What is the amount of limiting reagent in this chemical reaction?
$4.59$ is not "something" (you haven't specified the unit) of an $\ce{Al}$ mole, but your amount of $124\mathrm{~g}$ of $\ce{Al}$ is exactly $4.59$ moles of $\ce{Al}$, it's just a unit change. In reaction you wrote, every two moles of $\ce{Al}$ reacts with one mole of $\ce{Fe2O3}$. This may help you to find the solution.