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):
4fe(oh)+(aq)+ 4oh-(aq)+o2(g)+2h2o(l) = 4fe(oh)3(s)
how many grams of o2 are consumed to precipitate all of the iron in 85ml of 0.075 m fe(ii).
explain step by step

0,051g of O₂

Explanation:

The reaction of precipitation of Fe is:

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

-Where the Fe(OH)⁺ is Fe(II) and Fe(OH)₃ is Fe(III)-

This reaction is showing you need 1 mol of O₂(g) per 4 moles of Fe(II) for a complete reaction.

85mL= 0,085L of 0,075M Fe(II) are:

0,085L*0,075M = 6,34x10⁻³ moles of Fe(II)

For a complete reaction of 6,34x10⁻³ moles of Fe(II) you need:

6,34x10⁻³ moles of Fe(II)× =

1.59x10⁻³ moles of O₂. In grams:

1.59x10⁻³ moles of O₂× = 0,051g of O₂

I hope it helps!

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