Calculate the molar enthelpies of reaction occuring in parts A and B

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SUMMARY

The discussion focuses on calculating the molar enthalpies of reaction for magnesium (Mg) and magnesium oxide (MgO) using calorimetry data. In Part A, the enthalpy change for 0.31 g of Mg is calculated to be -8 kJ, while in Part B, the enthalpy change for 1.22 g of MgO is determined to be -18 kJ. The calculations utilize the formula for enthalpy change based on the mass of the reactants and their respective molar enthalpies, specifically -601.1 kJ/mol for Mg. Additionally, a suggestion to use the formula q=mcΔT for heat transfer calculations is mentioned.

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Homework Statement


Part A:
Mass of Mg (s) = 0.31 g
Initial temperature of calorimeter contents = 24.1 C
Finial temperature of calorimeter contents = 36.8 C

Part B:
Mass of MgO (S) = 1.22 g
Initial temperature of calorimeter contents = 24.0 C
Finial temperature of calorimeter contents = 31. 9 C

100.0 ML of HCL used and -601.6 KJ/mol Mg reacted

Homework Equations


Calculate the molar enthalpies of reaction occurring Parts A and B.

The Attempt at a Solution


Part A (answer):
Enthalpy change of reaction = (chemical amount)(molar enthalpy of reaction)
=Enthalpy change or reaction= (0.31g)(1mol/24.31)(-601.1 kJ/mol/1mol)
= -7.665 kJ
= -8 kJ

Part B (answer):
Enthalpy change of reaction = (chemical amount)(molar enthalpy of reaction)
=Enthalpy change of reaction= (1.22g)(1mol/40.31g)(-601.1 kJ/mol/1mol)
= -18.19 kJ
= -18 kJ

Homework Statement


Homework Equations


The Attempt at a Solution

 
Last edited:
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Shouldn't you use q=mcΔT?
 

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