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11 years ago
calculate the molar enthalpies of reaction for the reactions occurring in Parts A and B

reacting Mg and MgO with 100mL of HCl (aq)

Part A (results):
Mass of Mg(s) 0.31 g
Initial temperature of calorimeter contents 24.1 ºC
Final temperature of calorimeter contents 36.8 ºC

Part B (results):
Mass of MgO(s) 1.22 g
Initial temperature of calorimeter contents 24.0 ºC
Final temperature of calorimeter contents 31.9 ºC

 -601.6 KJ/mol Mg (accepted value)
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#1
Staff Member
Educator
11 years ago
The heat, Q, would be given by:
Q = -(mass of mixture in grams) * (heat capacity of mixture in joules per gram per degree Celsius) * (change in temperature in degrees Celsius)

The molar enthalpy is then Q/number of moles of solute.

You can calculate the change in temperature and the moles of solute, but you need the mass of solute plus HCl and the heat capacity of the mixture. You should have measured the former and the latter would have to be given to you or be looked up.
Mastering in Nutritional Biology
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EA Author
wrote...
#2
11 years ago
Quote from: padre (11 years ago)
The heat, Q, would be given by:
Q = -(mass of mixture in grams) * (heat capacity of mixture in joules per gram per degree Celsius) * (change in temperature in degrees Celsius)

The molar enthalpy is then Q/number of moles of solute.
 
You can calculate the change in temperature and the moles of solute, but you need the mass of solute plus HCl and the heat capacity of the mixture. You should have measured the former and the latter would have to be given to you or be looked up.
So for part A, is this correct?
Q=
   =-(100.31g)(4.19)(36.5C-24C)=5253.74J
H=Q/n
  =5253.74J/1mol
  =5253.74J/mol
   
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