[ARGOS++]

Dear Hwgw, Dear Reader;

This initial Question deserves a better “End”/Answer!

So I will show you an easy way how to solve this, but also little more difficult Problems:
Your Scheme/Diagram for your first Question should look like:
A.)    Reaction:             Fe³⁺   +     C₂O₄ ^2-     =         [Fe(C₂O₄) ₃]^3-
B.)    Balancing:         1 Fe³⁺   +  3 C₂O₄ ^2-      =     1 [Fe(C₂O₄) ₃]^3-

The lines: C₁.) and C_m.) can be omitted in your case, because from the ratio you can set:  1 part = 1.0 Mole.  That gives you:
C₂.)   Moles:                1.0 m          4.0 m       
To get the Multipliers you have to divide the known values from C₂.) by the corresponding Indices from B.):   
D.)   Multiplier:             1.00            1.3333

As the “Multiplier” is telling you how many times you can fulfil the balanced Equation B.) with your Amount of this Reagent, so the lowest “Multiplier” of all is indicating the “Limiting Reagent”. 
All other Reagents are in Excess, or just enough.

In the same manner you can solve the second part of your Question.

For the whole Recipe (incl. Excesses) and future explanations you may visit:   "Stoichiometry Problem”.

I hope it may be of help to you all.

Dear Mr. AWK:   Did you realise what you wrote in the second half of your last sentens?
                       It is really worth to be remembered!:

     .. ... , if this attempt is useful

Good Luck!
                    ARGOS⁺⁺