Question

How much energy is evolved during the formation of 197 g of Fe, according to the reaction below?

Fe2O3(s) + 2 Al(s) → Al2O3(s) + 2 Fe(s) ΔH°rxn = -852 kJ

A) 964 kJ


B) 1.52 x 103 kJ


C) 3.02 x 103 kJ


D) 8.40 x 103 kJ


E) 482 kJ

Answer 1

ΔH for formation of 197g Fe⁰ = 1.503 x 10³ Kj => Answer choice 'B'

Step-by-step explanation:

Given Fe₂O₃(s) + 2Al⁰(s) => Al₂O₃(s) + 2Fe⁰(s) + 852Kj

197g Fe⁰ = (197g/55.85g/mol) = 3.527 mol Fe⁰(s)

From balanced standard equation 2 moles Fe⁰(s) => 852Kj, then ...

3.527 mole yield (a higher mole value) => (3.527/2) x 852Kj = 1,503Kj (a higher enthalpy value).

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NOTE => If 2 moles Fe gives 852Kj (exo) as specified in equation, then a higher energy value would result if the moles of Fe⁰(s) is higher than 2 moles. The ratio of 3.638/2 will increase the listed equation heat value to a larger number because 197g Fe⁰(s) contains more than 2 moles of Fe⁰(s) => 3.527 mole Fe(s) in 197g. Had the problem asked for the heat loss from less than two moles Fe⁰(s) - say 100g Fe⁰(s) (=1.79mole Fe⁰(s)) - then one would use the fractional ratio (1.79/2) to reduce the enthalpy value less than 852Kj.