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 6.69-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 33.0 mL of a 0.115 M aqueous solution of KBrO_3(aq). The unbalanced equation for the reaction is BrO_3^-(aq) + Sb^3+ [aq] rightarrow Br^-(aq) + Sb^5+ (aq) (unbalanced) Calculate the amount of antimony in the sample and its percentage in the ore.

Explanation / Answer

Br03- --> Br-

balance the oxygen using H20

BrO3- ---> Br- + 3H20

now balance the Hydrogen atoms using H+

so

Br03- + 6H+ ---> Br- + 3H20

now

balance the charge using electrons

so

Br03- + 6H+ + 6e- ---> Br- + 3H20

now

Sb+3 ---> Sb+5 + 2e-

so

the overall balanced reaction is

Br03- + 6H+ + 3 Sb+3 ---> Br- + 3H20 + 3 Sb+5

we know that

moles = molarity x volume (L)

so

moles of Br03- = 0.115 x 33 x 10-3

moles of Br03- = 3.795 x 10-3

now

we can see that

moles of Sb+3 = 3 x moles of Br03-

so

moles of Sb+3 = 3 x 3.795 x 10-3

moles of Sb+3 = 11.385 x 10-3

now

mass = moles x molar mass

so

mass of Sb+3 = 11.385 x 10-3 x 121.76

mass of Sb+3 = 1.3862

so

amount of antimony in the sample is 1.3862 g

now

% = 1.3862 x 100 / 6.69

% = 20.72

so

% antimony in the ore is 20.72 %

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