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Question

Calculate ΔH for the following reaction:
8Al(s)+3Fe3O4(s)4Al2O3(s)+9Fe(s)
Given ΔHf for Al2O3(s) and Fe3O4(s) are -1669.8 kJ and -1120.9 kJ respectively.

A
-3316.5 kJ
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B
-3650.4 kJ
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C
-2650.4 kJ
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D
None of the above
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Solution

The correct option is A -3316.5 kJ
Given, ΔHf(Al2O3(s))=1669.8 kJ and ΔHf(Fe3O4(s))=1120.9 kJ

the reaction is 8Al(s)+3Fe3O4(s)4Al2O3(s)+9Fe(s)

as per convention
ΔHf(Fe(s))=0 kJ, ΔHf(Al(s))=0 kJ

We know that ΔH for a reaction =ΔHf(products)ΔHf(reactants)

=4ΔHf(Al2O3(s))+9ΔHf(Fe(s))3ΔHf(Fe3O4(s))8ΔHf(Al(s))

=4×1669.8+3×1120.9
=3316.5 kJ

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