How much energy is evolved during the formation of 197 g of fe, according to the reaction below? fe2o3(s) + al(s) → al2o3(s) + fe(s) δh°rxn = -852 kj?

Answer is: 1,51·10³ kJ is evolved during the formation.
Chemical reaction: Fe₂O₃(s) + Al(s) → Al₂O₃(s) + Fe(s); ΔH°rxn = -852 kJ.
m(Fe) = 197 g.
n(Fe) = m(Fe) ÷ M(Fe).
n(Fe) = 197 g ÷ 55,85 g/mol.
n(Fe) = 3,527 mol.
Use proportion: 2 mol(Fe) : -854 kJ = 3,527 mol(Fe) : ΔH°rxn.
ΔH°rxn = -854 kJ · 3,527 mol ÷ 2 mol.
ΔH°rxn = -1510 kJ = -1,51·10³ kJ.

Answer Link

Alleei Alleei · Answer 2 · 2019-09-24T13:23:49+02:00

Answer : The energy evolved in the reaction is 1499.52 kJ

Explanation : Given,

Mass of Fe = 197 g

Molar mass of Fe = 56 g/mole

Enthalpy of reaction = -852 kJ

First we have to calculate the moles of Fe.

[tex]\text{ Moles of Fe}=\frac{\text{ Mass of Fe}}{\text{ Molar mass of Fe}}=\frac{197g}{56g/mole}=3.52moles[/tex]

Now we have to calculate the energy evolved in the reaction.

From the balanced chemical reaction we conclude that,

As, 2 mole of Fe evolved energy = 852 kJ

So, 3.52 mole of Fe evolved energy = [tex]\frac{3.52}{2}\times 852kJ[/tex]

= 1499.52 kJ

Therefore, the energy evolved in the reaction is 1499.52 kJ