Question

Balance the equation

$\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2\to \mathrm{PbO}+{\mathrm{NO}}_2+{\mathrm{O}}_2$

Answer 1

The reaction can be solved in two steps:

1. Complete the reaction by writing the products.
2. Balance the equation.

1. Complete the reaction by writing the products:

$$\begin{gathered} \mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2\left(\mathrm{s}\right)\to \mathrm{PbO}\left(\mathrm{s}\right)+{\mathrm{NO}}_2\left(\mathrm{l}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right) \\ (\mathrm{lead}\mathrm{nitrate})\left(\mathrm{lead}\right)(\mathrm{nitrogen}\mathrm{dioxide})\left(\mathrm{oxygen}\right) \end{gathered}$$

2. Balancing of the reaction by hit and trial method:

A balanced equation is a chemical reaction equation in which the overall charge and the number of atoms for each element in the reaction are the same for both the reactants and the products.

To balance the chemical equation, begin by multiplying the element or compound which has less number of atoms with a number that would make the number equal on both sides.

$$\begin{gathered} \mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2\left(\mathrm{s}\right)\to \mathrm{PbO}\left(\mathrm{s}\right)+{\mathrm{NO}}_2\left(\mathrm{l}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right) \\ (\mathrm{lead}\mathrm{nitrate})\left(\mathrm{lead}\right)(\mathrm{nitrogen}\mathrm{dioxide})\left(\mathrm{oxygen}\right) \end{gathered}$$

Types of atoms	Number of atoms on reactant side	Number of atoms on product side
Lead	1	1
Nitrogen	2	1
Oxygen	6	5

Balance the equation by multiplying lead oxide with 2 on product side as the number of oxygen atoms are less on product side.

$$\begin{gathered} \mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2\left(\mathrm{s}\right)\to 2\mathrm{PbO}\left(\mathrm{s}\right)+{\mathrm{NO}}_2\left(\mathrm{l}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right) \\ (\mathrm{lead}\mathrm{nitrate})\left(\mathrm{lead}\right)(\mathrm{nitrogen}\mathrm{dioxide})\left(\mathrm{oxygen}\right) \end{gathered}$$

Types of atoms	Number of atoms on reactant side	Number of atoms on product side
Lead	1	2
Nitrogen	2	1
Oxygen	6	6

As the number of atoms are still unequal on both side, multiply lead nitrate with 2 and nitrogen dioxide with 4.

$$\begin{gathered} 2\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2\left(\mathrm{s}\right)\to 2\mathrm{PbO}\left(\mathrm{s}\right)+4{\mathrm{NO}}_2\left(\mathrm{l}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right) \\ (\mathrm{lead}\mathrm{nitrate})\left(\mathrm{lead}\right)(\mathrm{nitrogen}\mathrm{dioxide})\left(\mathrm{oxygen}\right) \end{gathered}$$

Types of atoms	Number of atoms on reactant side	Number of atoms on product side
Lead	2	2
Nitrogen	4	4
Oxygen	12	12

As the number of all the atoms is equal on both sides, the chemical equation is now balanced.

So, the overall complete and balanced chemical equation is

$$\begin{gathered} 2\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2\left(\mathrm{s}\right)\to 2\mathrm{PbO}\left(\mathrm{s}\right)+4{\mathrm{NO}}_2\left(\mathrm{l}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right) \\ (\mathrm{lead}\mathrm{nitrate})\left(\mathrm{lead}\right)(\mathrm{nitrogen}\mathrm{dioxide})\left(\mathrm{oxygen}\right) \end{gathered}$$