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)
<em>-Where the Fe(OH)⁺ is Fe(II) and Fe(OH)₃ is Fe(III)-</em>
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× = <em>0,051g of O₂</em>
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I hope it helps!
