3 FeO(l) + 2 Al(l) → 3 Fe(l) + Al2O3(s)
This is the best I could do for cutting and pasting a useful equation from my PowerPoint slides. The superscripts and subscripts do not carry over.
If you have 25.0 grams of each reactant, which is limiting?
How do you approach this problem? Divide 25.0 grams by the molar mass of each reactant. (71.85 g/mole for FeO and 26.98 g/mole for Al). This gives you moles of each reactant. You cannot compare grams to grams (as given in this case) because it is a nonsensical relationship in chemistry terms. You must compare moles to moles. Now, use moles of each reactant to figure out which gives the LEAST amount of product. This is where you use your balanced chemical equation to find the molar ratio of reactant to product. In this case you have 3 moles of Fe(l) for 3 moles of FeO(l). Essentially this is a 1:1 relationship. So your moles of reactant for FeO is the same as moles of product of Fe(l). For Aluminum, you have 3 moles of FeO (from the coefficient of the balanced equation) for every 2 moles of aluminum. Therefore it is a 3/2 relationship of moles of product to moles of reactant.
Although you can compare moles of product from one reactant to moles of product from another reactant to determine the limiting reagent, many times this number is converted back into grams by multiplying by the molar mass of the product you are dealing with.
The comparison in this case is between 19.4 grams of Fe(l) (from 25.0 grams of FeO) versus 77.6 grams of Fe(l) from 25.0 grams of Aluminum. By definition, the limiting reactant is the one that gives the least amount of product. Therefore, the Iron II oxide is the limiting reactant. The 19.4 grams of product is your maximum amount of product from these two given amounts of reactant, or in other words, it is your theoretical yield.