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 7.63-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 37.7 mL of a 0.115 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.

Explanation / Answer

BrO3{-} + 3 Sb{3+} + 6 H{+} Br{-} + 3 Sb{5+} + 3 H2O

(0.0377 L) x (0.115 mol/L BrO3{-}) x (3 mol Sb / 1 mol BrO3{-}) x (121.760 g Sb/mol) = 1.584g Sb

(1.584 g Sb) / (7.63 g) = 0.2075 = 20.75% Sb in the ore.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.