Fe(II) can be precipitated from a slightly basic aqueous solution by bubbling oxygen through the solution, which converts Fe(II) to insoluble Fe(III):

Question

4Fe(OH)⁺(aq) + 4OH^-(aq) + O₂(g) + 2H₂O (l) —> 4Fe(OH)₃

 How many grams of O₂ are consumed to precipitate all of the iron in 75.0 mL of 0.0350 M Fe(II)?

Answer 1

4Fe(OH)⁺(aq) + 4OH^-(aq) + O₂(g) + 2H₂O (l) ==> 4Fe(OH)₃(s)

Find moles Fe(OH)⁺ and then find moles O₂ needed to react with it. Finally, convert moles O₂ to grams.

moles Fe(OH)⁺ = 75.0 ml x 1 L/1000 ml x 0.0350 mol/L = 0.002625 moles Fe(OH)⁺

moles O₂ consumed = 0.002625 mol Fe(OH)⁺ x 1 mol O₂ / 4 mol Fe(OH)⁺ = 0.0006563 moles O₂

Grams O₂ consumed = 0.0006563 moles O₂ x 32 g/mol = 0.0210 moles O₂ consumed