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 8.85-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 43.7 mL of a 0.115 M aqueous solution of KBrO3(aq). The unbalanced equation for the reaction is

BrO3^- (aq) + Sb^3+ (aq) yields Br^- (aq) + Sb^5+ (aq)

Calculate the amount of antimony in the sample and its percentage in the ore.

Explanation / Answer

Balanced Reaction:-

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

moles of KBrO3 used = molarity*volume of solution in litres = 0.115*0.0437 = 0.00503

Thus, moles of Sb3+ = 3*moles of BrO3- = 0.0151

Now, molar mass of Sb = 121.76 g/mole

Thus, mass of Sb in the ore = moles*molar mass = 1.836 g

% Sb = (mass of Sb/mass of sample)*100 = (1.836/8.85)*100 = 20.743 %

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