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 7.91-g sample of stibnite, an ore of antimony, is dissolved in hot, concentrated over a reducing agent so that all the antimony is in the form Sb (aq). The Sb (aq) is completely oxidized by 39.0 mL of a 0.115 M aqueous solution of KBrOs(aq). The unbalanced equation for the reaction is H+(aq) +BrOS (aq) + su" (aq) Br.(aq) + Sb5+(aq)+H,olg Calculate the amount of antimony in the sample and its percentage in the ore (unbalanced) Number Number

Explanation / Answer

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

The Balanced equation is expressed as

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

In the above balanced reaction equation;

1 mole of BrO3- oxidizes 3 moles of Sb3+.

Now,

39.0 mL of a 0.115 M aqueous solution of KBrO3(aq)

Moles = Molarity x Liter

So, moles of KBrO3 (or BrO3-) = 0.115 M x (39/1000) L
                                                = 0.115 M x 0.039 L
                                                = 0.004485 moles

Since, 1 mole of BrO3- oxidizes 3 moles of Sb3+.

0.004485 mole of BrO3- oxidizes (3 x 0.004485) moles of Sb3+.

0.004485 mole of BrO3- oxidizes 0.013455 moles of Sb3+.

Now, Molar mass of Sb = 121.76 g/mol

So, 1 mole of Sb = 121.76 g

0.013455 moles of Sb = 0.013455 x 121.76 g
                                    = 1.64 g

Sample mass = 7.91 g

So, % of Sb in the sample = (1.64 g / 7.91 g) x 100
                                           = 20.73 %

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