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 7.33-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 34.7 mL of a 0.120 M aqueous solution of KBrO_3(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 BrO3 - = 34.7 x 0.120 / 1000 = 4.164 x 10^-3

balanced equation:

BrO3^- + 6H^+ + 3Sb^3+ -----------> Br^- + 3Sb^5+ + 3H2O

1mol                     3 mol

4.164 x 10^-3        x mol

moles of Sb+3 = 0.012492

mass of Sb = 0.012492 x 121.76 = 1.52 g

amount of antimony = 1.52 g

antimony percent in ore = (1.52 / 7.33 ) x 100

                                      = 20.75 %

antimony percent in ore = 20.75 %

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