0.737 g of a sample contains iron. all amount of iron is oxidied to Fe2+ on...

0.737 g of a sample contains iron. all amount of iron is oxidied to Fe2+ on Pt anode which has constant potential at 1.0 V against SCE. an iodide chemical coulometer is used to measure the amont of current (Q), the amount of released iodide needs 27.mL of thiosulfate (0.0217 N). calculate the percentage of Fe3O4 in the sample. (current 10 mA)

Expert Solution

From the data,

moles of thiosulfate used = 0.0217 N x 27 ml

= 0.586 mmol

amount of released iodide = 0.586/2

= 0.293 mmol

moles of Fe in sample = 3 x 0.293 mmol x 55.845 g/mol/1000

= 0.05 g

percentage of Fe3O4 in the sample = 0.05 x 100/0.737

= 6.78%

queen_honey_blossom answered 2 years ago

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