Question 1. Record the following for each of the three trials: · Trial 1 a Mass of the empty calorimeter (g) 18.6g b Initial temperature of the calorimeter (°C) 21.5°C c Maximum temperature in the calorimeter from the reaction (°C) 35.0°C d Calculate ΔT by subtracting (b) from (c) (°C) ΔT = Tmaximum – Tinitial 35.0°C-21.5°C=13.5°C e Mass of the calorimeter and its contents after the reaction (g) 68.74g f Calculate the mass of the contents of the calorimeter (g) by subtracting (a) from (e) 68.74g-18.6g=50.14g g Calculate 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 2 a Mass of the empty calorimeter (g) 18.6g b Initial temperature of the calorimeter (°C) 21.5°C c Maximum temperature in the calorimeter from the reaction (°C) 44.0°C d Calculate ΔT by subtracting (b) from (c) (°C) ΔT = Tmaximum – Tinitial 44.0°C-21.5°C=22.5°C e Mass of the calorimeter and its contents after the reaction (g) 68.83g f Calculate the mass of the contents of the calorimeter (g) by subtracting (a) from (e) 68.83-18.6=50.23 g Calculate the moles of Mg reacted (MW = 24.305 g/mole) .25/24.305=.010g/mole · Trial 3 a Mass of the empty calorimeter (g) 18.6g b Initial temperature of the calorimeter (°C) 21.5°C c Maximum temperature in the calorimeter from the reaction (°C) 52.9°C d Calculate ΔT by subtracting (b) from (c) (°C) ΔT = Tmaximum – Tinitial 52.9°C-21.5°C=31.4°C e Mass of the calorimeter and its contents after the reaction (g) 68.92g f Calculate the mass of the contents of the calorimeter (g) by subtracting (a) from (e) 68.92g-18.6g=50.32 g Calculate the moles of Mg reacted (MW = 24.305 g/mole) .35/24.305=.014g/mole 2. Calculate the heat released into the solution for the 3 reactions, according to the formula: qreaction= (Ccal * ΔT) + (mcontents* Cpcontents* ΔT) 3. Assume Cpcontents= Cpwater= 4.18 J/g °C a Trial 1 (J) (4.18J/g°C)(68.74g)(35.0°C-21.5°C)= 3878.9982 J/g°C b Trial 2 (J) (4.18J/g°C)(68.83g)(44.0°C-21.5°C)= 6473.4615 J/g°C c Trial 3 (J) (4.18J/g°C)(68.92g)(52.9°C-21.5°C)= 9045.88784 J/g°C 4. Find the molar heat of reaction for each experiment in units of kilojoules / (mole of Mg) by dividing the heat of reaction (converted to kJ by dividing by 1000) by the moles of Mg used. a Trial 1 (kJ/mol) b Trial 2 (kJ/mol) c Trial 3 (kJ/mol) 5. Calculate and record the average molar heat of reaction from the three results:
Question
1. Record the following for each of the three trials:
· Trial
1
| a | Mass of the empty calorimeter (g)18.6g |
| b | Initial temperature of the calorimeter (°C)21.5°C |
| c | Maximum temperature in the calorimeter from the reaction (°C)35.0°C |
| d | Calculate ΔT by subtracting (b) from (c) (°C) ΔT = Tmaximum – Tinitial35.0°C-21.5°C=13.5°C |
| e | Mass of the calorimeter and its contents after the reaction (g)68.74g |
| f | Calculate the mass of the contents of the calorimeter (g) by subtracting (a) from (e)68.74g-18.6g=50.14g |
| g | Calculate 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 2
| a | Mass of the empty calorimeter (g)18.6g |
| b | Initial temperature of the calorimeter (°C)21.5°C |
| c | Maximum temperature in the calorimeter from the reaction (°C)44.0°C |
| d | Calculate ΔT by subtracting (b) from (c) (°C) ΔT = Tmaximum – Tinitial44.0°C-21.5°C=22.5°C |
| e | Mass of the calorimeter and its contents after the reaction (g)68.83g |
| f | Calculate the mass of the contents of the calorimeter (g) by subtracting (a) from (e)68.83-18.6=50.23 |
| g | Calculate the moles of Mg reacted (MW = 24.305 g/mole).25/24.305=.010g/mole |
· Trial 3
| a | Mass of the empty calorimeter (g)18.6g |
| b | Initial temperature of the calorimeter (°C)21.5°C |
| c | Maximum temperature in the calorimeter from the reaction (°C)52.9°C |
| d | Calculate ΔT by subtracting (b) from (c) (°C) ΔT = Tmaximum – Tinitial52.9°C-21.5°C=31.4°C |
| e | Mass of the calorimeter and its contents after the reaction (g)68.92g |
| f | Calculate the mass of the contents of the calorimeter (g) by subtracting (a) from (e)68.92g-18.6g=50.32 |
| g | Calculate the moles of Mg reacted (MW = 24.305 g/mole).35/24.305=.014g/mole |
2. Calculate the heat released into the solution for the 3 reactions, according to the formula:
| qreaction= (Ccal * ΔT) + (mcontents* Cpcontents* ΔT) |
3.
Assume Cpcontents= Cpwater= 4.18 J/g °C
| a | Trial 1 (J)(4.18J/g°C)(68.74g)(35.0°C-21.5°C)=3878.9982 J/g°C |
| b | Trial 2 (J)(4.18J/g°C)(68.83g)(44.0°C-21.5°C)=6473.4615 J/g°C |
| c | Trial 3 (J)(4.18J/g°C)(68.92g)(52.9°C-21.5°C)=9045.88784 J/g°C |
4. Find the molar heat of reaction for each experiment in units of kilojoules / (mole of Mg) by dividing the heat of reaction (converted to kJ by dividing by 1000) by the moles of Mg used.
| a | Trial 1 (kJ/mol) |
| b | Trial 2 (kJ/mol) |
| c | Trial 3 (kJ/mol) |
5. Calculate and record the average molar heat of reaction from the three results:
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