First, we need to determine the number of moles of magnesium oxide MgO in the given mass:moles of MgO = mass / molar massmoles of MgO = 4.50 g / 40.31 g/molmoles of MgO = 0.1117 molNow, we can use the stoichiometry of the reaction and the given standard enthalpy change to calculate the enthalpy change for the reaction with 0.1117 mol of MgO:H = -120.3 kJ/mol 0.1117 molH = -13.43 kJThe standard enthalpy change for the reaction with 4.50 grams of magnesium oxide and excess hydrochloric acid is -13.43 kJ rounded to two decimal places .