Preparation of a complex iron salt

Anonymous
timer Asked: Jul 7th, 2014
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Question Description

calculate the number of grams (theoretical yield)of the final iron complex that can be produced from 10.00 g of ferrous ammonium sulfate hexahydrate

Tutor Answer

dark knight
School: UT Austin

Thank you for the opportunity to help you with your question!

this is the chemical euaqtion for preparation of salt

2Fe(NH4)2(SO4)2 * 6H2O(aq) + 3H2C2O4(aq) + H2O2(aq) + 3K2C2O4(aq) ----- 2K3Fe(C2O4)3 *

+ 3H2O(s) + 8H2O(L) + 2(NH4)2SO4(aq) + 2H2SO4(aq) 


Look at the equation: 2 moles complex reactant produces 2 moles complex product.

The number of moles of the product will be the same as the number of moles reactant.

moles reactant = mass / molar mass = 10 g / 284.049 g/mol = 0.0252.

mass product = moles * molar mass = 0.0252 mol * 491.2 g/mol = 12.378 = 12.38 g.

so the number of grams (theoretical yield)of the final iron complex that can be produced from 10.00 g of ferrous ammonium sulfate hexahydrate is 12.38 gram.

Please let me know if you need any clarification. I'm always happy to answer your questions.

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Anonymous
Excellent job

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