dHrxn = (n*dHf products) - (n*dHf reatants).
That gives you dH/mol Al2O3, per 3 mols Fe, per 3 mols FeO
, or per 2 mols Al.
And note that 250 C is not 298K. I assume that 0 is a typo and you meant for it to be a degree symbol.
Calculate the standard heat of reaction at 250C (298K) and 1 atm pressure of
3FeO(s) + 2Al(s) → Al2O3(s) + 3Fe(s)
in terms of per mole of Al2O3 formed, per mole of Fe formed, per mole of FeO reacted, per mole of Al reacted and per g of Fe formed.
ΔH298K,FeO(s) = -63.3 kcal/mol (-264.84 kJ/mol)
ΔH298K,Al2O3(s) = -400 kcal/mol (-1673.6kJ/mol)
Aw,Fe(s) = 56 g/mol
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