It is known that in chemistry, we have to deal with the chemical equations since these chemical equations not only help us to determine the identity of substances that are being reacted (reactants) but also the substances which are being formed by their reactions (products).
It must be surprising to you that you can predict the product of the following reaction just by seeing the L.H.S of the equation. All you have to do is to remember some key points.
How to predict the product of the following reaction?
To predict the nature of substances being formed during a chemical reaction, always pay attention to the nature of reacting species and the type of reaction occurring.

A reaction between Metal and halogens always results in the formation of salt. The most common example is the formation of NaCl such as
2Na(metal) + Cl_{2}(halogen) → 2NaCl(salt)

The reaction between an acid and base always results in the formation of salt and water due to neutralization reaction such as
HCl + NaOH → NaCL + H_{2}O

In ordinary synthesis reactions, the final product would be the combination of reactants taking part in the chemical reaction such as follows
2Al + 3Cl_{2} → 2AlCl_{3}

In simple displacement reactions, one part of a compounds gets replace by the other part such as follows
PbCl_{4} + 2F_{2} → PbF_{4} + 2Cl_{2}

To predict the product of a reaction, the equation balancer is also available on the internet where all you have to do is to enter the reactants and the complete final equation would be displayed in a few seconds.
Apart from the identity of substances, the chemical equation also helps us to determine the quantity ( the number of atoms of each element) of each substance taking part in the reaction.
However, for the correct determination of quantity, the chemical equations must be completely balanced. Before moving towards the methods of balancing chemical equations, let us recall the basic definitions of balanced and unbalanced chemical equations.
What is an unbalanced equation?
An unbalanced chemical equation possesses an unequal number of atoms at both sides of the equations. For illustration of equation balancing, let us consider a general reaction
2A + 2B → AB
It can be seen from the above equation that in the reaction mixture, there are two atoms of A and two atoms of B. Both react to form a product AB which contains only one atom of A and one atom of B.
It is not be possible according to the law of mass conservation. Thus, the given equation is unbalanced. Let us consider a chemical equation this time.
Na_{2}+Cl_{2} → NaCl(unbalanced equation)
In the given example, there are two atoms of sodium and two atoms of chlorine at L.H.S. There are also atoms of sodium and one atom of chlorine at the R.H.S which makes the equation unbalanced.
Whenever we study chemical reactions in chemistry, we have to deal with unbalanced chemical equations which need to be balanced. Otherwise, it can ruin the entire research work due to wrong observations. Therefore, it is necessarily important to learn how equation balancer works & how to balance chemical equations.
What is chemical equation balancing?
A balanced chemical equation possesses an equal number of atoms at both sides of the equation. Such as
2A + 2B → 2AB
It can be seen that there are two atoms of A and two atoms of B which on reacting with each other, changes into two molecules of AB (2AB:2A_{1}=2•1=2 atoms, 2B_{1}=2•1=2) Now, let us consider a chemical equation to clear the concerns
Na_{2} + Cl_{2} → 2NaCl
Just as general equation, there are two atoms of sodium (1Na_{2}=1•2 = 2) and two atoms of chlorine (1Cl_{2}=1•2=2) which reacts with each other to form 2 molecules of NaCl (2NaCl:2Na=2•1=2, 2C_{1}=2•1=2)
Hence, proved that the process of balancing chemical equations involves balancing the equal number of atoms of reactants and products.
How to balance chemical equations?
There are many ways of balancing equations however, each method is specific to a specific type of reaction. The most common method used to balance chemical equations is the inspection method in which you have to count the number of atoms of each element at both sides of equations. This method is mainly being used to balance the number of atoms at each side of equations.
To determine the number of atoms, you have to multiply the subscript value of each element present in the compound with the coefficient value such as
AB_{4} + 2C_{2} → AC + BC(unbalanced)
To balance this equation, first determine the number of atoms taking part in the reaction such as
1A_{1} = 1•1 = one atom of A
1B_{4} = 1•4 = four atoms of B
2C_{2} = 2•2 = four atoms of C
 Since there is only one atom of A at L.H.S, there is no need to add any coefficient or subscript value with A at R.H.S.
 Since there are four atoms of B and four atoms of C, the product BC would be converted to 2B_{2}C. We cannot write the product as 2B_{2}C_{2} because it would show the four atoms of B and five atoms of C since there is another atom of C present in product AC (2B_{2}C_{2}+AC=4+1=5 unbalanced).
 To balance the value of C, change the subscript value of C in AC as AC_{2}
Thus, the final equation would be such as
AB_{4} + 2C_{2} → AC_{2} + 2B_{2}C(Balanced)
To balance the number of atoms at both sides of atoms, the chemical equation balancer is also available on the internet. You just have to write the unbalanced equations, and in a few seconds, it will balance equations for you.
However, since you cannot use it in the examination hall or during the unavailability of the internet, it is advised not to depend on an equation balancer completely.
Apart from the chemical equation product calculator, the limiting reactant calculator is also being greatly used by people. The method of using such chemical reaction calculator is the same as others, however, the final equation is shown with the exact value of coefficient and subscript.
So you can predict the limiting reactant easily by seeing the equation. Due to this reason, such a balance chemical equation calculator is more preferred by the researchers.
Test your skills with balancing chemical equations practice
How to balance equation?
HNO_{3} + Ca(OH)_{2} → Ca(NO_{3})_{2} +H_{2}O
First of all, compare the total number of atoms of each element taking part in the chemical reaction.

