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:
Answer:
Br2O + NaOH → NaBrO + NaBr + H2O
