Question

Having trouble balancing a chemical equation, how would I balance this?
$N{H_3} + {O_2} \to NO + {H_2}O$

Answer 1

We will balance this chemical reaction by Oxidation number method. After balancing the oxidation numbers of the atoms at both sides, we need to balance the remaining atoms O and H by hit and trial method. We should follow the procedure of balancing stepwise in order to arrive at the correct result.

Complete step-by-step answer:
We will balance this chemical reaction by Oxidation number method.
The balancing is done in following steps:$4N{H_3} + 5{O_2} \to 4NO + 5{H_2}O$
i) We will first write the oxidation number of each atom in the equation
${N^{ - 3}}{H^{^{ + 1}}}_3\, + \,{O^0}_{2\,} \to \,{N^{ + 2}}{O^{ - 2}}\, + \,{H^{ + 1}}_2{O^{ - 2}}$ $- (1)$

ii) We will then identify the atoms which undergo change in oxidation number.
${N^{ - 3}}{H_3}\, + \,{O_2}\, \to \,{N^{ + 2}}O\, + \,{H_2}O$
$N{H_3}\, + \,{O^0}_2\, \to \,N{O^{ - 2}}\, + \,{H_2}O$
We can easily figure out from equation $- (1)$ that nitrogen undergoes an increase in oxidation number while oxygen undergoes a decrease in oxidation number.

iii) We will then calculate the increase and decrease in oxidation number with respect to reactant atoms.
${N^{ - 3}}{H^{^{ + 1}}}_3\, + \,{O^0}_{2\,} \to \,{N^{ + 2}}{O^{ - 2}}\, + \,{H^{ + 1}}_2{O^{ - 2}}$
Increase in oxidation number of Nitrogen is from $- 3\,in\,N{H_3}\,to\, + 2\,in\,NO$
Decrease in oxidation number of Oxygen is from $0\,in\,{O_{2\,}}\,to\, - 2\,in\,{H_2}O$
Since, two oxygen atoms are there, we should multiply it by 2 to account for the total change in the oxidation number.

iv) The, we need to equate the increase and decrease in oxidation number.
Increase in oxidation number $= 2 + 3 = 5$
Decrease in oxidation number of Oxygen$= 2(0 + 2) = 4$
Multiplying oxidation half by $4$ and reduction half by $5$.
$4N{H_3} + 5{O_2} \to 4NO + 5{H_2}O$ $- (2)$
Now, the oxidation numbers are balanced. Our next job is to balance the atoms present in the reaction.

v) Now, our job is to balance the number of N, H and O atoms by hit and trial method.
We are writing the total number of N,H and O on both sides.

No. of atoms	Left hand side	Right hand side
Nitrogen	$4$	$4$
Hydrogen	$12$	$10$
Oxygen	$10$	$9$

The no. of atoms of each element should be the same at both the sides. We can easily figure that there is one less water molecule on the right side than the left side. Hence, we need to add one water molecule in equation $- (2)$

Now, the balanced chemical equation after adding one water molecule in equation $- (2)$in the right hand side becomes as follows:
$4N{H_3}\, + \,5{O_2} \to \,4NO + \,6{H_2}O$
This is the required balanced equation.
$4N{H_3}\, + \,5{O_2} \to \,4NO + \,6{H_2}O$

Note: The balancing of a chemical reaction is done to equalize the atoms of different elements or compounds which are involved in it. It is done to fulfil the requirement of the law of conservation of mass. We need to identify carefully the oxidation number of the reactants and products. Also identify carefully the elements undergoing increase and decrease in oxidation number.