Question

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)how many grams of O2 are consumed to precipitate all of the iron in 85ml of 0.075 M Fe(II).

Answer 1

Answer:

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)×$\frac{1molO_2}{4molesFe(II)}$ = 1,6x10⁻³ moles of O₂(g)

These moles are:

1,6x10⁻³ moles of O₂(g)×$\frac{32g}{1mol}$ = 0,051g of O₂ are consumed

I hope it helps!