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 5.01-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 23.7 mL of a 0.120 M aqueous solution of KBrO_3(aq). The unbalanced equation for the reaction is H^+(aq) + BrO_3^-(aq) + Sb^3+(aq) rightarrow Br^-(aq) + Sb^5+(aq) + H_2O(l) (unbalanced) Calculate the amount of antimony in the sample and its percentage in the ore.

Explanation / Answer

6H+(aq)+BrO3-(aq)+3Sb3+(aq)--->Br-(aq)+3Sb5+(aq)+ 3H2O(l)

From the stoichiometry , it is known that 3 mol of Sb3+ react with 1 mol of KBrO3

Molarity of KBrO3 = 0.120

Volume of KBrO3 = 23.7ml

No of mol of KBrO3 = (0.120mol/1000ml)×23.7ml = 0.00284

Therefore, No of mol of Sb3+ = 3×0.00284 = 0.00852

Molar mass of Sb = 121.76g/mol

Mass of Sb = 121.76 g/mol×0.00852mol = 1.04g

Percentage of Sb = (1.04g/5.01g)×100 = 20.76

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.