Fe(ii) can be precipitated from a slightly basic (aq) solution by bubbling oxygen through the solution, which converts fe(ii) to insoluble fe(iii): 4fe(oh)+(aq)+ 4oh-(aq)+o2(g)+2h2o(l)=4fe(oh)3(s)h ow many grams of o2 are consumed to precipitate all of the iron in 85ml of 0.075 m fe(ii).

0,051g of O₂ are consumed

Explanation:

The reaction of precipitation of Fe(II) is:

4Fe(OH)⁺(aq) + 4OH⁻(aq) + O₂(g) + 2H₂O(l) → 4Fe(OH)₃(s)

The moles of Fe(II) you have in 85ml of 0.075 M Fe(II) are:

0,085L×0,075M = 6,4x10⁻³ moles of Fe(II)

For the reaction, you require 4 moles of Fe(II) and 1 mol of O₂(g) to precipitate the iron. Thus, moles of O₂(g) you require are:

6,4x10⁻³ moles of Fe(II)× = 1,6x10⁻³ moles of O₂(g)

These moles are:

1,6x10⁻³ moles of O₂(g)× = 0,051g of O₂ are consumed

I hope it helps!