Question

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

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

207 kJ

241 kJ

415 kJ

769 kJ

130 kJ

Answer 1

Let us write the reaction given

Fe₂O₃(s) + 2 Al(s) → Al₂O₃(s) + 2 Fe(s)

Heat of reaction = -852 kJ

Weight of Al = 48.7 g

Molar mass of Al = 27 gram/mole

Moles of Al = weight of Al/ molar mass of Al = 48.7 /27 = 1.8037 moles

As per reaction stoichiometry 1mole Fe2CO3 = 2 moles Al

moles of Fe2CO3 = (1/2)*1.8037 = 0.90185 moles

Heat of reaction per mole of Fe2CO3 = 852 KJ

therefore for 0.90185 moles = 0.90185 *852 =768.3762 KJ

Hence answer should be option (4) 769 KJ