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 7.83-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 31.7 mL of a 0.140 M aqueous solution of KBrO3(aq). The unbalanced equation for the reaction is

The quantity of antimony in a sample can be determined by an oxidation-reduction titration with an oxidizing agent. A 7.83-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 31.7 mL of a 0.140 M aqueous solution of KBrOs(aq). The unbalanced equation for the reaction is BrO,(aq)-Sb"(aq) -> Br-(aq)+5b5+(aq) (unbalanced) Calculate the amount of antimony in the sample and its percentage in the ore. Number 1.70 Number 21.7

Explanation / Answer

Here you cannot directly equate moles of antimony to moles of KBrO3.

First you should know the balanced equation.

The balanced equation will be

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

Molar mass of antimony is 121.75 gm

equate the equivalents of Sb^3+ and BrO3-

(w /121.75) x 2 = (0.140 x 31.7 / 1000 ) x 6

mass w = 1.62 g

Percentage of antimony in the ore is = (1.62 / 7.83) x 100 = 20.71%

percentage is = 20.7%

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