Problem L.35

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Problem L.35

Postby Daniel_Frees_1L » Tue Oct 02, 2018 11:59 am

Was anyone actually able to balance the third chemical equation in problem L.35? I got stuck after balancing Fe, then NA, then BR, then C, the oxygens would not balance. I even plugged the problem into a chemical equation balancer online after being stuck for a while and it said the equation cannot be balanced? Is there a typo in the book or am I just messing something up?

the equation to be balanced was: FeBr2 +Na2CO3 -> NaBr + CO2 + Fe3O4

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Re: Problem L.35

Postby megansardina2G » Tue Oct 02, 2018 3:04 pm

Hi! The third equation in L35 in my textbook was given as:
Fe3Br8 + Na2CO3 -----> NaBr + CO2 + Fe3O4
(I have the sixth edition textbook, but I think the equation should be the same for this problem, even if you have the seventh)

The equation that you wrote down in your question had FeBr2 in place of the Fe3Br8, so I think that was your problem in balancing the equation, but I will still post the process to balance this equation correctly below just in case.

Start by balancing the Br.
Fe3Br8 + Na2CO3 -----> 8NaBr + CO2 + Fe3O4

Now, since there are 8 Na on the right to balance the Br, you need to fix the left side of the equation to also provide 8 Na molecules, so the stoichiometric coefficient for Na2CO3 should be 4.
Fe3Br8 + 4Na2CO3 -----> 8NaBr + CO2 + Fe3O4

Finally, balance the carbon and oxygen. In the equation above, there are four carbon on the left and only one on the right. The stoichiometric coefficient for the carbon dioxide (CO2) should be four, which balances both the carbon and oxygen.
Fe3Br8 + 4Na2CO3 -----> 8NaBr + 4CO2 + Fe3O4

To check: 3 Fe, 8 Br, 8 Na, and 12 O on both sides, meaning that the equation is balanced.

Hope this helped!

Matthew Mar 1J
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Re: Problem L.35

Postby Matthew Mar 1J » Tue Oct 02, 2018 3:18 pm

I got stuck on the same problem. I'm pretty sure that there is a typo in the 7th edition of the textbook because the question includes a string of chemical equations, each with a product that is used as a reactant in the next reaction.
The second equation FeBr2 + Br2 ---> Fe3Br8 has Fe3Br8 as it's product, but the next equation listed doesn't include this compound and instead uses FeBr2.
Fe + Br2 ---> FeBr2
____________FeBr2 + Br2 ---> Fe3Br8
___________________________FeBr2 + Na2CO3 ---> NaBr + CO2 +Fe3O4
Which is a complicated way to explain how I spent half an hour figuring out a typo.

Ashley Zhu 1A
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Re: Problem L.35

Postby Ashley Zhu 1A » Tue Oct 02, 2018 5:14 pm

I have the 7th edition and had the same problem where I spent forever on the problem because it wouldn't balance. I'm pretty sure it's a typo as well because in the solutions manual they also give the answer by using Fe3Br8.

Cole Elsner 2J
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Re: Problem L.35

Postby Cole Elsner 2J » Tue Oct 02, 2018 7:15 pm

The solutions manual has Fe3Br8 listed. That's the only way it works out. I guess it's a typo in the Seventh edition :/

Leslie Cheng 4B
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Re: Problem L.35

Postby Leslie Cheng 4B » Tue Oct 02, 2018 8:08 pm

I have the seventh edition too but no solutions manual...I definitely did not think about it being a typo and thought I was just horrible at balancing equations. Thank you all for catching that :)

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