Calculating chemical equations can be tricky especially when the equations are complex i.e when two compounds give rise to three or more compounds. To help you out in such cases, Testbook provides you with a high-speed chemical equation calculator with steps. Along with it, the article also focuses on tricks and tips for balancing chemical equations and some brainstorming FAQs.

Check out the Balancing Chemical Equations Calculator Step by Step Guide below

The following steps should be followed to use the balancing chemical equation calculator:

Step 1: First, enter the chemical formula in the text box.

Step 2: To obtain a balanced equation, now click on "Balance".

Step 3: A new window will open to display the balanced equation, structure, and equilibrium constant for the specified chemical equation.

Why use the Balancing Chemical Equation Calculator?

Atoms only rearrange themselves during a chemical reaction. During a chemical reaction, it cannot be created or destroyed. To comply with the law of conservation of matter, which stipulates that matter cannot be created or destroyed in a closed system, chemical equations must be balanced.

The balance of a chemical equation is controlled by the rule of conservation of mass. This law states that in a chemical reaction, mass cannot be created or destroyed. As a result, the total mass of the elements or molecules present on the reactant side must match the total mass of the elements or molecules present on the product side. The law of conservation is not applicable if the chemical equation is unbalanced.

Hence, the balanced chemical equation calculator gives the balanced chemical equation.

What is a Balanced Chemical Equation?

A chemical equation is a way to depict a chemical reaction using different substances.

An equation for a chemical reaction is said to be balanced if both the reactants and the products have the same number of atoms and the total charge for each component of the reaction. In other words, both sides of the reaction have an equal balance of mass and charge.

Unbalanced Equation

The reactants and products of a chemical reaction are listed in an imbalanced chemical equation, but the amounts necessary to meet the conservation of mass are not specified.

The components that initiate a reaction are called reactants. Products are substances that remain after a reaction. Subscripts are a component of the chemical formulas for the reactants and products that show how many atoms of the previous element exist.

In a balanced chemical equation, a little whole number called a coefficient is placed in front of the formula.

For instance, the mass balance of the following equation for the reaction between Iron Oxide and Carbon to produce iron and Carbon Dioxide is:

\(\mathrm{Fe}_{2} \mathrm{O}_{3}+\mathrm{C} \longrightarrow \mathrm{Fe}+\mathrm{CO}_{2}\)

Because there are no ions on either side of the equation, the equation is balanced for charge (net neutral charge).

Two iron atoms are present in the equation's reactants side (to the left of the arrow), yet only one iron atom is present in the equation's products side (right of the arrow). You can know the equation is unbalanced even if you don't add up the quantities of the other elements.

The equation must be balanced so that each type of atom appears in equal amounts on both the left and right sides of the arrow. This is accomplished by altering the compounds' coefficients (numbers placed in front of compound formulas). In this example, the subscripts (tiny numbers to the right of some atoms, including iron and oxygen) are never altered. The chemical identification of the compound would change if the subscripts were changed.

The balanced equation is:

\(2 \mathrm{Fe}_{2} \mathrm{O}_{3}+3 \mathrm{C} \rightarrow 4 \mathrm{Fe}+3 \mathrm{CO}_{2}\)

There are 4 Fe, 6 O, and 3 C atoms on both the left and right sides of the equation. When balancing equations, it's a good practice to multiply the subscript of each atom by the coefficient to double-check your work.

Algebraic Balancing Method

By allocating algebraic variables as stoichiometric coefficients to each species in the imbalanced chemical equation, chemical equations can be balanced. To determine the values of each stoichiometric coefficient, these variables are used as variables in mathematical equations that are then solved. The reaction between glucose and oxygen that produces carbon dioxide and water has been used as an example to further understand this process.

Step 1: The chemical formulas for the reactants and products must be written down in order to produce the imbalanced chemical equation.

In this instance, glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) and oxygen (O2) are the reactants, whereas carbon dioxide (CO2) and water are the products (H2O)

C6H12O6 + O2 → CO2 + H2O is the unbalanced chemical equation.

Step 2: In the imbalanced chemical equation, algebraic variables are now assigned to each species (as stoichiometric coefficients). The equation in this illustration can be expressed as follows.

aC6H12O6 + bO2 → cCO2 + dH2O

To balance each component of the reaction, a set of equations must now be created (between the reactant and product sides).

The following equations can be created using this example.

The Carbon Equation: 'a' C6H12O6 molecules on the reactant side will have '6a' carbon atoms. 'c' molecules of CO2 will have 'c' carbon atoms on the product side.

The only species that contain carbon in this equation are C6H12O6 and CO2.

Consequently, the equation for carbon can be written as follows:

6a = c

The Hydrogen Equatio: C6H12O6 and H2 'a' molecules of C6H12O6 contain '12a' hydrogen atoms each, whereas 'd' H2O molecules contain '2d' hydrogen atoms.

These are the species that contain hydrogen in this equation.

As a result, 12a = 2d is the equation for hydrogen.

By multiplying both sides of this equation by 2, the result is:

6a = d

The Oxygen Equation: In this chemical equation, oxygen is present in every species. As a result, the equation for oxygen can be obtained by applying the following relations:

There are '6a' oxygen atoms for each of the 'a' C6H12O6 molecules.

O2 has a total of "2b" oxygens in each of its "b" molecules.

CO2 molecules have '2c' oxygen atoms in each molecule.

D water molecules each contain d oxygen atoms.

Consequently, the oxygen equation is

6a + 2b = 2c+ d

Step 3: A system of equations is created by listing all of the equations for each element. The set of equations, in this case, is as follows:

6a=c and 6a=d for carbon and hydrogen respectively and 6a + 2b = 2c+ d

The solution with the fewest possible values of the variables is necessary for this system of equations. One of the coefficients is given a value in order to arrive at this solution. In this instance, it is assumed that a has a value of 1. As a result, the equation system is changed as follows:

a = 1

c = 6a = 6*1 = 6

d = 6a = 6

The value of "b" can be found by substituting the values of a, c, and d in the formula

6a + 2b = 2c + d.

6*1 + 2b = 2*6 + 6

2b = 12; b = 6

It's vital to remember that when solving these equations, each variable must be a positive integer. If fractional values are received, each variable must be multiplied by the lowest common denominator across all the variables. Because the variables store the stoichiometric coefficient values, which must be positive integers, this is required.

Step 4: The values of the variables can now be inserted into the chemical equation discovered in step 2 since the smallest value of each variable has been discovered.

Thus, the formula becomes:

C6H12O6 + 6O2 → 6CO2 + 6H2O instead of aC6H12O6 + bO2 → cCO2 + dH2O.

We hope the free online tool in the form of the Chemical Equation Calculator has been useful to you. For more informative and educational content like this, you can download the free Testbook App. Here you get preparation assistance for various competitive examinations in the form of Test Series, Online Classes, Quizzes, and many more.