Question

500 g of Fe reacted with water to yield 4.2 g of H2. Fe3O4 was also produced in the reaction. What is the %yield of H2?

Answer 1

2Fe + 3H2O ==> 3H2 + Fe2O3
mols Fe = grams/molar mass = approx 9 but that's an estimate. You need a more accurate number than that.
Using the coefficients in the balanced equation, convert mols Fe to mols H2. That's approx 9 mols Fe x (3 mols H2/2 mols Fe) = approx 13.5
Convert mols H2 to grams. g = mols x molar mass = 13.5 x 2 = approx 27 grams. This is the theoretical yield (TY). The actual yield (AY) is 4.2g
%yield = apprx (4.2/27)*100 = ?

Answer 2

How would i go about finding the theoretical yield of Fe3O4? Would I just plug in one of those numbers into the % yield equation?

Answer 3

That's just another stoichiometric problem isn't it?
You have mols Fe, use the coefficients in the balanced equation to convert mols Fe to mols Fe2O3 and convert that to grams. OR since you already have mols H2, you can use mols H2 and convert using coefficients to mols Fe2O3 and from there to grams. All of these stoichiometry problems are worked the same way.
1. Write and balance the equation.
2. Change grams of what you have to mols. mols = grams/molar mass
3. Use the coefficients in the balanced equation to convert mols of what you have to mols of what you want.
4. Then convert mols of what you want to grams. g = mols x molar mass. This is the theoretical yield