Question

How‌ ‌would‌ ‌you‌ ‌balance‌ ‌a‌ ‌reaction‌ ‌with‌ ‌a‌ ‌polyatomic‌ ‌ion‌ ‌only‌ ‌on‌ ‌one‌ ‌side?‌ ‌
Example‌ ‌would‌ ‌be‌ ‌having‌ ‌ammonia‌‌ ‌‌${{\text{(N}}{{\text{H}}_{\text{4}}}{\text{)}}_{\text{2}}}$‌‌ ‌‌on‌ ‌one‌ ‌
side‌ ‌and‌ ‌then‌ ‌having‌ ‌nitrogen‌‌ ‌‌${{\text{N}}_{\text{2}}}$‌‌ ‌‌+‌‌ ‌‌water‌‌ ‌‌${{\text{H}}_{\text{2}}}{\text{O}}$‌‌ ‌‌on‌ ‌
the‌ ‌other‌ ‌side.‌ ‌Would‌ ‌you‌ ‌balance‌ ‌each‌ ‌individual‌ ‌atom,‌ ‌being‌ ‌that‌ ‌the‌ ‌polyatomic‌ ‌ion‌ ‌is‌ ‌only‌ ‌on‌ ‌one‌ ‌
side?‌ ‌

Answer 1

To determine the answer we should know what balancing the equation means. Balancing the equation means we have to determine the stoichiometry coefficients of each compound or molecule. The balanced equation follows the law of conversion of mass. The total mass of reactant will be equal to the total mass of the product.

Complete step-by-step answer: We have ammonia ${{\text{(N}}{{\text{H}}_{\text{4}}}{\text{)}}_{\text{2}}}$ on one side and then having nitrogen ${{\text{N}}_{\text{2}}}$ + water ${{\text{H}}_{\text{2}}}{\text{O}}$ so, we can write a reaction as follows:
${{\text{(N}}{{\text{H}}_{\text{4}}}{\text{)}}_{\text{2}}}\, + \,{{\text{O}}_{\text{2}}}\, \to {{\text{N}}_2}\,{\text{ + }}\,\,{{\text{H}}_{\text{2}}}{\text{O}}$
According to law of conversion of mass, the mass cannot be created or destroyed. We can only convert it from one form to another. To follow the law of conversion of mass we have to balance the number of atoms on both sides of the reactions. Polyatomic ions also contains the atoms so, it does not matter that we have polyatomic ions we have to balance the number of atoms.
Nitrogen is two on both sides of the arrow so, it is balanced. Hydrogen atoms are eight on the reactant side and one on the product side so, we will add coefficient four in front of ${{\text{H}}_{\text{2}}}{\text{O}}$ at product side to balance the hydrogen.
${{\text{(N}}{{\text{H}}_{\text{4}}}{\text{)}}_{\text{2}}}\, + \,{{\text{O}}_{\text{2}}}\, \to {{\text{N}}_2}\,{\text{ + }}\,4\,{{\text{H}}_{\text{2}}}{\text{O}}$
Now we have four oxygen atoms on the product side and two on the reactant side so, we will add coefficient $2$ in front of ${{\text{O}}_{\text{2}}}$ on the reactant side to balance the oxygen.
${{\text{(N}}{{\text{H}}_{\text{4}}}{\text{)}}_{\text{2}}}\, + \,2\,{{\text{O}}_{\text{2}}}\, \to {{\text{N}}_2}\,{\text{ + }}\,4\,{{\text{H}}_{\text{2}}}{\text{O}}$

Note: The main purpose to balance a reaction is to follow the law of conversion of mass. Presence of polyatomic ions on one side or both side of the reaction does not affect the way of balancing a reaction. We add the coefficients in front of the atoms or molecule, the coefficients are not added as subscript or superscript. We add the coefficient at the side where fewer atoms are present. The subscript present outside of the bracket represents the number of the atoms present indie the bracket.