Question

The quantity of antimony in a sample can be determined by an oxidation-reduction titration with an oxidizing agent. A 9.61-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 42.0 mL of a 0.130 M aqueous solution of KBrO3(aq). The unbalanced equation for the reaction is BrO3(aq) + Sb3+(aq) -> Br-(aq) +Sb5+(aq) (unbalanced)

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

*Find Number in grams & percentage

Answer 1

BrO3{-} + 3 Sb{3+} + 6 H{+} → Br{-} + 3 Sb{5+} + 3 H2O
moles BrO3- added = M BrO3- x L BrO3-
= (0.130)(0.0420) = 0.00546 moles BrO3-

The balanced equation tells us that 1 mole of BrO3- reacts with 3 moles of Sb3+.

0.00546 moles BrO3- x (3 mole Sb3+ / 1 mole BrO3-) = 0.01638 moles Sb3+

From the periodic table, the molar mass of Sb (or Sb3+; the 3 missing electrons have very little effect on the mass ) = 121.8 g.

0.01638 moles Sb x (121.8 g Sb / 1 mole Sb) = 1.995 g Sb

%Sb = (g Sb / g ore) x 100
%Sb=(1.99 g / 7.99 g) ×100= 24.96% Sb