Balancing Chemical Equations - Definition, Method, Formulas, Examples, FAQs

Chemical Equation

A chemical equation is a symbol for a chemical reaction in which the reactants and products are represented by their chemical formulae.

$2 \mathrm{H}_2+\mathrm{O}_2 \rightarrow 2 \mathrm{H}_2 \mathrm{O}$ is an example of a chemical equation that depicts the reaction between Hydrogen and oxygen to generate water.

The part of the chemical equation to the left of the ‘→' sign is the reactant side, while the part to the right of the arrow symbol is the product side.

Stoichiometric Coefficient

The total number of molecules of a chemical species that participate in a chemical reaction is described by a stoichiometric coefficient.

It calculates the ratio between the responding species and the reaction products.

The stoichiometric coefficients of O₂ and H₂O in the reaction $\mathrm{CH}_4+2 \mathrm{O}_2 \rightarrow \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}$ are 2 and 1, respectively, in the equation $\mathrm{CH}_4+2 \mathrm{O}_2 \rightarrow \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}$.

Stoichiometric coefficients are assigned in a way that balances the total number of atoms of an element on the reactant and product sides when balancing chemical equations.

Balancing Of Chemical Equation Method

When it comes to balancing chemical equations, the first step is to acquire the entire imbalanced equation. The combustion process between propane and oxygen is used as an example to demonstrate this method.

Step 1: Create an unbalanced equation using the chemical formulae of the reactants and products (if it is not already provided).

The molecular formula for propane is C₃H₈. It creates carbon dioxide (CO₂) and water when it comes into touch with oxygen (O₂) (H₂O).

$\mathrm{C}_3 \mathrm{H}_8+\mathrm{O}_2 \rightarrow 3 \mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}$

Step 2: On the reactant and product sides, compare the total number of atoms of each element. In this situation, the number of atoms on each side can be tabulated as follows.

Step 3: On the reactant and product sides, stoichiometric factors are now applied to compounds containing an element with a specific number of atoms.

The coefficient must balance the number of atoms on each side.

The hydrogen and oxygen atoms' stoichiometric coefficients are normally assigned last.

It's time to adjust the number of atoms in the reactant and product elements.

It's important to note that the total number of atoms of an element in one molecule of a species is calculated by multiplying the stoichiometric coefficient by the total number of atoms of that element in that molecule.

When the CO₂ molecule is assigned the coefficient 3, the total number of oxygen atoms in CO₂ becomes 6. The coefficient is first assigned to in this case.

Step 4: Step 3 is repeated until the reactant and product sides of the reactive elements have the same number of atoms. In this scenario, hydrogen is balanced next. The chemical equation is altered in this manner.

Now that the hydrogen atoms are balanced, the next element to be balanced is oxygen. There are 10 oxygen atoms on the product side, implying that the reactant side must also contain 10 oxygen atoms.

Each O₂ molecule contains 2 oxygen atoms. Therefore, the stoichiometric coefficient that must be assigned to the O₂ molecule is 5. The updated chemical equation is tabulated below.

Step 5: After all of the individual elements have been balanced, the total number of atoms in each element on the reactant and product sides is compared once more.

The chemical equation is said to be balanced if there are no inequalities.

Every element now has the same number of atoms on the reactant and product sides in this example.

As a result, $\mathrm{C}_3 \mathrm{H}_8+5 \mathrm{O}_2 \rightarrow 3 \mathrm{CO}_2+4 \mathrm{H}_2 \mathrm{O}$ is the balanced chemical equation.

EXAMPLE:

$\mathrm{Al}+\mathrm{O}_2 \rightarrow \mathrm{Al}_2 \mathrm{O}_3$ is an unbalanced chemical equation.

Traditional method

The reaction can be balanced using the old method as follows:

The aluminium atoms are balanced first.

$\mathrm{Al}+\mathrm{O}_2 \rightarrow \mathrm{Al}_2 \mathrm{O}_3$

There are two oxygen atoms on the reactant side and three on the product side, so the oxygen atoms must be balanced now. As a result, there must be three O2 molecules that produce two Al2O3 atoms.

$2 \mathrm{Al}+3 \mathrm{O}_2 \rightarrow 2 \mathrm{Al}_2 \mathrm{O}_3$ is the chemical conversion equation.

Because the number of aluminium atoms on the product side has doubled, the number on the reactant side must have doubled as well

Balanced chemical equation is found to be $2 \mathrm{Al}+3 \mathrm{O}_2 \rightarrow 2 \mathrm{Al}_2 \mathrm{O}_3$

Some solved Examples

Question.1 The number of moles of Br₂ produced when two moles of potassium permanganate are treated with excess potassium bromide in an aqueous acid medium is:

1) 1

2) 3

3) 2

4) 4

Solution:

$
\mathrm{MnO}_4^{-}+\mathrm{Br}^{-} \rightarrow \mathrm{Br}_2+\mathrm{Mn}^{2+}
$
Oxidation half

$
\left[2 \mathrm{Br}^{-} \rightarrow \mathrm{Br}_2+2 e^{-}\right] \times 5
$
Reduction half

$
\left[\mathrm{SH}^{+}+\mathrm{MnO}_4^{-}+\mathrm{Se}^{-} \rightarrow \mathrm{Mn}^{+2}+4 \mathrm{H}_2 \mathrm{O}\right] \times 2
$
Adding both half reaction

$
\begin{gathered}
2 \mathrm{MnO}_4^{-}+10 \mathrm{Br}^{-}+16 \mathrm{H}^{+} \rightarrow 5 \mathrm{Br}_2+2 \mathrm{Mn}^{2+}+\mathrm{SH}_2 \mathrm{O} \\
2
\end{gathered}
$
2 moles of $\mathrm{MnO}_4^{-}$will produce 5 moles of $\mathrm{Br}_2$
Hence answer is 5 .

Hence, the answer is the option (2).

Question.2 In button cells widely used in watches, number of correct statements from the followings:
a) Redox reaction is observed.
b) 2 mole electrons are exchanged for 1 mole of $Z n$ metal
c) Ag is converted to $\mathrm{Ag}_2 \mathrm{O}$
d) $Z n$ is converted to $\mathrm{Zn}^{2+}$

Solution:

$
\mathrm{Zn}(s)+\mathrm{Ag}_2 \mathrm{O}(s)+\mathrm{H}_2 \mathrm{O}(l) \longrightarrow \mathrm{Zn}^{2+}+2 \mathrm{Ag}(\mathrm{~s})+2 \mathrm{OH}^{-}
$
2 mole $e^{-}$are utilized.
Redox reaction is observed.
2 mole electrons are exchanged for 1 mole of $Z n$ metal
$Z n$ is converted to $\mathrm{Zn}^{2+}$
Hence, the answer is (3).

Question.3 What is the number of electrons used in the reduction of 1mol of nitrate in the presence of acid?

1) 2

2) 5

3) 10

4) 9

Solution:

$
2 \mathrm{NO}^{-3}+10 \mathrm{e}^{-}+12 \mathrm{H}^{+} \rightarrow \mathrm{N}_2+6 \mathrm{H}_2 \mathrm{O}
$
Hence, the answer is option (3).

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