Asked by Mandy
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?
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?
Answers
Answered by
DrBob222
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.
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.
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