Fe(II) can be precipitated from a slightly basic aqueous solution by bubbling oxygen through the solution,...

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Fe(II) can be precipitated from a slightly basic aqueoussolution by bubbling oxygen through the solution, wheich convertsFe(II) to insoluble Fe(III):

4Fe(OH)+(aq)+4OH-(aq)+O2(g)+2H2O(l)arrow 4Fe(OH)3(s)

How many grams of O2 are consumed to precipitate allof the iron in 5.0 x 101 mL of 0.065 M Fe(II)?

Please explain!

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3.9 Ratings (380 Votes)

4 moles of Fe(II) takes up 1 mole of O2 to form 4 moles of Fe(III)

moles of Fe(II) present = molarity x volume

                                      = 0.065 M x 5 x 10^1 ml

                                      = 3.25 mmol

So,

moles of O2 needed = 3.25/4 = 0.8125 mmol

grams of O2 consumed = 0.8125 mmol x 32 g/mol/1000

                                       = 0.026 g


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