The quantity of antimony in a sample can be determined by an oxidation-reduction PLEASE HELP?

The amount of antimony in a sample can be determined by redox titration with an oxidizing agent. A 7.99 g sample of stibine, an ore of antimony, is dissolved in hot concentrated HCl(aq) and passed through a reducing agent so that all the antimony is in the Sb3(aq) form. Sb3(aq) is completely oxidized by 32.4 mL of a 0.140 M aqueous solution of KBrO3(aq). The unbalanced equation for the reaction is

BrO3{-} + 3 Sb{3+} + 6 H{+} → Br{-} + 3 Sb{5+} + 3 H2O

(0.0324 L) x (0.140 mol/L KBrO3) x (3 mol Sb / 1 mol KBrO) x (121.76 g Sb/mol) = 1.66 g Sb

(1.66g) / (7.99g) = 0.208 = 20.8% Sb

I never thought about it much

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