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

The quantity of antimony in a sample can be determined by an oxidation-reduction titration with an oxidizing agent. A 7.99-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 32.4 mL of a 0.140 M aqueous solution of KBrO3(aq). The unbalanced equation for the reaction is

Br3^-+Sb^3+——>Br^-+Sb^5+

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

Please help so confused

2 Answers

  • 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

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

  • Never gave too much thought about that

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