Question

The quantity of antimony in a sample can be determined by an oxidation-reduction titration with an oxidizing agent. A 8.65-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 37.8 mL of a 0.130 M aqueous solution of KBrO3(aq). The unbalanced equation for the reaction is

BrO3 - (aq) + Sb 3+ (aq) -> Br - (aq) + Sb 5+ (aq) (unbalanced)

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

_ g

_ %

Help please? Thank you.

Answer 1

Solution-

The balanced equation can be obtained as below

BrO3- ==> Br- . . .Add 3 H2O to the right side to balance O.
BrO3- ==> Br- + 3H2O . . .add 6H+ (acidic solution) to the left side to balance H.
BrO3- + 6H+ ==> Br- + 3H2O . . .add 6e- 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-

Now we 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)(0.0378) = 0.004914 moles BrO3-

From the balanced equation, we get that 1 mole of BrO3- reacts with 3 moles of Sb3+.

0.004914 moles BrO3- x (3 mole Sb3+ / 1 mole BrO3-) = 0.0147 moles Sb3+

the molar mass of Sb= 121.8 g.

0.0147 moles Sb x (121.8 g Sb / 1 mole Sb) = 1.79 g Sb

%Sb = (g Sb / g ore) x 100 = (1.79 / 8.65) *100 = 20.69 %Sb