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 7.75-g sample of stibnite, an ore of antimony, is dissolved in hot, concentrated CI(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 33.8 mL of a 0.130 M aqueous solution of KBro3(aq). The unbalanced equation for the reaction is (unbalanced) Calculate the amount of antimony in the sample and its percentage in the ore. Number Number

Explanation / Answer

.First we should know the balanced equation.We cannot directly equate moles of antimony to moles of KBrO3The balanced equation is
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)*2=(33.8ml/1000)(0.130 M)*6
(w/121.75)= 0.013182

=>w= 1.604

Percentage of antimony in the ore is (1.604/7.75)*100
=20.70%
Hence the amount of antimony in the sample is 1.604 gm and percentage is 20.70%

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