Question

The quantity of antimony in a sample can be determined by an oxidation-reduction titration with an oxidizing agent. A 9.95-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 41.8 mL of a 0.135 M aqueous solution of KBrO3(aq). The unbalanced equation for the reaction is BrO3-(aq)+Sb3+(aq)->Br-(aq)+Sb5+(aq) UNBALANCED Calculate the amount of antimony in the sample and its percentage in the ore.

Answer 1

First we need to balance the reaction

Oxidation Half

Reduction Half

Hence the overall reaction will be

Number of moles of KBrO3 = volume of solution in L * molarity

=> 41.8/1000 * 0.135

=> 0.005643 moles

Moles of Sb will be 0.005643 * 3 = 0.016929 moles

Molar mass of antimony = 121.76 gm/mol

Mass of antimony = 0.016929 mol * 121.76 gm/mol = 2.061 grams

Percent of antimony in the ore = Mass of antimony/ Mass of stibnite sample * 100

=> 2.061/9.95 * 100

=> 20.713%