[saiyanmx89]

How many grams of product (complex) would be obtained with 1.00g of ethylenediamine?
1NiCl₂*6H₂O(aq) + 3C₂H₈N₂(aq) -> 1[Ni(C₂H₈N₂)₃]*Cl₂(s) + 6H₂O(l)

I have no idea where to start.

[macman104]

ethylenediamine is your second reactant (the C₂H₈N₂).  So assume that you have an excess of 1NiCl₂*6H₂O(aq), and that your limiting reagent is C₂H₈N₂.  They want to know how many moles of 1[Ni(C₂H₈N₂)₃]*Cl₂(s) can you make if you have 1.00 grams of C₂H₈N₂.

So start by knowing that your C₂H₈N₂ is going to be the limiting reagent, so it will dictate how many moles of your complex you will make.  Now, make sure your equation is balanced, and work the problem the same way as before (except you will have different molar masses and ratios).

[saiyanmx89]

1.00gC₂H₈N₂ x (1 mol C₂H₈N₂/60.12g C₂H₈N₂) x (1 mol Ni(C₂H₈N₂)₃/3 mol C₂H₈N₂) x (239.06g Ni(C₂H₈N₂)₃/1 mol Ni(C₂H₈N₂)₃  = 1.33g Ni(C₂H₈N₂)₃ ?

[macman104]

Careful the complex is Ni(C2H8N2)3*Cl2 (I don't see the Cl in your equation), so you need to make sure your molar mass is correct.  However, your setup looks correct, as long as you've done the math correctly.