Question

How do you balance $KCl{{O}_{3}}\to KCl+{{O}_{2}}$ ? Is Cl chlorine?

Answer 1

Hint: To balance the given chemical reaction, we need to adjust and find suitable stoichiometric coefficients which will satisfy the equation. The stoichiometric coefficients can be either a whole number or a fraction.

Complete step by step answer:
- Balancing an equation is the prime most task when we further solve an illustration related to the equation. Proper stoichiometric coefficients will easily balance any chemical reaction; we just need to apply hit and trial methods.
- Balancing a chemical reaction is similar to that of solving any linear mathematical expression. We calculate the number of atoms on the both sides of the equation and assign the variables to unknown quantities; solving this gives us stoichiometric coefficients.
- Let us now balance the given equation;
$KCl{{O}_{3}}\to KCl+{{O}_{2}}$
We have,
LHS: One K, one Cl and 3 O
RHS: One K, one Cl and 2 O
Assign the variables to the unknown coefficients;
$xKCl{{O}_{3}}\to yKCl+z{{O}_{2}}$
- Thus, we obtain linear system of equation as;
K: x = y gives, x – y = 0
Cl: x = y gives, x – y = 0
O: 3x = 2z gives, 3x – 2z = 0
Solving these equations, we get,
x = $\dfrac{2}{3}$ , y = $\dfrac{2}{3}$ and z = 1
Multiplying by common multiple we get,
$2KCl{{O}_{3}}\to 2KCl+3{{O}_{2}}$
And, yes Cl is chlorine.

Note: There is a method which gives a direction to balance a chemical reaction which describes that we need to balance everything in an equation firstly before O and H atoms; then O and lastly H.
- As, the given reaction is the redox reaction we can use either the oxidation number method or ion electron method to balance the same.