I can't confirm those dH formation values but I'll assume you have them right. In calculating, however, when you finish isn't that 3344.1 kJ that many kJ for the reaction as shown and not kJ/mol.
It appears you used dHrxn = (n*dHformation products) - (n*dH formation reactants). Having said all of that, however, you did not multiply by 3 and you used the 3344.1 correctly with the 0.228. Other than that I don't see anything wrong with this.
Calculate enthalpy change when 35.0 grams barium oxide reacts with 5.00 grams of aluminum.
0.228 mol BaO ( limiting reagent)
0.1853 mol Al
Molar Ratio 1.23
-1669.8 - 3(-558.1) = 3344.1 kJ/mol
0.228 BaO reacts therefore:
(0.228/3) x 3344.1 = 254.2 kJ
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