The enthalpy of formation of the solid Al2O3(s) is -1675.6 kJ/mol .
Using the relation;
ΔH°rxn = ∑ΔH°fproducts - ∑ΔH°freactants
We know that;
ΔH°f H2O(l) = -285.8 kJ/mol
ΔH°fAl(OH)3(s) = -1315 kJ/mol
ΔH°fAl2O3(s) = ?
Substituting values;
21.0kJ/mol =[ (ΔH°fAl2O3(s) ) + 3( -285.8 kJ/mol)] - 2 ( −1277kJ/mol)
21.0kJ/mol = (ΔH°fAl2O3(s) ) + 1696.6 kJ/mol
ΔH°fAl2O3(s) = 21.0kJ/mol - 1696.6 kJ/mol
ΔH°fAl2O3(s) = -1675.6 kJ/mol
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