The quantity of an antimony in a sample can be determined by and oxidation reduc

Question

The quantity of an antimony in a sample can be determined by and oxidation reduction titration with an oxidizing agent. A 6.91- g sample of stibnite, an ore of antimony, is dissolved in hot, concentrated HCl(aq) and passed over a reducing agent so that all antimony is in the form Sb3+(aq). The Sb3+ (aq) is completely oxidized by 34.1 mL of a 0.115 M aqueous solution KBrO(3) (aq).The unbalanced eqaution for the reaction is.

H+(aq) + BrO3-(aq)+Sb3+(aq)=Br-(aq)+Sb5+(aq)+H2O(I) (unbalanced)

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

g

%

Explanation / Answer

m = 6.91 g of Sb

balance:

12H+(aq) + 2BrO3-(aq)+5Sb3+(aq)=2Br-(aq)+5Sb5+(aq)+6H2O(I)

now...

ratio is 2 mol o BrO3- = 5 mol of Sb

calculate

mol of KBrO3 = MV = (0.115*34.1*10^-3) = 0.0039215

ratio is 5/2 so

mol of sb = 5/2*0.0039215 =0.0098 mol of Sb

mass = mol*MW = 0.0098 *121.76 = 1.193 g of Sb

%Sb = mass of Sb / Total mass * 100% = 1.193/6.91*100 = 17.264 %

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