How do you do limiting reactant, theoretical yield, percent yield chemistry problems?

you follow these 6 steps.

1) write a balanced equation
2) convert everything given to MOLES
3) determine limiting reagent
4) convert moles limiting reagent to moles other species
5) convert moles back to mass. this is theoretical mass.  aka theoretical yield
6) % yield = actual recovered mass / theoretical mass x 100%

the idea is that since a balanced equation has coefficients in MOLE ratios, we can use those ratios to convert between moles of different species.  ok?  watch...

*** 1 ***
Fe2O3(s) + 3CO(g) ---> 3 Fe(s) + 3 CO2(g)... is NOT correctly balanced is it?  

Fe2O3(s) + 3CO(g) ---> 2 Fe(s) + 3 CO2(g)... is correctly balanced
......... ........ ........ ........ ....?..
....... ......... ...... ......this is correct

*** 2 ***
150g Fe2O3 x (1 mole Fe2O3 / 159.7g Fe2O3) = 0.939 moles Fe2O3

no other masses, volumes, atoms, whatever were given

*** 3 ***
Fe2O3 was limiting. that was given.

what you usually do is this.. calculate the amount of the other reactant needed to consume all of the first and compare that to how much you actually have available...

example...
from the balanced equation, 1 mole Fe2O3 reacts with 3 moles CO.. therefore the amount of CO we need to consume all the Fe2O3 we have (from step 2) is...

0.939 moles Fe2O3 x (3 moles CO / 1 mole Fe2O3) = 2.82 moles CO

if we had exactly that much available (let's say the problem said 150g Fe2O3 reacted with 79.0g CO), then the reactants are in a "stoichiometric" ratio.  either could be considered limiting.  if we had less than 2.82 moles CO available, then CO would be limiting (we didn't have enough CO).  if we had more than 2.82 moles CO, then CO would be in XS (eXcesS) and Fe2O3 would be limiting.

*** 4 ***
again, we convert moles limiting reagent to moles product via the coefficients of the balanced equation...

from the balanced equation, 1 mole Fe2O3 ---> 2 moles Fe... therefore...
0.939 moles Fe2O3 x (2 moles Fe / 1 mole Fe2Oe) = 1.88 moles Fe

*** 5 ***
1.88 moles Fe x (55.84g Fe / 1 mole Fe) = 104.9g Fe... this is theoretical yield..

*** 6 ***
% yield = actual / theoretical x 100% = 87.9g / 104.9g x 100% = 83.8%