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 8.69-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 Sb (aq). The Sb(aq) is completely oxidized by 39.5 mL of a 0.125 M aqueous solution of KBrO3(aq). The unbalanced equation for the reaction is Br0, ag)+sb(a))+Sb(g) (unbalanced) Calculate the amount of antimony in the sample and its percentage in the ore. Number click to edit Number

Explanation / Answer

ANS:

Balanced equation for the reaction would be,

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

So 1 mole of BrO3- react with 3 moles of Sb3+

moles of BrO3- = 0.125 M x 39.5 ml = 4.93 mmol

moles of Sb3+ present = 3 x 4.93 = 14.79 mmol

amount of Sb present = 14.79 mmol x 121.76 g/mol = 1.801 g

mass% of Sb in the sample = 1.801 x 100/8.69 = 20.72%

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