I don't have MgCO3 delta H.
I have -393.5 kJ for CO2 and -601.8 kJ for MgO. I found -1112 kJ on the web for MgCO3 but I don't know how reliable that is. If we take that number, then
delta Hrxn = (products)-(reactants) =
(MgO)+(CO2)-(MgCO3)
(-601.8) + (-393.5) -(-1112) = 116.7 kJ/mol MgCO3.
We have 54.7 grams = 54.7/84.31 = 0.649 moles. Then 116.7 x 0.649 = 75.7 kJ
which is close. You must have them listed in your text or in the problem Just substitute the numbers you have.
Calculate the enthalpy change in kilojoules when 54.7g of MgCO3 decomposes according to the following equation:
MgCO3 (s) into MgO (s) + CO2 (g)
My delta H answer was 100.6 KJ but the real answer is 76.1 KJ. Can someone please help?
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