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

Question

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 HCI(aq) and passed over a reducing agent so that all the antimony is in the form Sb^3+(aq). The Sb^3+(aq) is completely oxidized by 29.2 mL of a 0.120 M aqueous solution of KBrO_3 (aq). The unbalanced equation for the reaction is H^+ (aq) + BrO_3^- (aq) + Sb^3+ (aq) rightarrow Br^- (aq) + Sb^5+ (aq) + H_2 O (l) (unbalanced) Calculate the amount of antimony in the sample and its percentage in the ore.

Explanation / Answer

The balanced reaction is :

6H+ (aq) + BrO3^- (aq) + Sb^3+ (aq) -----> Sb^5+ (aq) + Br^- + 3H2O

So, 1 mol Sb^3+ reacts with 1 mol BrO3^-.

moles of BrO3^- added = 0.0292 mL * 0.12 M = 3.504 *10^-3 moles

Moles of Sb^3+ present in the sample = 3.504 *10^-3 moles

Mass of Sb^3 present in the sample = 121.76 g/mol * 3.504 *10^-3 moles = 0.43 gm

% Sb =(0.43 gm/6.18) *100 = 6.9 %

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