The quantity of antimony in a sample can be determined by an oxidation-reduction titration with an oxidizing agent. a 6.75-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 36.5 ml of a 0.105 m aqueous solution of kbro3 (aq). the unbalanced equation for the reaction is

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Smokey

Chemistry

8 June, 05:21

The quantity of antimony in a sample can be determined by an oxidation-reduction titration with an oxidizing agent. a 6.75-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 36.5 ml of a 0.105 m aqueous solution of kbro3 (aq). the unbalanced equation for the reaction is

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Norah · Answer 1

Mass of the stibnite = 6.75

Calculating the moles for bromate titaration = 0.105 x 0.0365 = 0.00365 moles Since it is 3Sb3, the moles would be 3 x 0.00365 = 0.01068.

Molar mass of the Sb is 121.7g/mol, so the mass would be Sb =.01068 x 121.7 g = 1.299g

% of the Sb = 1.299g / 6.75 = 0.1925.

So the percentage would be 19.25%