To calculate the standard enthalpy change for the fusion reaction between aluminum and oxygen, we need to use the following equation:H = n * Hfwhere H is the standard enthalpy change, n is the number of moles of the product, and Hf is the standard enthalpy of formation of the product.The balanced chemical equation for the reaction between aluminum and oxygen to form aluminum oxide is:4 Al s + 3 O2 g 2 Al2O3 s The standard enthalpy of formation Hf for aluminum oxide Al2O3 is -1675.7 kJ/mol.First, we need to find the number of moles of aluminum Al in 10 grams:Molar mass of Al = 26.98 g/molmoles of Al = mass / molar massmoles of Al = 10 g / 26.98 g/mol = 0.3707 molFrom the balanced equation, we can see that 4 moles of Al react to form 2 moles of Al2O3. Therefore, we can find the moles of Al2O3 formed:moles of Al2O3 = 0.3707 mol Al * 2 mol Al2O3 / 4 mol Al = 0.1854 mol Al2O3Now, we can calculate the standard enthalpy change H for the reaction:H = n * HfH = 0.1854 mol Al2O3 * -1675.7 kJ/mol H = -310.6 kJThe standard enthalpy change for the fusion reaction between aluminum and oxygen to form aluminum oxide when 10 grams of aluminum reacts completely with excess oxygen under standard conditions is approximately -310.6 kJ.