Problem L.35

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Swetha Sundaram 1E
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Problem L.35

Postby Swetha Sundaram 1E » Thu Oct 05, 2017 7:43 pm

Hey! Can anybody help me set up problem L.35 because I know I first have to balance the three reactions and then convert the tons into grams, but then afterwards do I basically work backwards from 8 mol NaBr to the mol of Fe using stoichiometry?

Jared Smith 1E
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Joined: Fri Sep 29, 2017 7:04 am

Re: Problem L.35

Postby Jared Smith 1E » Thu Oct 05, 2017 8:42 pm

Once you have converted the tons to grams, divide by the molar mass of NaBr which will give you the moles of NaBr that need to be produced. Then multiply the by the mole ratio of (1mol Fe(sub3)Br(sub8)/8mol NaBr). Then refer to the previous equation for the mol ratio of (3mol FeBr(sub2)/1mol Fesub3)Br(sub8) and multiply. Lastly, look to the first equation for the mol ratio of (1mol Fe/1mol FeBr(sub2) and multiply again. Then convert from moles to grams and then to kilograms.

jennywu
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Re: Problem L.35

Postby jennywu » Fri Oct 06, 2017 1:29 am

Firstly, balance the equations

Fe + Br --> FeBr2
3FeBr2 + Br2 --> FeBr8
Fe3Br8 + 4Na2CO3 --> 8NaBr + 4CO2 + Fe3O3

Then convert tonnes into kgs.

m= 2.5t= 2500kg

You can merge all the above equations into one big equation. Whereby all the reactants of all three reactions go on the left hand side and all products on the right.

You get:
3Fe + 3Br2 + Br2 + 4Na2CO3 --> 8NaBr + 4CO2 + Fe3O4

Simplify:
3Fe + 4Br2 + 4Na2CO3 --> 8NaBr + 4CO2 + Fe3O4

Fe: NaBr = 3:8 ratio

N(NaBr) = 2500 / ( 22.99 + 79.90) = 24.29779

N(Fe) = (3/8) * 24.29779 = 9.1116
m(Fe) = 9.1116 * 55.84 = 508.795 = 509. 3sf kgs

Hope this helps :)


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