L35 Fundamentals Textbook Problem

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L35 Fundamentals Textbook Problem

Postby Minahil_Tufail_3I » Thu Oct 08, 2020 9:18 pm

Hi, I was hoping to get some help with this problem:
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
Fe3Br8 + 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 was able to balance the equations, but I got stuck afterwards with what to do next.

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Re: L35 Fundamentals Textbook Problem

Postby Ryan_Page_1J » Thu Oct 08, 2020 9:30 pm

Note: I researched this question and found this on this platform from 2015.

First you convert the 2.50 t of NaBr to grams using the conversion factor 1.0 x 10^6 grams= 1 ton. Then you find the molar mass of NaBr (102.89g) and divide the grams of NaBr by NaBr's molar mass. Next, find the mole ration between Fe and NaBr which is 3 mol Fe: 8 mol NaBr. Lastly, multiply by the molar mass of Fe (55.84g) which will result in 5.09x10^5g of Fe which is equivalent to 509 kg of Fe.

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Re: L35 Fundamentals Textbook Problem

Postby rhettfarmer-3H » Fri Oct 09, 2020 12:44 pm

oh wow, this is a long one. First balance the equation we are using. By doing so should look like Fe3Br8 + 4Na2CO3 =8NaBr +4CO2 +Fe3O4. We know we have 2.5 tons of NaBr so we must work backwards to Fe. Get out of tons. We want grams! So 1 ton is 10^6 grams. So convert. Then convert to moles so we can use the molar ratio to get to Fe. Hence, the molar mass is 102.84 NaBr. Get moles. Then convert to Fe3Br8 from NaBr moles. This is simple molar ratio is 1:8 so divide by 8 then within the Fe3Br8. The molar ratio of iron to Br is 3:8 so multiple the moles of Fe3Br8 to Fe by a factor of 3. then lastly we multiple the molar mass of Fe. I need a glass of water after that. see work attached.
IMG_2757 2.pdf
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