Write the balanced equation for the following:

Phosphorous burns in air to give phosphorous pentoxide.

Balanced chemical equation:

A chemical equation in which the number of atoms of each element on both sides of the equation is equal is called a balanced chemical equation.

Step: 1: Representation of the chemical reaction:

• When phosphorous reacts with air, it forms phosphorous pentoxide.
• Reaction can be written as:

$\underset{\mathrm{Phosphorous}}{\mathrm{P}\left(\mathrm{s}\right)}+\underset{\mathrm{Oxygen}}{{\mathrm{O}}_2\left(\mathrm{s}\right)}\to \underset{\mathrm{Phosphorous}\mathrm{pentaoxide}}{{\mathrm{P}}_2{\mathrm{O}}_5\left(\mathrm{s}\right)}$

Step: 2: Calculating the number of atoms on both sides of the reaction:

Elements	Number of atoms on the reactant side	Number of atoms on the product side
P	1	2
O	2	5

Step: 3: Balancing the uneven numbers of atoms on the reactants and product side:

1. As the number of atoms on both reactant and product sides is unequal. So, balancing of Phosphorous and oxygen can be done by multiplying the required coefficient for them in the reaction.
2. This, can be achieved by multiplying “4” to Phosphorous, and “5” to Oxygen i.e., the reactant side. Also, we need to multiply the product side i.e., Phosphorous pentoxide by 2, to make both the sides equal.

$\underset{\mathrm{Phosphorous}}{4\mathrm{P}\left(\mathrm{s}\right)}+\underset{\mathrm{Oxygen}}{5{\mathrm{O}}_2\left(\mathrm{g}\right)}\to \underset{\mathrm{Phosphorous}\mathrm{pentaoxide}}{2{\mathrm{P}}_2{\mathrm{O}}_5\left(\mathrm{s}\right)}$

Elements	Number of atoms on the reactant side	Number of atoms on the product side
P	4	4
O	10	10

We can see that phosphorous and oxygen atoms are same on both sides.

Now, the equation is balanced.

$\underset{\mathrm{Phosphorous}}{4\mathrm{P}\left(\mathrm{s}\right)}+\underset{\mathrm{Oxygen}}{5{\mathrm{O}}_2\left(\mathrm{g}\right)}\to \underset{\mathrm{Phosphorous}\mathrm{pentaoxide}}{2{\mathrm{P}}_2{\mathrm{O}}_5\left(\mathrm{s}\right)}$