Question

How much energy is evolved during the reaction of 48.7 g of al, according to the reaction below? assume that there is excess fe2o3. fe2o3(s) + 2 al(s) → al2o3(s) + 2 fe(s) δh°rxn = -852 kj?

Answer 1

The standard DH for this reaction is -852 kJ/rxn. When attacking a problem like this, the first thing to always do is to convert grams (which is a unit of measurement) into moles (which is a unit of chemistry). The atomic weight of aluminum is 26.98 g/mol, so you can divide your mass by the molecular weight to find that you have 1.81 mol of aluminum. Note then that you use up two moles of aluminum per reaction cycle (as defined by the coefficient), so you can divide your number of moles of aluminum by 2 mol Al/rxn. So now you know that you have enough aluminum for .903 reaction cycles, so you can multiply your standard enthalpy by the number of reaction cycles that can progress, and you should end up with a total of -769 kJ of energy, which equates to a release or evolution (rather than an absorption) of 769 kJ of energy.