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### 2013 Final Practice Exam, Question 4a; n=?

Posted: Thu Mar 16, 2017 11:55 am
In the 2013 Final Practice Exam, for question 4a how do you know that n=4?

### Re: 2013 Final Practice Exam, Question 4a; n=?

Posted: Thu Mar 16, 2017 12:09 pm
The iron goes from an oxidation state of 0 to +2 (and the oxygen goes from 0 to -2). And since there are two moles of iron and oxygen, 2*2electrons=4 electrons. Hope that helps!

### Re: 2013 Final Practice Exam, Question 4a; n=?

Posted: Fri Mar 17, 2017 12:20 pm
I believe each oxygen in the O2 in the reactants initially has an oxidation number of 0, while in the product all 4 oxygens have an oxidation number of -2. This would make each Fe in the product have an oxidation number of +2, while in the reactants each Fe had an oxidation number of 0. Thus, each Fe was oxidized from 0 to 2+ and each oxygen atom in the O2 reactant was reduced from 0 to 2-. Therefore, since there were 2 Fe that were oxidized from 0 to 2+ and 2 oxygens that were reduced from 0 to 2-, the number of electrons transferred would be 2 atoms x 2 electrons = n = 4.

### Re: 2013 Final Practice Exam, Question 4a; n=?

Posted: Fri Mar 17, 2017 9:14 pm
Why does 2Fe(s) in the reactants have an oxidation number of zero again? Is it because it is in it's most stable form or something like that?

### Re: 2013 Final Practice Exam, Question 4a; n=?

Posted: Fri Mar 17, 2017 9:35 pm