Question

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).
please explain step by step

Answer 1

Answer:

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)×$\frac{1molO_2}{4moles Fe(II)}$ =

1.59x10⁻³ moles of O₂. In grams:

1.59x10⁻³ moles of O₂×$\frac{32g}{1molO_2}$ = 0,051g of O₂

I hope it helps!