For part A B & C do calculations and write the missing numbers in the tables **Part D: Application of Hess’s Law**a. **Write in your experimentally-derived enthalpy values in the following thermochemical equations.** 1. \( \text{Mg(s)} + 2 \text{HCl(aq)} \rightarrow \text{MgCl}_2\text{(aq)} + \text{H}_2\text{(g)} \) \hspace{10pt} \( \Delta H_1 = \underline{\hspace{50pt}} \) 2. \( \text{MgO(s)} + 2 \text{HCl(aq)} \rightarrow \text{MgCl}_2\text{(aq)} + \text{H}_2\text{O(l)} \) \hspace{10pt} \( \Delta H_2 = \underline{\hspace{50pt}} \)b. **Use Hess’s Law to get \( \Delta H_3 \) for reaction (3) from \( \Delta H_1 \) and \( \Delta H_2 \).** 3. \( \text{Mg(s)} + \text{H}_2\text{O(l)} \rightarrow \text{MgO(s)} + \text{H}_2\text{(g)} \) \hspace{10pt} \( \Delta H_3 = \underline{\hspace{50pt}} \)c. **Now calculate \( \Delta H_3 \) from standard molar enthalpies of formation found in the thermodynamic tables in the appendix.**d. **Discussion: Compare your experimental value of \( \Delta H_3 \) with the value calculated from the thermodynamic tables. Can you explain why they shouldn’t be exactly the same?**In this exercise, you will apply Hess's Law to calculate the enthalpy of a reaction (\( \Delta H_3 \)) by using experimentally-derived values (\( \Delta H_1 \) and \( \Delta H_2 \)) and then compare these values with those calculated from standard thermodynamic tables. This process involves analyzing potential discrepancies and understanding potential sources of experimental error. ## Calorimetry Experiment Data### Part A: Calorimeter Constant**Trials: 1**- **Temperature of 50.0 mL cool water (Tc):** 18.8°C- **Temperature of 50.0 mL warm water (Th):** 80.0°C- **Maximum temperature on mixing (T2):** 52.8°C- **Heat capacity of calorimeter (J/°C):** - **Average heat capacity of calorimeter (J/°C):** *Instruction:* Show a sample calculation of the heat capacity of the calorimeter for one run. Show both values and the average in the table above. In the calculations for Parts B and C, use C_cal = 0 if your average heat capacity is negative.### Part B: Mg plus HCl**Trials: 1**- **Mass of magnesium:** 0.1490g- **Initial temperature of HCl (T1):** 19.4°C- **Final (maximum) temperature (T2):** 35.1°C- **ΔT = T2 - T1:** - **ΔH (kJ/mol Mg):** - **Average value of ΔH (kJ/mol Mg):** ### Part C: MgO plus HCl**Trials: 1**- **Mass of weighing dish plus MgO:** 0.8182g- **Mass weighing dish after use:** 0.5691g- **Net mass of MgO taken for reaction:** 0.2491g- **Initial temperature of HCl (T1):** 18.5°C- **Final (maximum) temperature (T2):** - **ΔT = T2 - T1:** - **ΔH (kJ/mol MgO):** - **Average value of ΔH (kJ/mol MgO):***Instruction:* In your notebook, show a sample calculation of the heat of reaction for one run for Parts B and C. Show both values and the average in the tables above. Assume 50.0g, with c_solution = 4.18 J/g°C.

The enthalpy of reaction, ΔH0rxn is: ΔH0rxn= ΔH0f(products)- ΔH0f(reactants)Part D a. (1) For the reaction, Mg(s)+2HCl(aq) → MgCl2(aq)+H2(g) The enthalpy value for the above thermochemical equation can be calculated as follows: HCl(aq)=-167.2 kJ/molMg(s)=0H2(g)=0MgCl2(aq)=-796.9 kJ/mol Therefore, ΔH0rxn= ΔH0f(MgCl2(aq))+ΔH0f(H2(g))- [ΔH0f(Mg(s))+ΔH0f(HCl(aq))]ΔH1= -796.9+0- [0+2(-167.2)] kJ/mol=-462.5 kJ/mol (2) For the reaction, MgO(s)+2HCl(aq) → MgCl2(aq)+H2O(l) The enthalpy value for the above thermochemical equation can be calculated as follows: HCl(aq)=-167.2 kJ/molMgO(s)=-601.7H2O(l)=-285.820MgCl2(aq)=-796.9 kJ/mol Therefore, ΔH0rxn= ΔH0f(MgCl2(aq))+ΔH0f(H2O(l))- [ΔH0f(MgO(s))+ΔH0f(HCl(aq))]ΔH2= -796.9-285.82- [-601.7-2×167.2)] kJ/mol=-146.72 kJ/mol b. Hess's law states that if a reaction takes place in several steps, then its ΔHrxn is the sum of the enthalpies of reactions for each step. Therefore, adding the first reaction and reversing/adding the second reaction we get, Mg(s)+2HCl(aq) → MgCl2(aq)+H2(g) ∆H1=-462.5 kJ/molMgCl2(aq)+H2O(l) → MgO(s)+2HCl(aq) ∆H2=-(-146.72 kJ/mol)__________________________________Mg(s)+H2O(l) → MgO(s)+H2(g) ∆H3=-315.78 c. The enthalpy value for Mg(s)+H2O(l) → MgO(s)+H2(g) can be calculated from standard molar enthalpies of formation as follows: Mg(s)=0H2O(l)=-285.820MgO(s)=-601.7H2(g)=0 Therefore, ΔH0rxn= ΔH0f(MgO(s))+ΔH0f(H2(g))- [ΔH0f(Mg(s))+ΔH0f(H2O(l))]ΔH3= -601.7+0- [0+-285.82)] kJ/mol=-315.88 kJ/mol