theoretical yield?

Iron Oxide (Ferric Oxide) + Aluminum ➜ Iron + Aluminum Oxide
Fe₂O₃ + 2Al ➜ 2Fe + Al₂O₃

molecular weights:
Fe = 55.85
O = 15.99
Al = 26.98
Fe₂O₃ = 2•55.85 + 3•15.99 = 111.7+47.97 = 159.67
2Al = 53.96
2Fe = 111.7
Al₂O₃ = 53.96+ 47.97 = 101.93
check: 159.67+ 53.96 = 111.7+ 101.93 = 213.63

159.67 g of Fe₂O₃ + 53.96 g of Al ➜ 111.7 g of Fe + 101.93 g of Al₂O₃


Given that you produce 171.7 g of iron from 500.0 g Al and 500.0 g of
Fe2O3, calculate the mass of iron that should be produced from this reaction

first which is the limiting reactant. 
from above, that is the Fe₂O₃ as about a third as much of Al is needed

ratio of Fe to Fe₂O₃ is 111.7/159.67
therefore 111.7/159.67 = x/500
x = 349.8 g of Fe

but you get only 171.7 g so your actual yield is much lower. 

Molar mass of Fe = 55.85 g/mol
Molar mass of Fe₂O₃ = 55.85×2 + 16.00×3 = 159.7 g/mol
Molar mass of Al = 26.98 g/mol

Initial moles of Al = (500.0 g) / (26.98 g/mol) = 18.532 mol
Initial moles of Fe₂O₃ = (500.0 g) / (159.7 g/mol) = 3.1309 mol

Balanced equation for the reaction:
2Al(s) + Fe₂O₃(s) → 2Fe(s) + Al₂O₃(s)

Mole ratio Al : Fe₂O₃ = 2 : 1
If Fe₂O₃ completely reacts, Al needed = (3.1309 mol) × 2 = 6.2618 mol < 18.532 mol
Al is in excess, and thus Fe₂O₃ is the limiting reactant (limiting reagent).

Mole ratio Fe₂O₃ : Fe = 1 : 2
Maximum moles of Fe produced = (3.1309 mol) × 2 = 6.2618 mol
Theoretical yield of Fe = (6.2618 mol) × (55.85 g/mol) = 349.7 g