The quantity of antimony in a sample can be determined by an oxidation-reduction titration with an oxidizing agent. A 6.33-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 Sb3 (aq). The Sb3 (aq) is completely oxidized by 29.9 mL of a 0.120 M aqueous solution of KBrO3(aq).

The unbalanced equation for the reaction is BrO3- (aq) + Sb^3+ (aq) > Br^3- (aq) + Sb^5+ (aq)

A) Calculate the amount of antimony in the sample (grams)

B) Calculate the percentage in the ore (%)

Explanation:

BrO3- (aq) + Sb^3+ (aq) > Br^3- (aq) + Sb^5+ (aq) is an unbalanced equation and needs to be balanced

BrO3- (aq) → Br^3- (aq

to balance it water must be added to the right side and H⁺ be added to the left side

BrO₃⁻ + 6 H⁺ + 8e⁻ → Br³⁻ + 3 H₂ O

Sb³⁺ (aq) → Sb⁵⁺ + 2e⁻

multiply the second equation by 4

BrO₃⁻ + 6 H⁺  + 8e⁻ → Br³⁻ + 3 H₂ O

4Sb³⁺ → 4Sb⁵⁺ + 8 e⁻

add the two equation together and cancel the 8 e electrons on both side

BrO₃⁻ + 4Sb³⁺ + 6 H⁺  → Br³⁻ +  4Sb⁵⁺ + 3 H₂ O

number of mole of BrO₃⁻  = volume in liters × molarity  = (29.9 / 1000) L × 0.120 M = 0.003588 moles

from the balanced equation of reaction;

one mole of BrO₃⁻  requires 4 moles of Sb³⁺

0.003588 moles of BrO₃⁻  will require = 0.003588 × 4 = 0.0144 moles of Sb³⁺

a) amount of antimony in grams in the sample =  0.0144 moles × 121.8 g ( molar mass of antimony) = 1.748 g

b ) percentage of antimony in the ore = 1.748 g / 6.33 g = 27.62 %