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
