This is an oxidation-reduction (redox) reaction:

* 6 Br0 + 6 e- → 6 Br-I (reduction)*

* 2 B0 - 6 e- → 2 BIII (oxidation)*

**Br2 **is an oxidizing agent, **B **is a reducing agent.

### Word equation

* boron + bromine → boron tribromide*

### Input interpretation

* 2B + 3Br2 → BBr3*

* Boron bromine tribromide*

### Balanced equation

Balance the chemical equation algebraically:

* 2B + 3Br2 → BBr3*

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

* 2B + 3Br2 → BBr3*

Set the number of atoms in the reactants equal to the number of atoms in the products for** B **and

*Br:**B: c1 = c3*

*Br: 2c2 = 3c3*

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 ** c1 = 1 **and solve the system of equations for the remaining coefficients:

*c1 = 1*

*c2 = 3/2*

*c3 =1*

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

*c1 = 2*

*c2 = 3 *

*c3 = 2*

Substitute the coefficients into the chemical reaction to obtain the balanced

equation:

** Answer**:

* 2B + 3Br2 → BBr3*