Question

51. Balance the following equation. Also name the products formed.
${\mathrm{NaNO}}_3\to {\mathrm{NaNO}}_2+{\mathrm{O}}_2$

Answer 1

Step 1:- List the number of atoms of different element present in the unbalanced equation

The given chemical equation is as follows:

${\mathrm{NaNO}}_3\to {\mathrm{NaNO}}_2+{\mathrm{O}}_2$

Element	Number of atoms on reactant side	Number of atom on product side
Na	1	1
N	1	1
O	3	4

Step 2:- Start balancing with the compound that contains maximum number of atoms (reactant or product)

• Here, NaNO₃ is having maximum number of atom and after that check which element have maximum number of atoms.
• There Oxygen (O₂) is having maximum number of atom. check if it balanced or not means number of atom is equal to the number of reactant.

Atoms of oxygen	In reactant	In product
Initial	3 (in NaNO3)	2 in (NaNO2) and 2 in O2 so, total 4 oxygen atoms
Final (Balanced)	4 NaNO3 = 12 Oxygen atoms	4 NaNO2 + 2O2 = 12 Oxygen atoms

Now oxygen is balanced.

Step 3:- Balance the other atoms present in the equation

• Nitrogen (N₂) in LHS is 1 and RHS is also 1. so, Nitrogen is balanced.
• Sodium(Na) in LHS is 1 in RHS is also 1 so, sodium is balanced.
• To equalize the number of atoms we can only add coefficients for e.g. “4NaNO₃” instead of “NaNO_3×4”.

Therefore, the balanced chemical equation is:

$\underset{\mathbf{Sodium}\mathbf{}\mathbf{Nitrate}\mathbf{}}{4{\mathrm{NaNO}}_3}\to \underset{\mathbf{Sodium}\mathbf{}\mathbf{nitride}}{4{\mathrm{NaNO}}_2}+\underset{\mathbf{Oxygen}}{2{\mathrm{O}}_2}$

The product formed in the reaction is sodium nitride (NaNO₂).