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 oxidzing agent.

A 7.05 g of stibnite, an ore antimony, is dissolved in hot, concentrated HCL and passed over a reducing agent so that all the antimony is in form Sb^3+ (aq). The Sb^3+ (aq) is completely oxidized by 32.0 mLof a 0.125 M (aq) 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 it percentage in ore.

SHow all the work if you would like but I really only need the answers! Please and Thank you!

Explanation / Answer

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

moles of BrO3^- = (32 x 0.125)/1000 = 0.004 moles

moles of antimony = 3 x 0.004 = 0.012 moles

mass of antimony in sample = 0,012 x 121.76 = 1.46 gm

% in or = (1.46/7.05) x 100 = 20.72 %

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