Question

How many kilojoules of heat will be released when exactly 1 mole of aluminum, Al, is burned to form Al2O3(s) at standard state conditions?

Answer 1

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Answer 2

Answer 3

The standard enthalpy of formation (ΔHf°) of Al2O3(s) is the heat released when 1 mole of Al2O3 is formed from its elements in their standard states at standard state conditions.

The balanced chemical equation for the formation of Al2O3 from Al is: 4 Al(s) + 3 O2(g) → 2 Al2O3(s)

The standard enthalpy of formation of Al2O3 is -1675 kJ/mol.

Since the stoichiometric coefficient of Al is 4 in the balanced equation, we need to find the heat released when 4 moles of Al react to form 2 moles of Al2O3.

Heat released for 4 moles of Al reacting to form 2 moles of Al2O3: = (2 moles of Al2O3 / 4 moles of Al) * (-1675 kJ/mol) = -837.5 kJ/mol

Therefore, when exactly 1 mole of aluminum (Al) is burned to form Al2O3(s) at standard state conditions, 837.5 kilojoules (kJ) of heat will be released.