0. You haven't asked a question; I assume the problem is to determine the %Sb in the sample.
1. Balance the equation.
2. mols KBrO3 = M x L = ?
3. Using the coefficients in the balanced equation, convert mols KBrO3 to mols Sb^3+
4. g Sb = mols Sb x atomic mass Sb.
5. %Sb = (g Sb/g sample)*100 = ?
Post your work if you get stuck.
The quantity of antimony in a sample can be determined by an oxidation-reduction titration with an oxidizing agent. A 7.79-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 35.4 mL of a 0.125 M aqueous solution of KBrO3(aq). The unbalanced equation for the reaction is
H^+ + BrO3^- +Sb^3- ->Br^-
+ Sb^5+ +H2O
Calculate the amount of antimony in the sample and its percentage in the ore.
I am extremly confused by this problem, and do not know where to start.
Thank you so much for all of your help!
1 answer
2 answers