Since there are two N atoms at the product side, multiply HNO3 with 2. The equation would become
2HNO_{3} + Ca(OH)_{2} → Ca(NO_{3})_{2} + H_{2}O(unbalanced)

Since the reactant side contains four atoms of hydrogen, balance the number of atoms at the product side by multiplying H2O with 2. The equation would become
2HNO_{3} + Ca(OH)_{2} → Ca(NO_{3})_{2} + 2H_{2}O(unbalanced)
Since both sides are containing equal numbers of atoms now, thus the given equation is balanced.
Can you balance the equation with the fraction?
While balancing the number of atoms in an unbalanced chemical equation, you might be subjected to the cases where you have to use a fractional coefficient other than a whole number coefficient.
This is being done in cases when there is no way to balance chemical equations completely. Moreover, this fractional value is being used temporarily and never mentioned in the final equation. To clarify the point, let us consider an example
C_{4}H_{10} → H_{2}O + CO_{2}(unbalanced)
First, we have to balance the number of atoms on both sides  starting with the hydrogen atom.
Since there are ten atoms of hydrogen at L.H.S, there would be five molecules of H2O. Moreover, since there are four atoms of Carbon at L.H.S, there would be four molecules of CO_{2}
The final equation would be written such as
C_{4}H_{10} + O_{2} → 5H_{2}O + 4CO_{2}(unbalanced)
Since there are a total of 13 atoms of oxygen at R.H.S, there is not any whole number coefficient that could be multiplied by the subscript 2 to form 13, so we wrote the coefficient in fraction 13/2 such as follows
C_{4}H_{10} + 13/2 O_{2} → 5H_{2}O + 4CO_{2}(balanced)
It is a rule that the final equation cannot be written in fractional coefficient, thus, to solve this fractional quantity, we multiply the entire equation with 2 to convert the fraction into the whole number which results in the final equation such as
2C_{4}H_{10} + 13O_{2} → 10H_{2}O + 8CO_{2}(unbalanced)
Since putting the fractional quantity in the equation is an intermediate step, whenever you would use an online equation solver, this intermediate step would not be shown there.
How to balance the equation by oxidation number method?
This method is mainly being used to balance redox equations on the basis of oxidation numbers. The method is as follows

