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 8.69-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 34.0 mL of a 0.145 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 8.69-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 34.0 mL of a 0.145 M aqueous solution of KBrO3(aq). The unbalanced equation for the reaction is Bro;(aq)+5b3+(aq) Br-(aq)+5b5+(aq) (unbalanced) Calculate the amount of antimony in the sample and its percentage in the ore. Number Number

Explanation / Answer

Balanced equation for the reaction would be,

BrO3-(aq) + 3Sb3+(aq) + 6H+(aq) ----> Br-(aq) + 3Sb5+(aq) + 3H2O(l)

So 1 mole of BrO3- react with 3 moles of Sb3+

moles of BrO3- = 0.145 M x 34 ml = 4.93 mmol

moles of Sb3+ present = 3 x 4.93 = 14.79 mmol

amount of Sb present = 14.79 mmol x 121.76 g/mol = 1.801 g

mass% of Sb in the sample = 1.801 x 100/8.69 = 20.72%

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