.e quantity of antimony in a sample can be determined by an oxidation-reduction

Question

.e quantity of antimony in a sample can be determined by an oxidation-reduction titration with an oxidizing agent. A 6.07-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 Sb3t(aq). The Sb3 (aq) is completely oxidized by 27.6 mL of a 0.125 M aqueous solution of KBrO3(aq). The unbalanced equation for the reaction is H+(aq) + Bro, (aq)+Sb3 + (aq) Br-(aq)+Sb5+ (aq)+H2O(1) (unbalanced) Calculate the amount of antimony in the sample and its percentage in the ore. Number Number

Explanation / Answer

The balanced equation is

6H+(aq) + BrO3^-(aq) + 3Sb^3+(aq) = Br^-(aq) + 3Sb^5+(aq) + 3H2o(l)

Determine the bromate used :

(0.125 mol/L) * (0.0276L) = 0.00345 mol bromate

Determine the SB that reacted :

0.00345 mols bromate times( 3mol SB / 1 mol bromate) = 0.01035 mol Sb

Determine the grams of antinomy:

0.01035 mol * 121.760 g/mol = 1.26 g

Calculate the percentage of antimony in the sample

1.26 g / 6.07 g * 100% = 20.7%

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