Solve Fe(NH4)2(SO4)2. xH2O: Calculate Moles of Iron & Sulfate

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The discussion centers on calculating the moles of sulfate and iron in the compound Fe(NH4)2(SO4)2. xH2O. A sample weighing 0.982g was dissolved, and after adding excess BaCl2, a precipitate of BaSO4 weighing 1.17g was obtained. The moles of sulfate were calculated to be 0.00501, while the moles of iron were determined to be 0.00250, which is half of the sulfate amount due to the stoichiometry of the compound.

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Hi, I would be very grateful if anyone could show me how to answer the following question:

The value of x in Fe(NH4)2(SO4)2. xH2 O can be found by determining the amount in moles of sulfate in the compound.
A 0.982g sample was dissolved in water and excess BaCl2 (aq) was added.
The precipitate of BaSO4 was separated and dried and found to weigh 1.17g.

(a) Calculate the amount, in moles, of sulfate in the 0.982g sample of Fe(NH4)2(SO4)2. xH2O

(according the the answers it's 0.00501 which I do not understand)

(b) Calculate the amount, in moles, of iron in the 0.982g sample of Fe(NH4)2(SO4)2. xH2O

(apparently the answer is 0.00250 which is half of the previous answer - why?!)

Thanks in advance.
 
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You have asked the same question at CF and you were already given hints there, nobody is going to solve the question for you.
 
Borek said:
You have asked the same question at CF and you were already given hints there, nobody is going to solve the question for you.

The hints were wrong though. Its ok though, I understand it now.
 
CSG18 said:
The hints were wrong though.

They were perfect.
 

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