First of all, write the complete equation along with the oxidation states of each element of reactants and product such as follow
^{+1*2}K_{2} ^{+6*2}Cr_{2}^{2*7}O_{7} + ^{1}H^{1}CL
→
^{+3}Cr^{1*3}Cl_{3} + ^{1}K^{1}Cl + ^{+1*2}H_{2}^{2}O 
Now consider the atoms which do not undergo any kind of change in their oxidation numbers. From the above equation, it can be seen that the oxidation state of Cl in +1H1Cl at reactant side is 1 which changes into zero in the product side 0Cl2. However in +1K1Cl, the oxidation state of Cl is again being mentioned as 1. It predicts that there are two types of Cl molecules present in the reactant mixture  one which undergo change in oxidation states and other which do not undergo change in their oxidation state. Thus, we can write equation such as
^{+1*2}K_{2} ^{+6*2}Cr_{2}^{2*7}O_{7} + ^{1}H^{1}CL + HCL
→
^{+3}Cr^{1*3}Cl_{3} + ^{0}Cl_{2} + ^{1}K^{1}Cl + ^{+1*2}H_{2}^{2}O
Since there are two atoms of Cr at reactant side which possess oxidation state of +6 (+1×2K_{2}^{+6×2}Cr_{2}^{2×7}O_{7}). After the reaction, the oxidation state of Cr changes to +3 in ^{+3}Cr^{1×3}Cl_{3}. Thus to balance equation, multiply CrCl3 with 2 which result in equation as follows
^{+1*2}K_{2} ^{+6*2}Cr_{2}^{2*7}O_{7} + ^{1}H^{1}CL + HCL
→
2^{+3}Cr^{1*3}Cl_{3} + ^{0}Cl_{2} + ^{1}K^{1}Cl + ^{+1*2}H_{2}^{2}O


The above equation predicts that there should be 6 HCl atoms at L.H.S which undergo change in oxidation state by losing six electrons and obtain an oxidation state of zero in 0Cl2. To balance this, multiply HCl by six such as
^{+1*2}K_{2} ^{+6*2}Cr_{2}^{2*7}O_{7} + 6^{1}H^{1}CL + HCL
→
2^{+3}Cr^{1*3}Cl_{3} + ^{0}Cl_{2} + ^{1}K^{1}Cl + ^{+1*2}H_{2}^{2}O 
Since there are two atoms of potassium at reactant side ^{+1×2}K_{2} ^{+6×2}Cr_{2}^{2×7}O_{7} , multiple KCl with 2 to balance number of K at both sides of equation such as
^{+1*2}K_{2} ^{+6*2}Cr_{2}^{2*7}O_{7} + 6^{1}H^{1}CL + HCL
→
2^{+3}Cr^{1*3}Cl_{3} + ^{0}Cl_{2} + 2^{1}K^{1}Cl + ^{+1*2}H_{2}^{2}O 
To count the total number of oxidized Cl atoms at the reactant side, sum up the total number of Cl in 2KCl and 2CrCl3 which are equal to eight. Thus, put this figure in equation such as
^{+1*2}K_{2} ^{+6*2}Cr_{2}^{2*7}O_{7} + 6^{1}H^{1}CL + 6HCL
→
2^{+3}Cr^{1*3}Cl_{3} + ^{0}Cl_{2} + 2^{1}K^{1}Cl + ^{+1*2}H_{2}^{2}O 
Since the reactant side contains seven oxygen atoms, multiply H2O with seven.
^{+1*2}K_{2} ^{+6*2}Cr_{2}^{2*7}O_{7} + 6^{1}H^{1}CL + 6HCL
→
2^{+3}Cr^{1*3}Cl_{3} + ^{0}Cl_{2} + 2^{1}K^{1}Cl + 7^{+1*2}H_{2}^{2}O 
Finally, simplify the equation by removing the oxidation states and summing up the total unoxidized and oxidized Cl molecules such as
K_{2}Cr_{2}O_{7} + 12HCl
→
2CrCl_{3} + C_{l2} +2KCl + 7H_{2}O
The balancing chemical equations calculator can be used for this purpose as well as to ensure the correctness of the final equation.
How to balance chemical equation by ionelectron method
This method is also being used to balance redox equations which possess ions and an aqueous medium. The basic principle is to eliminate those ions which do not undergo a change in oxidation states by giving or gaining valence electrons.
We consider only those which show the change in oxidation states either by giving or gaining electrons in the valence shell. Therefore, the method is being called the ionelectron method. To understand the method, let us discuss the redox reaction between HCl and KMnO4 in which Cl and MnO4 change oxidation state such as
Cl^{} + MnO_{4}^{} → Cl_{2}^{0} + Mn^{2+}

