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!

the number of valence electrons for each molecule or ion is shown beneath the structure. remember hydrogen will not have more than two electrons. this hydrogen is part of a covalent bond (sharing two electrons). a double bond here would cause hydrogen to share four electrons with phosphorus.

the answer is true. hope it .