How much energy is evolved during the reaction of 51.2 g of al, according to the reaction below? assume that there is excess fe2o3. fe2o3(s) + 2 al(s) â al2o3(s) + 2 fe(s) δh°rxn = -852 kj?

Respuesta :

Answer : The amount of energy evolved during the reaction is, [tex]-807.696KJ[/tex].

Solution : Given,

[tex]\Delta H=-852KJ[/tex]

Mass of Al = 51.2 g

Molar mass of Al = 27 g/mole

First we have to calculate the moles of Al.

[tex]\text{ Moles of Al}=\frac{\text{ Mass of Al}}{\text{ Molar mass of Al}}=\frac{51.2g}{27g/mole}=1.896moles[/tex]

The balanced combustion reaction is,

[tex]Fe_2O_3+2Al\rightarrow Al_2O_3+2Fe[/tex]

From the given reaction, we conclude that

As, 2 moles of Al evolved energy = [tex]-852KJ[/tex]

So, 1.896 moles of Al evolved energy = [tex]\frac{-852KJ}{2moles}\times 1.896moles=-807.696KJ[/tex]

Therefore, the amount of energy evolved during the reaction is, [tex]-807.696KJ[/tex].

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