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 5.83-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 22.8 mL of a 0 145 M aqueous solution of KBrO_3 (aq). The unbalanced equation for the reaction is Calculate the amount of antimony in the sample and its percentage in the ore.

Explanation / Answer

balanced equation :

BrO3^- + 6H^+ + 3Sb^3+ ------------------> Br^- + 3Sb^5+ + 3H2O

moles of BrO3- = 22.8 x 0.145 / 1000 = 3.306 x 10^-3

for 1 mol BrO3- -------------> 3 mol Sb+3

3.306 x 10^-3 Bro3- ----------->   ??

moles of Sb+3 = 3.306 x 10^-3 x 3 / 1 = 9.918 x 10^-3

molar mass of Sb+3 = 121.76 g/mol

mass = moles x molar mass = 9.918 x 10^-3 x 121.76

         = 1.21 g

mass of Sb+3 = 1.21 g

percentage = (mass of sample/ mass of ore ) x 100

                  = (1.21 / 5.83) x 100

percentage = 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.