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 9.17-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 35.9 mL of a 0.145 M aqueous solution of KBrO3(aq). The unbalanced equation for the reaction is

BrO3- + Sb3+ --> Br- +Sb5+

Calculate the amount of antimony in the sample and its percentage in the or

Map eneral Chemistry University Science Books presented by Sapling Learning 4th Edition Donald McQuarrie Peter A. Rock Ethan Gallogly The quantity of antimony in a sample can be determined by an oxidation-reduction titration with an oxidizing agent. A 9.17-g sample of stibnite, an ore of antimony, is dissolved in hot, concentrated HCI(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 35.9 mL of a 0.145 M aqueous solution of KBrO3(aq). The unbalanced equation for the reaction is Br03 (aq) + Sb3 + (aq) Br(aq) + Sb5+ (aq) (unbalanced) Calculate the amount of antimony in the sample and its percentage in the ore Number Number Previous Give Up & View Solution Check Answer Next Exit er Hint

Explanation / Answer

Balance of the reaction:

Half reactions:

BrO3- -> Br-

Sb3+ -> Sb5+

Balance oxygens with water:

BrO3- -> Br- + 3H2O

Sb3+ -> Sb5+

Add missing hydrogens:

6H+ + BrO3- -> Br- + 3H2O

Sb3+ -> Sb5+

Balance charges:

6H+ + BrO3- + 4e- -> Br- + 3H2O

Sb3+ -> Sb5+ + 2e-

Equalize charges:

6H+ + BrO3- + 6e- -> Br- + 3H2O

3Sb3+ -> 3Sb5+ + 6e-

Equalize reactions:

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

After balancing, we get the moles of KBrO3 used:

0.0359 L * 0.145 mol/L = 0.0052 moles of KBrO3

With the moles gotten, we get moles of Sb3+:

0.0052 moles of KBrO3 * (3 moles of Sb3+ / 1 mole of KBrO3) = 0.0156165 moles of Sb3+

We now convert these moles to grams:

0.0156165 moles * (121.76 g/mol) = 1.9 grams of antimony

Percentage of antimony on stibnite = 1.9 / 9.17 = 20.72 %

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