P}}{{\rm{O}}_4} \to {\rm{C}}{{\rm{a}}_3}{\left( {{\rm{P}}{{\rm{O}}_4}} \right)_2} + 6{{\rm{H}}_2}{\rm{O}}\)

Step V: To check the correctness of the balanced equation, we count atoms of each element on both sides of the equation.

Element	Number of Atoms in Reactants (LHS)	Number of Atoms in Products (RHS)
\({\rm{Ca}}\)	\(3\)	\(3\)
\({\rm{P}}\)	\(2\)	\(2\)
\({\rm{O}}\)	\(14\)	\(14\)
\({\rm{H}}\)	\(12\)	\(12\)

It’s now a balanced chemical equation. 

Summary

The balancing of the chemical equation is based on the Law of conservation of mass. According to the law, the number of atoms of an element is equal to the on both sides of the equation. The balancing is done by placing numbers on the left side of the chemical compounds. These numbers are known as coefficients and change for the same compound from the equation to the equation. In this article, we discussed the law that governs the balancing of the chemical equation, the steps that need to be followed, and a solved example.

FAQs on Balancing a Chemical Equation

Q.1. What is a simple, balanced equation?
Ans: A balanced equation is a chemical equation in which the number of atoms for each element in the reactant and product side is the same. The total charge is also the same for both the reactants and the products. 

Q.2. How do you balance a given chemical equation?
Ans: Balancing is done by changing the coefficients (the numbers in front of the compound or molecule) so that the number of atoms of the element is the same on each side of the equation.

Q.3. What are coefficients and subscripts?
Ans: The subscripts are a part of the formulas, and once the formulas for the reactants and products are framed, they cannot change. The coefficients indicate the number of each substance involved in the reaction and may differ from equation to equation as per balancing requirements. 

Q.4. How is an equation written?
Ans: A chemical equation consists of the chemical compounds represented by chemical formulas on the left that comprise the reactants and chemical compounds represented by chemical formulas that comprise the products. The arrow symbol \((“→”)\) separates the two. A plus symbol \((+)\) separates the substances in the reactants and products. Each chemical’s or molecule’s state of matter is denoted by an abbreviation in parenthesis in subscript next to the compound. A chemical in the gaseous state, for example, would be expressed by \(({\rm{g}})\). Similarly, for solid, liquid and aqueous state \(({\rm{s}}),\,({\rm{l}})\) and \(({\rm{aq}})\) is used. Aqueous substances are dissolved in water.

Q.5. Give an example of a balanced chemical equation.
Ans: A balanced chemical equation is an equation in which the number of atoms of different elements on the reactants side (left side) is the same as those on the product side (right side). For example, in the chemical equation \(2 \mathrm{Cu}+\mathrm{O}_{2} \rightarrow 2 \mathrm{CuO}\), the number of \({\rm{Cu}}\) and \({\rm{O}}\) atoms on both sides of the equation are equal; hence it is a balanced chemical equation.

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