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 6.67-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 36.0 mL of a 0.105 M aqueous solution of KBrO3(aq). The unbalanced equation for the reaction is

Br [aq) + Sb3+ [aq] rightarrow Br- (aq) + Sb5+(aq) (unbalanced) Calculate the amount of antimony in the sample and its percentage in the ore.

Explanation / Answer

The balanced equation is:

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


Moles of BrO3- = volume x concentration of KBrO3

= 36.0/1000 x 0.105 = 0.00378 mol


Moles of Sb = Moles of Sb3+ = 3 x moles of BrO3-

= 3 x 0.00378 = 0.01134 mol


Mass of Sb = moles x molar mass of Sb

= 0.01134 x 121.76

= 1.38 g


Percentage of Sb = mass of Sb/mass of sample x 100%

= 1.38/6.67 x 100%

= 20.7%

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