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 6.65-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 Sb3 (aq). The Sb3*(aq) is completely oxidized by 29.0 mL of a 0.130 M aqueous solution of KBrO3(aq). The unbalanced equation for the reaction is Bro,aq)+ Sb (aq)-> Br(aq)+Sb (unbalanced) Calculate the amount of antimony in the sample and its percentage in the ore. Number Number

Explanation / Answer

BrO3- => Br- ... add 3 H2O to the right side to balance O

BrO3- => Br- + 3 H2O.. add 6 H+ to the left side to balance H

BrO3- + 6H+ => Br- + 3 H2O ... Add 6 e- to the left side to balance the charge

BrO3- + 6H+ + 6e- ==> Br- + 3H2O

Sb3+ ==> Sb5+ . . .Add 2e- to the right side to balance the charge.
Sb3+ ==> Sb5+ + 2e-

Multiply the Sb equation by 3 to give 6e- on the right side. Then add the new equation to the BrO3- equation; the 6e- will cancel.

.3Sb3+ ==> 3Sb5+ + 6e-
+BrO3- + 6H+ + 6e- ==> Br- + 3H2O
===================================
3Sb3+ + BrO3- + 6H+ ==> 3Sb5+ + Br- + 3H2O


moles BrO3- added = M BrO3- x L BrO3- = (0.130)(0.0290) = 0.00377 moles BrO3-

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

0.00377 moles BrO3- x (3 mole Sb3+ / 1 mole BrO3-) = 0.01131 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.

0.01131 moles Sb x (121.8 g Sb / 1 mole Sb) = 1.37756 g Sb

%Sb = (g Sb / g ore) x 100 = (1.37756/ 6.65) x 100 = 20.71 %Sb

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