A quantity of antimony in a sample can be determined by an oxidation-reduction t

Question

A quantity of antimony in a sample can be determined by an oxidation-reduction titration with an oxidizing agent. A 9.62 gram sample of stibnite, an ore of antimony, is dissolved in hot, concentrated HCl (ag) and passed over a reducing agent so that all the antimony is in the form Sb"3. The Sb3 (aa) is completely oxidized by 43.7 mL of a 0.1250 M solution of KBrO3. Calculate the amount of antimony in the sample and its percentage in the ore. 2. The unbalanced equation for the reaction is: Bro,- (ag) + Sb+3 (ag) Br (ag) + Sb+5 (ag)

Explanation / Answer

The balanced reaction is given as :

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

Number of mol of KBrO3 = 0.1250 x 0.0437 = 0.00546 mol

So the number of mol of antimony present will also be 0.00546 mol

Amount of antimony in sample = 0.00546 x molar mass

= 0.00546 x 121.76

= 0.665 grams

Percentage in ore = ( 0.665 / 9.62) x 100

= 6.91 %

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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