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
                                    Â
= 0.065 M x 5 x 10^1 ml
                                    Â
= 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
                                     Â
= 0.026 g
