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Author Topic: Calculate the theo. Yield of the Iron Complex  (Read 2502 times)

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GreenAssailant

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Calculate the theo. Yield of the Iron Complex
« on: November 27, 2011, 02:59:10 AM »

4.0g of ferrous ammonium sulphate, FeSO4 ∙ (NH4)2SO4 ∙ 6H2O is used. Since the oxalate is in excess, calculate the theoretical yield of the iron complex.

NOTE: We assume 100% conversion efficiency, going from 1 reaction to the next.
3 separate equations in a reaction series:

[1] FeSO4∙(NH4)2SO4∙ 6H2O + H2C2O4∙ 2H2O ---> FeC2O4 + (NH4)2SO4 + H2SO4 + 8H2O

[2] 6 FeC2O4 + 3H2O2 + 6K2C2O4∙ H2O ---> 4 K3[Fe(C2O4)3]∙ 3 H2O + 2 Fe(OH)3 + 6 H2O

[3] 2 Fe(OH)3 + 3 H2C2O4∙ 2H2O + 3 K2C2O4∙ H2O ----> 2 K3[Fe(C2O4)3]∙ 3H2O + 9 H2O

Attempt:
I found the molar mass of ammonium sulphate hydrate compound to be:

m[FeSO4∙(NH4)2 SO4∙6H2O] = 392.21 g/mol
Because we assume that the oxalate is in excess, that means the ammonium sulphate hydrate compound is obviously the limiting reagent. So I use molar mass and the grams provided to get the moles, which is:

mol[FeSO4∙(NH4)2 SO4∙6H2O] = 0.010198618 mol

Now, I get the moles of each product of the r'xn using the mol[FeSO4∙(NH4)2 SO4∙6H2O], but I am experiencing dilemma:

mol[FeC2O4] = mol[(NH4)2SO4] = mol[H2SO4]
^ All of these products have the same amount of moles produced.

My question is: which one do I use to proceed to the next reaction?
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Nobby

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Re: Calculate the theo. Yield of the Iron Complex
« Reply #1 on: November 27, 2011, 05:06:41 AM »

The moles are correct. But  write 0.01 mol is enough.

Now check how many mole  ammonia  iron sulfate correspond to the Oxalate complex.   The equations between are not important. Write on the right side your product.
« Last Edit: November 27, 2011, 05:23:56 AM by Nobby »
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