If I perform this reaction with 25 grams of iron (III) phosphate and an excess of sodium sulfate, how many grams of iron (III) sulfate can I make?

Question

2FePO4 + 3Na2SO4 --> Fe2(SO4)3 + 2Na3PO4

Answer 1

So, the first step in any stoichiometry question is to find the molecular weights of the compounds involved:

FePO4 = 55.845+ 30.974 + 16*4 = 150.819 g/mol

Fe2(SO4)3 = 55.845*2 + (32.065+16*4)*3= 399.885 g/mol

 

The next step is to find the number of moles of your reactant, FePO4:

grams of substance/molecular weight of substance= moles of substance

 

25g/150.819g/mol= 0.166 moles of FePO4

 

Then, think about the proportion of how many moles of your reactant there are for each mole of your product. In this case, there are 2 moles of FePO4 for every 1 mole of Fe2(SO4)3.

 

0.166 mole FePO4 * 1 mole Fe2(SO4)3/ 2 moles of FePO4 = .083 mole of Fe2(SO4)3

 

Now, multiply the number of moles of your product by its molecular weight:

0.083 moles of Fe2(SO4)3 * 399.885 g/mol = 33.190 grams of Fe2(SO4)3

Let me know if you have any further questions.