## Chemistry: The Molecular Science (5th Edition)

The energy transfer in this reaction is equal to $-205.4kJ$.
Molar Mass ($Al_2O_3$): 26.98* 2 + 16* 3 = 101.96g/mol 1. Calculate the amount in moles: $12.5g \times \frac{1 mol (Al_2O_3)}{ 101.96} = 0.1226 mol$ 2. Divide the amount by the coefficient of $Al_2O_3$: $\frac{0.1226}{ 1} = 0.1226 mol - reaction$ 3. Find the necessary heat: ** As we calculated in the last exercise $Δ_fH° = -1675.7 kJ/mol$ (for aluminum oxide) $0.1226mol-reaction \times -1675.7 kJ/mol-reaction = -205.4kJ$