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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 through an oxidation-reduction titration using an oxidizing agent. A 7.99-g sample of stibnite, which is an ore of antimony, is dissolved in hot, concentrated HCl(aq) and then passed over a reducing agent to ensure that all the antimony is in the form of Sb³⁺(aq). This 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⁻ + Sb³⁺ → Br⁻ + Sb⁵⁺.

Could you please help me calculate the amount of antimony in the sample and its percentage in the ore? I am feeling quite confused about this process.

2 Answers

A
Anonymous

Dec 31, 2024

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

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