The quantity of antimony in a sample can be determined by an oxidation-reducatio

Question

The quantity of antimony in a sample can be determined by an oxidation-reducation titration with an oxidizing agent. A 7.79-g sample of stibnite, and 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 35.4 mL of a 0.125M aqueous solution of KBrO3(aq). The unbalanced equation for the reaction is BrO3- (aq) + Sb3+ (aq) --> Br- (aq) +Sb5+ (aq). Calculate the amount of antimony in the sample and its percentage in the ore.

Explanation / Answer

2BrO(3)^- (aq) + 6Sb^3+ (aq) --> 2Br^- (aq) +6 Sb^5+ (aq)

let 'w' gm be the weight of antimony which reacted with KBrO3.

Molar mass of antimony is 121.75 gm

If you equate the equivalents of Sb^3+ and BrO3-

(w/121.75) x 2=(35.4 ml/1000)(0.125 M) x 6

w=1.616 g
Percentage of antimony in the ore is = (1.616 / 7.79)*100 = 20.75%

amount of antimony in the sample is 1.315 gm and percentage is 20.75%

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