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 6.05-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+ (ac is completely oxidized by 31.2 mL of a 0.110 M-aqueous solution of KBrO_3(aq). The unbalanced equation for the reaction is BrO^- _3 (aq) + Sb^3+ (aq) rightarrow Br^- (aq) Sb^5+ (aq) (unbalanced) Calculate the amount of antimony in the sample and its percentage in the ore.

Explanation / Answer

the balanced equation will be

BrO3- + 3 Sb3+ + 6 H+ --> Br{-} + 3 Sb5+ + 3 H2O

As per equation one mole of KBrO3 will be reduced by 3 moles of antimony ions

Moles of KBrO3 = Molarity X volume = 0.110 X 31.2 = 3.432 millmoles

Moles of Antimony used = 3 X 3.432 millomoles= 10.296 millimoles

Mass of antimony = moles X atomic weight = 10.296 X 121.76 = 1.2536 grams

Mass of ore = 6.05 grams

% of antimony = mass of antimony X 100 / mass of ore = 1.2536 X 100 / 6.05= 20.7%

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