How much energy is evolved during the reaction of 48.7 g of Al, according to the reaction below? Assume that there is excess Fe2O3.
Fe2O3(s) + 2 Al(s) → Al2O3(s) + 2 Fe(s), ΔH°rxn = -852 kJ
To determine the energy evolved during the reaction, we first need to calculate the number of moles of Al.
The molar mass of Al is 26.98 g/mol.
We can use the given mass of Al to calculate the number of moles:
Number of moles of Al = Mass of Al / Molar mass of Al
= 48.7 g / 26.98 g/mol
≈ 1.805 mol
Now, let's use the balanced equation to determine the energy evolved per mole of Al reacted. According to the balanced equation:
Fe2O3(s) + 2 Al(s) → Al2O3(s) + 2 Fe(s)
1 mole of Al produces -852 kJ of energy.
Therefore, 1.805 moles of Al would produce:
Energy evolved = (-852 kJ/mol) × (1.805 mol)
≈ -1538.66 kJ
Hence, approximately 1538.66 kJ of energy is evolved during the reaction of 48.7 g of Al.
To calculate the energy evolved during the reaction, we first need to determine the moles of aluminum (Al) used in the reaction. We can use the molar mass of aluminum to convert grams to moles.
Given:
Mass of Al = 48.7 g
Molar mass of Al = 26.98 g/mol
Number of moles of Al = Mass of Al / Molar mass of Al
Number of moles of Al = 48.7 g / 26.98 g/mol
Next, we need to use the stoichiometry of the balanced equation to determine the moles of Fe2O3 used in the reaction. According to the balanced equation, the ratio between Fe2O3 and Al is 1:2.
Number of moles of Fe2O3 = (Number of moles of Al) / 2
Now that we have the moles of Fe2O3 used, we can calculate the energy evolved during the reaction using the molar enthalpy change (ΔH°rxn).
Energy evolved = (Number of moles of Fe2O3) x (ΔH°rxn)
Plug in the values to calculate:
Energy evolved = [(48.7 g / 26.98 g/mol) / 2] x (-852 kJ)
Simplify the calculation:
Energy evolved = [(48.7 / 26.98) / 2] x (-852)
Therefore, the energy evolved during the reaction of 48.7 g of Al is equal to the calculated value.