This is an oxidation-reduction (redox) reaction:

4 BrI + 8 e- → 4 Br-I (reduction)

2 BrI - 8 e- → 2 BrV (oxidation)

Br2O is both an oxidizing and a reducing agent (disproportionation (dismutation)).

### Input interpretation

Br2O + NaOH → NaBrO + NaBr + H2O

Sodium hydroxide sodium bromate sodium bromide water

### Balanced equation

Balance the chemical equation algebraically:

Br2O + NaOH → NaBrO + NaBr + H2O

Add stoichiometric coefficients, c1, to the reactants and products:

c1Br2O + c2NaOH → c3NaBrO + c4NaBr + c5H2O

Set the number of atoms in the reactants equal to the number of atoms in the products for Br, O, H and Na:

Br: 2c1 = c3 + c4

O: c1 + c2 = 3c3 + c5

H: c2 = 2c5

Na: c2 = c3 + c4

Since the coefficients are relative quantities and underdetermined, choose a coefficient to set arbitrarily. To keep the coefficients small, the arbitrary value is ordinarily one. For instance, set c3 = 1 and solve the system of equations for the remaining coefficients:

c1 = 3/2

c2=3

c3=1

c4=2

c5 = 3/2

Multiply by the least common denominator, 2, to eliminate fractional coefficients:

c1 = 3

c2 = 6

c3 = 2

c4 = 4

c5 = 3

Substitute the coefficients into the chemical reaction to obtain the balanced equation: