presented by Sapling Learning Donald McQuarrie. Peter A. Rock Ethan Gallogly The

Question

presented by Sapling Learning Donald McQuarrie. Peter A. Rock Ethan Gallogly The quantity of antimony in a sample can be determined by an oxidation-reduction titration with an oxidizing agent. A 8.53-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 34.6 mL of a 0.140 M aqueous solution of KBro3(aq). The unbalanced equation for the reaction is Brou (aq) Sbst (aq) Br (aq) Sb5+ (aq) (unbalanced) Calculate the amount of antimony in the sample and its percentage in the ore. Number Number

Explanation / Answer

The balanced equation is as follows,

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

number of electron transfer in the reaction = 6

3 moles of Sb3+ reacts with 1 mol of BrO3-

moles of KBrO3 used to titrated Sb3+ solution = 34.6 x 0.140/ 1000 = 0.0048 moles

moles of Sb 3+ present in the solution = 3 x moles of KBrO3 used = 3x 0.0048 mol = 0.0145 mol

amount of Sb 3+ present in the solution = 0.0145 mol x 121.76 = 1.769 g

% of Sb in the ore = 1.769/ 8.53 x100 = 20.74 %

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