Since in this reaction, Cl is being oxidized by losing electrons while MnO4 is being reduced by gaining these electrons. Thus, we will divide the equation into two parts  oxidizing part and reducing part such as
Oxidizing part :
Cl^{} → Cl_{2}^{0}
Reducing part :
MnO_{4}^{} → Mn^{2+}

Since there are two atoms of Cl on the product side, the oxidation equation would become
2Cl^{} → Cl_{2}^{0}

Since the reaction is taking place in an acidic medium, add H^{+} at L.H.S and balance these H^{+} ions by adding H_{2}O at R.H.S such as
8H^{+} + MnO_{4}^{} → Mn^{2+} + 4H_{2}O

Add the number of electron lost and gained by each part such as
2Cl^{} → Cl_{2}^{0} + 2e
8H^{+} + MnO_{4}^{} + 5e → Mn^{2+} + 4H_{2}O

To balance the number of electrons lost and gained in both parts, multiply both parts with a specific coefficient to get the final equation. Since the reduction part is getting 5e, multiply oxidation part with 5 and since oxidation part is losing 2 electrons, multiply reduction part with 2.
(2Cl^{} → Cl_{2}^{0} + 2e)*5
(8H^{+} + MnO_{4}^{} + 5e → Mn^{2+} + 4H_{2}O) * 2

The final equation would be obtained by summing up both parts.
10Cl^{} + 16H + 2MnO_{4}^{} → 5Cl_{2} + 2Mn^{2+} + 8H_{2}O
The same method is being used for a reaction occurring in basic media. However, you have to add OH ions to balance both sides other than H+ ions. For further convenience, let us discuss an example of balancing equations in the basic medium as well.
How to balance equation in the basic medium?
Letus consider a redox reaction taking place in a basic medium such as follows
H_{2}O + MnO_{4}^{1} + C_{2}O_{4}^{2} → MnO_{2} + CO_{2} + OH^{1}

Since
MnO_{4}^{1}
and
C_{2}O_{4}^{2}
are showing changes in oxidation state, we would consider only them.

Since
C_{2}O_{4}^{2}
is being oxidized by giving electrons, it would constitute the oxidation part. Moreover, since
MnO_{4}^{1}
is getting reduced by gaining electrons, it would constitute a reduction part.

To balance the number of C atoms at both sides, change the oxidation part into
C_{2}O_{4}^{2} → 2CO_{2}

To balance the reduction part, add two H_{2}O molecules on the reactant side and balance it on the product side by adding four OH ions such as
MnO_{4}^{1} + 2H_{2}O → MnO + 4OH^{}

Add the number of electrons gained and lose in both parts such as
C_{2} O^{2}_{4} → 2CO_{2} + 2e
MnO_{4}^{1} + 2H_{2}O + 3e → MnO + 4OH^{}

Balance the number of electrons at both sides by multiplying the reduction part with 2 and the oxidation part with 3. Then add both parts to get the final equation such as
2MnO_{4}^{1} + 3C_{2}O_{4}^{2} +4H_{2}O → 2MnO + 6CO_{2} + 8OH^{}
To get accurate and fast results, you can use an advanced balancing chemical equation calculator to balance such redox equations as well.
How to use Equation Balancer?
We have made this balance chemical equations calculator to balance chemical equations instantly online. If you are using balancing chemical equations calculator with steps, you'll know how easy it is to operate.
There are many sample equations in this chemical equation balance calculator so that you can practice and balance equations. Also there is a chemistry periodic table under the chemical equation balancer calculator so that you can add values from there.
Once you entered your equation in the field, the balancing equations calculator will balance your equation immediately. Share this with your class mates & others so that they could also get benefit from it.