The Enthalpy Change of a ChemicalReactionExperiment 11.&nbspRec

The Enthalpy Change of a ChemicalReactionExperiment 11. Record the following for each of thethree trials:· Trial 1aMass of the empty calorimeter (g)18.6gbInitial temperature of the calorimeter (°C)21.5°CcMaximum temperature in the calorimeter from the reaction (°C)35.0°C dCalculate ΔT by subtracting (b) from (c) (°C) ΔT = Tmaximum- Tinitial35.0°C-21.5°C=13.5°CeMass of the calorimeter and its contents after the reaction (g)68.74gfCalculate the mass of the contents of the calorimeter (g) bysubtracting (a) from (e)68.74g-18.6g=50.14ggCalculate the moles of Mg reacted (MW = 24.305 g/mole)?50mL of 1M HCI is 50/1000×1=0.05.15/24.305=.006g/mole· Trial 2aMass of the empty calorimeter (g)18.6gbInitial temperature of the calorimeter (°C)21.5°CcMaximum temperature in the calorimeter from the reaction (°C)44.0°CdCalculate ΔT by subtracting (b) from (c) (°C) ΔT = Tmaximum- Tinitial44.0°C-21.5°C=22.5°CeMass of the calorimeter and its contents after the reaction (g)68.83gfCalculate the mass of the contents of the calorimeter (g) bysubtracting (a) from (e)68.83-18.6=50.23gCalculate the moles of Mg reacted (MW = 24.305 g/mole).25/24.305=.010g/mole· Trial 3aMass of the empty calorimeter (g)18.6gbInitial temperature of the calorimeter (°C)21.5°CcMaximum temperature in the calorimeter from the reaction (°C)52.9°CdCalculate ΔT by subtracting (b) from (c) (°C) ΔT = Tmaximum- Tinitial52.9°C-21.5°C=31.4°CeMass of the calorimeter and its contents after the reaction (g)68.92gfCalculate the mass of the contents of the calorimeter (g) bysubtracting (a) from (e)68.92g-18.6g=50.32gCalculate the moles of Mg reacted (MW = 24.305 g/mole).35/24.305=.014g/mole2. Calculate the heat releasedinto the solution for the 3 reactions, according to the formula_qreaction= (Ccal * ΔT) + (mcontents* Cpcontents* ΔT)3. Assume Cpcontents= Cpwater= 4.18 J/g °CaTrial 1 (J)(4.18J/g°C)(68.74g)(35.0°C-21.5°C)=3878.9982 J/g°CbTrial 2 (J)(4.18J/g°C)(68.83g)(44.0°C-21.5°C)=6473.4615 J/g°CcTrial 3 (J)(4.18J/g°C)(68.92g)(52.9°C-21.5°C)=9045.88784 J/g°C4. Find the molar heat ofreaction for each experiment in units of kilojoules / (mole of Mg) by dividingthe heat of reaction (converted to kJ by dividing by 1000) by the moles of Mgused.aTrial 1 (kJ/mol)bTrial 2 (kJ/mol)cTrial 3 (kJ/mol)5. Calculate andrecord the average molar heat of reaction from the three results: