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#
