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 6.31-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 31.1 mL of a 0.115 M aqueous solution of KBrO3(aq). The unbalanced equation for the reaction is

BrO3^-(aq)+Sb^3+(aq)----------->Br^-(aq)+Sb^5+(aq)

Calculate the amount of antimony in the sample and its percentage in the ore.

Number______g Number_______%

Explanation / Answer

THE BALANCED EQUATION IS

BrO3^- + 6H^+ + 3Sb^3+ ---> Br^- + 3Sb^5+ + 3H2O

now it is given 31.1 mL of a 0.115 M aqueous solution of KBrO3 oxidizes complete antimony

we also know one mol of KBrO3 oxidizes 1 mol of antimony

now we will calculate moles of KBrO3 in 31.1 mL of a 0.115 M

= 0.115 * 31.1 /1000

= 0.0035765 moles

or same moles of antimony are present in 6.31-g sample of stibnite

now I mol of antimony = 121.76 gram

so 0.0035765 moles of antimoney = 121.76 * 0.0035765

= 0.43547464 grams

% = 0.43547464 * 100/ 6.31

= 6.901 %

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