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### 7th edition L.35

Posted: Sun Oct 07, 2018 11:05 pm
"Sodium bromide, NaBr, which is used to produce AgBr for use in photographic film, can itself be prepared as follows:
Fe + Br2 --> FeBr2
FeBr2 + Br2 --> Fe3Br8
FeBr2 + Na2CO3 --> NaBr + CO2 + Fe3O4
What mass of iron, in kilograms, is needed to produce 2.50 t of NaBr? Note that these equations must first be balanced"

I think I understand the concept of this question and how to solve it but first I am having trouble balancing the third equation: FeBr2 + Na2CO3 --> NaBr + CO2 + Fe3O4. When I try to balance it, I have an extra oxygen in the products.

### Re: 7th edition L.35

Posted: Sun Oct 07, 2018 11:34 pm
I think Dr. Lavelle mentioned during a lecture that O2 is sometimes left out of the reactants if it's an unbalanced equation, so you would have to add that in as a reactant. The balanced equation would then be 6FeBr2+ 6Na2CO3 + O2--> 12NaBr + 6CO2 +2Fe3O4.

### Re: 7th edition L.35

Posted: Sun Oct 07, 2018 11:44 pm
the final balanced equation should be Fe3Br8 + 4Na2CO3 --> 8NaBr + 4CO2 + Fe3O4

### Re: 7th edition L.35

Posted: Sun Oct 07, 2018 11:45 pm
How do you know when a reaction cannot be balanced without adding the O2?

### Re: 7th edition L.35

Posted: Tue Oct 09, 2018 1:58 pm
I'm also having the same issue with the third equation. In the original problem, the reactants contain FeBr2 but in the solution manual it states that the FeBr2 is actually Fe3Br8, which is then balanced accordingly. How does FeBr2 turn into Fe3Br8?

### Re: 7th edition L.35

Posted: Tue Oct 09, 2018 9:22 pm
There's a typo in the third equation. The reactant should be the same as the product of the second reaction: Fe3Br8. Hope this helps!