Section L, #35

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Soumya Ravichandran 4H
Posts: 62
Joined: Fri Sep 28, 2018 12:19 am

Section L, #35

Postby Soumya Ravichandran 4H » Sat Oct 06, 2018 2:28 pm

Sodium bromide, NaBr, which is used to produce AgBr for use in photographic lm, can itself be prepared as follows.
Fe + Br2 --> FeBr2
FeBr2 + Br2 --> Fe3Br8
Fe3Br8 + Na2CO3 --> NaBr + CO2 + Fe3O4
What mass of iron, in kilograms, is needed to produce 2.50 t of NaBr?

I understand the stoichometry for converting the tons --> grams --> moles, but I am unsure how to get the molar ratio between NaBr and Fe as they are in different equations after balancing them?

Soyoung Park 1H
Posts: 30
Joined: Sat Sep 29, 2018 12:15 am

Re: Section L, #35

Postby Soyoung Park 1H » Sat Oct 06, 2018 2:44 pm

I was working on that problem too and I got wondering about the same thing! Can anyone help us out?

Posts: 30
Joined: Fri Sep 28, 2018 12:25 am

Re: Section L, #35

Postby 604656370 » Sun Oct 07, 2018 10:01 pm

I was stuck on this one too but figured out that you can add the three reactions. The result will give you:
Fe + 2Br2 + Na2CO3 ---> NaBr + CO2 + Fe3O4

Using this equation, you would need to:
1. Balance the equation
2. Calculate the moles of NaBr given you have 2.5t
3. Using the moles of NaBr and the ratio from the balanced equation, determine the moles of Fe you need
4. Convert moles Fe to kg

Hope that helps!

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