Question

The quantity of antimony in a sample can be determined by an oxidation-reduction titration with an oxidizing agent. A 9.57-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 49.4 mL of a 0.110 M aqueous solution of KBrO3(aq). The unbalanced equation for the reaction is BrO3-(aq) + Sb3+(aq) = Br-(aq) + Sb5+(aq). calculate the amount of antimony in the sample and its percentage in the ore

Answer 1

First, let's balance the equation by the redox method in acidic solution:
6H⁺ + BrO₃^- + 6e^- --------> Br^- + 3H₂O
3Sb³⁺ -----------> 3Sb⁵⁺ + 6e^-

Final balanced reaction:
6H⁺ + BrO₃^- + 3Sb³⁺ ----------> Br^- + 3Sb⁵⁺ + H₂O

This means that 1 mol of BrO3 reacts with 4 moles of Sb3+, so:

moles Sb³⁺ = 3moles BrO₃^-
moles Sb³⁺ = 3 * 0.110 mol/L * 0.0494 L
moles Sb³⁺ = 0.0163 moles

Let's calculate the mass of Sb:
mass = 0.0163 mol * 121.76 g/mol = 1.9849 g

Finally the % in the ore:
%= (1.9849/9.57) * 100 = 20.75%

Hope this helps