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

The quantity of antimony in a sample can be determined by an oxidation-reduction titration with an oxidizing agent. A 9.51-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 43.2 mL of a 0.125 M aqueous solution of KBrO3(aq). The unbalanced equation for the reaction is

BrO3-(aq)+Sb+3(aq)-->Br-(aq)+Sb+5(aq)

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

here is the balanced equation

6H+(aq) + BrO3¯(aq) + 3Sb3+(aq) ---> Br¯(aq) + 3Sb5+(aq) + 3H2O(l)

moles of bromate in the titration = .125 x .0432 moles = 0.00540 moles

these react with 3 x 0.00540 moles of Sb at the eq point = 0.0162 moles of Sb3+

Molar mass of Sb is 121.7g/mol so mass of Sb =

.0162 x 121.7 g = 1.972 g

% Sb = 1.972 / 9.51 = 20.7%