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 6.33-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 27.6 mL of a 0.130 M aqueous solution of KBrO_3(aq). The unbalanced equation for the reaction is BrO_3^- (aq) + Sb^3+ (aq) right arrow Br^- (aq) + Sb^5+ (aq) (unbalanced) Calculate the amount of antimony in the sample and its percentage in the ore.

Explanation / Answer

Balanced Equation :

Br+5 (aq) + 3Sb3+ (aq) -----------> Br- (aq) + 3Sb5+ (aq)

Millimoles of KBrO3 used = 27.6*0.13 = 3.588 mmoles

1 mole of KBrO3 oxidizes 3 moles of Sb3+.

So milli moles of Sb3+ neutralized = 3.588*3 = 10.764 mmoles

Atomic mass of antimony = 121.76 g/mol

amount of antimony = 121.76*10.764 g = 1.31 g

therefore % in the ore = (1.31/6.33) * 100 = 20.7%

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