To determine the change in enthalpy when 36.00 g of aluminum reacts with excess ammonium nitrate (NH4NO3) according to the given equation, we need to use the molar mass of aluminum and the stoichiometry of the reaction.
The molar mass of aluminum (Al) is 26.98 g/mol.
First, we need to calculate the moles of aluminum (Al) in 36.00 g of aluminum:
moles of Al = mass of Al / molar mass of Al
moles of Al = 36.00 g / 26.98 g/mol ≈ 1.334 mol
From the balanced equation, we can see that the stoichiometric ratio between aluminum (Al) and the change in enthalpy is 2: -2030 kJ. This means that for every 2 moles of aluminum reacting, the change in enthalpy is -2030 kJ.
Next, we can use the stoichiometry to calculate the change in enthalpy for the given amount of aluminum:
change in enthalpy = moles of Al * (change in enthalpy / stoichiometric coefficient of Al)
change in enthalpy = 1.334 mol * (-2030 kJ / 2) ≈ -1362.68 kJ
Therefore, the change in enthalpy when 36.00 g of aluminum reacts with excess ammonium nitrate (NH4NO3) is approximately -1362.68 kJ.