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 7.85-g sample of stibnite, an ore of antimony, is dissolved in hot, concentrated HCI(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 37.1 mL of a 0.120 M aqueous solution of KBrO3(aq). The unbalanced equation for the reaction is Calculate the amount of antimony in the sample and its percentage in the ore.

Explanation / Answer

BrO3- + 3 Sb3+ + 6 H+ => Br- + 3 Sb5+ + 3 H2O

Moles of BrO3- = volume x concentration of KBrO3

= 37.1/1000 x 0.120 = 0.004452 mol


Moles of Sb = Moles of Sb3+ = 3 x moles of BrO3-

= 3 x 0.004452 = 0.013356 mol


Mass of Sb = moles x molar mass of Sb

= 0.013356 x 121.76

= 1.626 g = 1.63 g


Mass% of Sb = mass of Sb/mass of sample x 100%

= 1.626/7.85 x 100%

= 20.72% = 20.7%

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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