Question 14.2: Calculate the energy released as heat when 60.0 grams of iro......
Calculate the energy released as heat when 60.0 grams of iron(III) oxide, Fe_{2}O_{3}(s), is combined with 15.0 grams of Al(s) in the thermite reaction (Figure 14.7) at 25°C and a constant pressure of one bar. The equation for the reaction is
Fe_{2}O_{3}(s) + 2\ Al(s) → Al_{2}O_{3}(s) + 2\ Fe(s) \qquad ΔH_{rxn}^{\circ } = −851.5\ kJ·mol^{–1}
Because the masses of both reactants are given, we must check to see if either is a limiting reactant. The number of moles of each reactant is given by
moles\ of\ Al = (15.0\ g\ Al)\left(\frac{1\ mol\ Al}{26.98\ g\ Al} \right) = 0.556\ mol\ Al
moles\ of\ Fe_{2}O_{3}= (60.0\ g\ Fe_{2}O_{3})\left(\frac{1\ mol\ Fe_{2}O_{3}}{159.7\ g\ Fe_{2}O_{3}} \right) = 0.376\ mol\ Fe_{2}O_{3}
Because 0.556 moles of Al(s) requires (0.556/2 =) 0.278 moles of Fe_{2}O_{3}(s) to react, we see that aluminum is the limiting reactant and that Fe_{2}O_{3}(s) is in excess by (0.376 – 0.278 =) 0.098 moles.
From the stoichiometric coefficients of the balanced chemical equation, we see that -851.5 kJ of energy is produced as heat for every two moles of aluminum consumed. Therefore, the energy liberated as heat by the combustion of 15.0 grams of aluminum at one bar is
q = (15.0\ g\ Al)\left(\frac{1\ mol\ A}{26.98\ g\ Al} \right) \left(\frac{–851.5\ kJ}{2\ mol\ Al} \right)= –237\ kJThis is about -16 kJ per gram of aluminum and is one of the most exothermic reactions known.