Question

How to balance chemical equations using algebraic methods?

Answer 1

Hint: Using simple algebra and the concept of conservation of mass or charge, we can balance any chemical equation. Use ‘conservation of mass’ for molecular equations and ‘conservation of charge’ for ionic equations.

Complete answer:
We will take this unbalanced molecular chemical equation as an example:
\[{\text{KMn}}{{\text{O}}_4} + \,{\text{N}}{{\text{H}}_3}{\text{ }} \to {\text{ Mn}}{{\text{O}}_2} + {\text{ KOH}} + {\text{ }}{{\text{N}}_2} + {\text{ }}{{\text{H}}_2}{\text{O}}\]
Assign six different coefficients (\[a\] to \[f\]) to all six chemical species. Now the above equation becomes:
\[a\,{\text{KMn}}{{\text{O}}_4} + \,b\,{\text{N}}{{\text{H}}_3}{\text{ }} \to {\text{ }}c\,{\text{Mn}}{{\text{O}}_2} + {\text{ }}d\,{\text{KOH}} + {\text{ }}e\,{{\text{N}}_2} + {\text{ }}f\,{{\text{H}}_2}{\text{O}}\]
Since, this is a molecular equation, we will use ‘conservation of mass’ here.
According to conservation of mass, the number of each atom in both sides of any chemical equation must be equal. Now, using this concept, we will equate the numbers of each atom one by one.
\[{\text{K: }}a = d\]
\[{\text{Mn: }}a = c\]
\[{\text{O: }}4a = 2c + d + f\]
\[{\text{N: }}b = 2e\]
\[{\text{H: 3}}b = d + 2f\]
To solve these equations, we have to reduce the number of variables.
It is clear that, \[a = c = d\] ... (i)
Substituting this in the equation for \[{\text{O}}\]atom, we get:
\[4a = 2a + a + f \Rightarrow a = f\] ... (ii)
Substituting \[d\]and \[f\]with \[a\] in the equation for \[{\text{H}}\] we get:
\[{\text{3}}b = a + 2a \Rightarrow b = a\] ... (iii)
Substituting this in the equation for \[{\text{N}}\] atom, we get:
\[a = 2e\] ... (iv)
From algebraic equations (i), (ii), (iii) and (iv):
\[a = b = c = d = 2e = f\]
By putting the smallest possible integer for coefficient, \[e = 1\], we can get the values of all other required coefficient as:
\[a = 2,\,b = 2,\,c = 2,\,d = 2,\,f = 2\]
Hence, the balanced chemical equation becomes:
\[2\,{\text{KMn}}{{\text{O}}_4} + \,2\,{\text{N}}{{\text{H}}_3}{\text{ }} \to {\text{ }}2\,{\text{Mn}}{{\text{O}}_2} + {\text{ }}2\,{\text{KOH}} + {\text{ }}{{\text{N}}_2} + {\text{ }}2\,{{\text{H}}_2}{\text{O}}\]

Note: If the equations in variables become very complex to solve, then we can reduce the number of variables by using our logic. Like in the above example, for atoms \[{\text{K}}\] and \[{\text{Mn}}\], \[a = c = d\]. Therefore, instead of three variables, we can take only one variable to simplify the calculations.