If 17.3g of powdered aluminium are allowed to react with excess Fe2O3, how much heat is produced?
2Al(s) + Fe2O3(s) --> 2Fe (s) + Al2O3 (s).
{(−1669)−(−822)}⋅kJ⋅mol-1
=−847⋅kJ⋅mol-1.
ΔH°rxn=−847⋅kJ⋅mol-1 .
=-847x (17.3 g/2*26.98g)
=-271.555J
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