Notice that we have a #1:2#mole ratio between #Fe_2O_3# and #Al#, and a #1:1#mole ratio between #Fe_2O_3# and #Al_2O_3#, since this will become useful later on.
Here are the standard state enthalpy value for the reactans and the products
For this reaction. #DeltaH# is equal to the sum of the #DeltaH_f^@#'s of the products minus the sum of the #DeltaH_f^@#'s of the reactants - each mutiplied by their stoichiometric coefficients
Since this reaction is exothermic, the heat given off will be equal to
#q = -DeltaH = 850kJ#
However, the heat given off is set to be #q = 650kJ#, less than what we've calculated so far; this means that #DeltaH# is bigger (since -650 is bigger than -850), which in turns means that fewer moles reacted.
Since the moles of #Al# and #Fe# do not influence the reaction's enthalpy, we'll focus on #Fe_2O_3#. Let's assume we have x moles of #Fe_2O_3#, instead of 1 mole, to start with. #DeltaH# will become