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*