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

amonunt in sample= 1.46 g

percentage in ore = 20.7%

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