Based on the chemical equation provided, 1 mole of Fe2O3 will react with 3 moles of Carbon to form 2 moles of Fe and 3 moles of CO (these numbers come from the coefficients).
Since we have excess carbon, that means the amount of Fe formed depends on how much Fe2O3 we are reacting. This is known as the "limiting reactant". 1 mole Fe2O3 forms 2 moles of Fe, so no matter how much Fe2O3 we react, we would expect to get twice that amount of Fe to maintain that 2:1 ratio.
This means that the theoretical yield from 0.313 mol of Fe2O3 is 0.626 mol of Fe. Work shown below.
2 mol Fe
0.313 mol Fe2O3 • 1 mol Fe2O3 = 0.626 mol Fe
We are expecting to form 0.626 mol Fe. We have 13.3 g. We need to convert the grams into moles to determine our actual yield. On the periodic table, Fe has a mass of 55.9. This means that 1 mole of Fe weighs 55.9 grams. We can use this as a conversion factor to calculate.
1 mol Fe
13.3 g Fe • 55.9 g Fe = 0.238 mol Fe
We only formed 0.238 mol Fe but theoretically we should have had 0.626 mol. We can calculate our percent recovery by dividing actual yield by theoretical yield.
0.238 mol Fe
0.626 mol Fe = 0.38 = 38%
Note how all the calculations are set up so that the starting units will cancel out to give you your new unit of measurement.
