The quantity of antimony in a sample can be determined by an oxidation-reduction titration with an oxidizing..

The quantity of antimony in a sample can be determined by an oxidation-reduction titration with an oxidizing agent. A 6.17-g sample of stibnite, an ore of antimony, is dissolved in hot, concentrated HCl(aq) and passed over a reducing agent so that all the antimony is in the form Sb3 (aq). The Sb3 (aq) is completely oxidized by 28.0 mL of a 0.125 M aqueous solution of KBrO3(aq). The unbalanced equation for the reaction is:

BrO3^-(aq) + Sb^3+ ==> Br^- (aq) + Sb^5+(aq) (Unbalanced)

Calculate the amount of antimony in the sample and its percentage in the ore.

Didn't learn this one in class either and the tutor only confused me more. Please help.
Thanks a bunch!

First we have to balance the equation. We do that by balancing each half-reaction separately, then combining them.

BrO3- ==> Br- . . .Add 3 H2O to the right side to balance O.
BrO3- ==> Br- + 3H2O . . .add 6H+ (acidic solution) to the left side to balance H.
BrO3- + 6H+ ==> Br- + 3H2O . . .add 6e- to the left side to balance the charge.
BrO3- + 6H+ + 6e- ==> Br- + 3H2O

Sb3+ ==> Sb5+ . . .Add 2e- to the right side to balance the charge.
Sb3+ ==> Sb5+ + 2e-

Multiply the Sb equation by 3 to give 6e- on the right side. Then add the new equation to the BrO3- equation; the 6e- will cancel.

.3Sb3+ ==> 3Sb5+ + 6e-
+BrO3- + 6H+ + 6e- ==> Br- + 3H2O
===================================
3Sb3+ + BrO3- + 6H+ ==> 3Sb5+ + Br- + 3H2O


moles BrO3- added = M BrO3- x L BrO3- = (0.125)(0.0280) = 0.00350 moles BrO3-

The balanced equation tells us that 1 mole of BrO3- reacts with 3 moles of Sb3+.

0.00350 moles BrO3- x (3 mole Sb3+ / 1 mole BrO3-) = 0.0105 moles Sb3+

From the periodic table, the molar mass of Sb (or Sb3+; the 3 missing electrons have very little effect on the mass ) = 121.8 g.

0.0105 moles Sb x (121.8 g Sb / 1 mole Sb) = 1.28 g Sb

%Sb = (g Sb / g ore) x 100 = (1.28 / 6.17) x 100 = 20.7 %Sb