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 9.73-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 Sb^3+(aq). The Sb^3+(aq) is completely oxidized by 44.2 mL of a 0.125 M aqueous solution of KBrO_3(aq). The unbalanced equation for the reaction is BrO_3^-(aq) + Sb^3+(aq) rightarrow Br^-(aq) + Sb^5+(aq) (unbalanced) Calculate the amount of antimony in the sample and its percentage in the ore.

Explanation / Answer

The balanced reaction is 3Sb3+ + BrO3- + 6H+ ==> 3Sb5+ + Br- + 3H2O rity

moles of KBrO3= molarity* Volume (L)= 0.125*44.2/1000 =0.005525

3 moles of Sb+3 requires 1 mole of BRO3-

0.005525 moles of BrO3- ( obtained from 0.005525 moles of KBrO3, KBrO3---------->K+ + BrO3-}

requires 0.005525*3= 0.016575 moles of Sb+3.

atomic weight of antiomony= 126, mass of Antimony= moles* atmoic weight= 0.016575*126= 2.088 gm

percentage of antimony= 100*2.088/9.73= 21.5%

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