step 1: balance redox reaction

reaction:

BrO3-(aq) + Br-(aq) → Br2(l)

half-reactions

BrO3-(aq) → Br2(l)

balance non-oxygen and non-hydrogen: 2 BrO3-(aq) → Br2(l)

balance oxygen: 2 BrO3-(aq) → Br2(l) + 6 H2O(l)

balance hydrogen: 2 BrO3-(aq) + 12 H+(aq) → Br2(l) + 6 H2O(l)

balance charges: 2 BrO3-(aq) + 12 H+(aq) + 10 e-→ Br2(l) + 6 H2O(l)

Br-(aq) → Br2(l)

balance non-oxygen and non-hydrogen: 2 Br-(aq) → Br2(l)

balance charges: 2 Br-(aq) → Br2(l) + 2 e-

balance electrons

2 BrO3-(aq) + 12 H+(aq) + 10 e-→ Br2(l) + 6 H2O(l)

5 x [2 Br-(aq) → Br2(l) + 2 e-]

combine reactions

2 BrO3-(aq) + 12 H+(aq) + 10 Br-(aq) → 6 Br2(l) + 6 H2O(l)

step 2:

total volume: 25.0 mL + 150.0 mL = 175.0 mL = 0.1750 L

initial moles

0.1750 L x (0.275 mol/L NaBrO₃ ) x (1 mol BrO3-/1 mol NaBrO3) = 0.048125 mol BrO₃-

0.1750 L x (0.175 mol/L KBr) x (1 mol Br-/1 mol KBr) = 0.030625 mol Br-

0.1750 L x (0.450 mol/L HClO₃) x (1 mol H+/1 mol HClO3) = 0.07875 mol H+

determine limiting reagent

0.048125 mol BrO₃- x (6 mol Br2/2 mol BrO3-) = 0.144375 mol Br2

0.030625 mol Br- x (6 mol Br2/10 mol Br-) = 0.018375 mol Br2

0.07875 mol H+ x (6 mol Br2/12 mol H+) = 0.039375 mol Br2

Br- is the limiting reagent

2 BrO3-(aq) + 12 H+(aq) + 10 Br-(aq) → 6 Br2(l) + 6 H2O(l)

I 0.048125 0.07875 0.030625

C -2x -12x -10x

since Br- is the limiting reagent, 10x = 0.030625 → x = 0.0030625

F 0.042 0.042 0

0.1750 L x (0.275 mol/L NaBrO₃ ) x (1 mol Na+/1 mol NaBrO3) = 0.048125 mol Na+

0.1750 L x (0.175 mol/L KBr) x (1 mol K+/1 mol KBr) = 0.030625 mol K+

0.1750 L x (0.450 mol/L HClO₃) x (1 mol ClO3-/1 mol HClO3) = 0.07875 mol ClO3-

check for charge neutrality

total moles of +charge = 0.048125 mol + 0.030625 mol + 0.042 mol = 0.12075

total moles of -charge = 0.07875 mol + 0.042 mol = 0.12075

The charges balance and there is electrical neutrality in the solution after the reaction.

0.048125 mol Na+/0.1750 L = 0.275 M

0.030625 mol K+/0.1750 L = 0.175 M

0.07875 mol ClO3-/0.1750 L = 0.450 M

0.042 mol BrO3-/0.1750 L = 0.240 M

0.042 mol H+/0.1750 L = 0.240 M

answer: 0.275 M Na+, 0.175 M K+, 0.450 M ClO3−, 0.240 M BrO3−, 0.240 M H+, 0 M Br−