First, write out the balanced equation.
2Al + Fe_2O_3 -> 2Fe + Al_2O_3
Second, convert the grams into moles for both substances.
124 g Al * (1 mol Al)/(13 g Al) = 124/13molAl ~~ 9.538
601 g Fe_2O_3 * (1 mol Fe_2O_3)/(160 g Fe_2O_3) = 601/160molFe_2O_3 ~~ 3.756
Third, find the limiting reactant(click here for a KhanAcademy video teaching this concept) by determining how many moles of Al_2O_3 can be created with each one.
124/13molAl * (1molAl_2O_3)/(2molAl) = 124/26molAl_2O_3 ~~ 4.769
601/160molFe_2O_3 * (1molAl_2O_3)/(1molFe_2O_3) = 601/320molAl_2O_3 ~~ 1.878
1.878 < 4.769 so Fe_2O_3 is the limiting reactant and Al is the excess reactant.
Finally, convert the moles of Al_2O_3 formed by the limiting reactant into grams.
601/320molAl_2O_3 * (102gAl_2O_3)/(1molAl_2O_3) = 30651/160gAl_2O_3 ~~ 191.569