o 4/29/2016 06:00 PM 4.1/1004/28/2016 11:03 PM -Print Calculator Periodic Table

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o 4/29/2016 06:00 PM 4.1/1004/28/2016 11:03 PM -Print Calculator Periodic Table Gradeb Print Calculator ndl Petriodic Table Question 22 of 23 Map University Science Boaks presented by Sapling Leam 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 9.77-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 Sb3 (aqg). The Sb (aq) is completely oxidized by 38.2 mL of a 0.145 M aqueous solution of KBrOslag). The unbalanced equation for the reaction is Bro; (aq) +Sb"(aq) Br-(aq)+5b5+ (aq) (unbalanced) Calculate the amount of antimony in the sample and its percentage in the ore. Number Number

Explanation / Answer

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

moles of BrO3- = 38.2 x 0.145 / 1000 = 5.54 x 10^-3

1 mol BrO3- -----------------> 3 mol Sb+3

5.54 x 10^-3 mol BrO3- -----------> 3 x  5.54 x 10^-3 mol Sb+3 = 0.0166 mol Sb+3

Sb+3 mass = 0.0166 x 121.76 = 2.02 g

amount of antimony = 2.02 g

percent of antimony = 2.02 x 100 / 9.77

= 20.71 %

percent of antimony = 20.71 %

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