Reaction Stoichiometry L.35

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Reaction Stoichiometry L.35

Postby derinceltik1K » Fri Oct 04, 2019 10:19 pm

Sodium bromide, NaBr, which is used to produce AgBr for use in photographic film, can itself be prepared as follows...
FeBr2+ Br2->Fe3Br8
Fe3Br8+ Na2CO3-> NaBr+ CO2+ Fe3O4

What is the mass of iron in kg is needed to produce 2.50 t of NaBr (equations must be first balanced)

- I keep getting the calculation wrong, can someone explain how to solve this?

Jessica Li 4F
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Joined: Fri Aug 09, 2019 12:16 am

Re: Reaction Stoichiometry L.35

Postby Jessica Li 4F » Fri Oct 04, 2019 11:41 pm

First, you should balance the equations to the following (states of matter not included):
Fe + Br2 -> FeBr2
3FeBr2 + Br2 -> Fe3Br8
Fe3Br8 + 4Na2CO3 -> 8NaBr + 4CO2 + Fe3O4

Then use dimensional analysis and mole ratios to solve, from the amount of NaBr given to iron:
2.50 t NaBr * (1000 kg NaBr/1 t NaBr) * (1000 g NaBr/1 kg NaBr) * (1 mol NaBr/102.9 g NaBr) * (1 mol Fe3Br8/8 mol NaBr) * (3 mol FeBr2/1 mol Fe3Br8) * (1 mol Fe/1 mol FeBr2) * (55.85 grams Fe/1 mol Fe) * (1 kg/1000 g) = 509 kg Fe

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Re: Reaction Stoichiometry L.35

Postby jeffreygong1I » Sat Oct 05, 2019 10:24 pm

Wait, so the book misprinted FeBr8 as FeBr2 in the last equation, right? Okay, that makes way more sense. I wasted so much time on this.

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