Question

The quantity of antimony in a sample can be determined by an oxidation-reduction titration with an oxidizing agent. A 6.35-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 27.7 mL of a 0.130 M aqueous solution of KBrO3(aq). The unbalanced equation for the reaction is
BrO3^- + Sb^3+= Br^- + Sb^5+

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

Answer 1

Balance the equation.
moles BrO3^- = M x L = ?
Use the coefficients in the balanced equation to convert moles BrO3^- to moles Sb.
Then grams Sb = mols x molar mass
%Sb = (grams Sb/mass sample)*100 = ?

Answer 2

BrO^3- + 3Sb^3+ + 6H^+ -----> Br^- + 3Sb^5+ +3H2o