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 9.31-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+(aq) is completely oxidized by 50.3 mL of a 0.105 M aqueous solution of KBr03(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

Moles of KBrO3 required for complete titration = MxV = 0.105M x 50.3 mL x (1L / 1000 mL) = 0.0052815 mol

The balanced chemical reaction is

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

3 mol, --- 1 mol, -- 6 mol, -------- 3 mol, -- 1 mol

In the above balanced reaction it is clear that

1 mol of BrO3- reacts with  3 moles of Sb3+ .

Hence 0.0052815 mol of BrO3- that will react with the moles of Sb3+

= 0.0052815 mol of BrO3- x (3 moles Sb3+ / 1 mol of BrO3-)

= 0.0158445 mol Sb3+  

Molecular(ionic) mass of Sb3+ = 121.76 g/mol

Hence mass of Sb3+ = 0.0158445 mol x 121.76 g/mol = 1.929 g (answer)

Hence percentage of antimony in the ore =  (1.929 g / 9.31 g)x100 = 20.72 % (answer)

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