Question

Fe(II) can be precipitated from a slightly basic aqueous solution by bubbling oxygen through the solution, wheich converts Fe(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 all of the iron in 5.0 x 10¹ mL of 0.065 M Fe(II)?

Please explain!

Answer 1

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