Given $$ 2 \mathrm{Al}_{2} \mathrm{O}_{3}(s) \longrightarrow 4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g) \quad \Delta H^{\circ}=3351.4 \mathrm{~kJ} $$ (a) What is the heat of formation of aluminum oxide? (b) What is \(\Delta H^{\circ}\) for the formation of \(12.50 \mathrm{~g}\) of aluminum oxide?

We are given the heat change for the decomposition of aluminum oxide: $$ 2 \mathrm{Al}_2 \mathrm{O}_3(s) \longrightarrow 4 \mathrm{Al}(s)+3 \mathrm{O}_2(g) \quad \Delta H^{\circ} = -3351.4 \mathrm{~kJ} $$ The above reaction is the decomposition of 2 moles of aluminum oxide into 4 moles of aluminum and 3 moles of oxygen. We can express the formation of aluminum oxide as: $$ 4 \mathrm{Al}(s) + 3 \mathrm{O}_2(g) \longrightarrow 2 \mathrm{Al}_2 \mathrm{O}_3(s) $$ From the given reaction, $$ \Delta H^{\circ}_\textrm{decomposition} = -3351.4 \mathrm{~kJ} $$ We know \(\Delta H^{\circ}_\textrm{formation} = -\Delta H^{\circ}_\textrm{decomposition}\), which gives us: $$ \Delta H^{\circ}_\textrm{formation} = -(-3351.4 \mathrm{~kJ}) = 3351.4 \mathrm{~kJ} $$ So, the heat of formation for 2 moles of aluminum oxide is 3351.4 kJ. For 1 mole of aluminum oxide, the heat of formation is: $$ \Delta H^{\circ}_\textrm{formation(1 mole)} = \frac{3351.4 \mathrm{~kJ}}{2} = 1675.7 \mathrm{~kJ} $$

Now, we need to calculate how many moles of aluminum oxide are represented by 12.50 grams. The molar mass of aluminum oxide is: $$ \mathrm{M_{Al_2O_3}} = (2 \times \mathrm{M_{Al}}) + (3 \times \mathrm{M_{O}}) = 2 \times 26.98 + 3 \times 16 = 101.96 \mathrm{\frac{g}{mol}} $$ The number of moles of aluminum oxide in 12.50 grams can be calculated: $$ \mathrm{moles_{Al_2O_3}} = \frac{\mathrm{mass_{Al_2O_3}}}{\mathrm{M_{Al_2O_3}}} = \frac{12.50 \mathrm{g}}{101.96 \mathrm{\frac{g}{mol}}} = 0.1225 \mathrm{mol} $$ Now, we can find the heat change for the formation of 12.50 grams (0.1225 moles) of aluminum oxide using the heat of formation for 1 mole of aluminum oxide: $$ \Delta H^{\circ}_\textrm{formation(12.50g)} = 0.1225 \mathrm{mol} \times 1675.7 \mathrm{\frac{kJ}{mol}} = 205 \mathrm{~kJ} $$ So, the heat change for the formation of 12.50 grams of aluminum oxide is approximately 205 kJ.