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divagothboi
16.04.2020 •
Chemistry
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 (%)
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Ответ:
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 %
Ответ: