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# Hess's Law. The experiment conducted was meant to determine the enthalpy of formation of MgO(s) and CaO(s)

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Introduction

Hess's Law Data Collection and Processing Table V: Heats of Formation Heat of Formation (kJ/mol) Substance Trial 1 Trial 2 Magnesium oxide - 101.5 � 0.5 - 81.1 � 0.5 Magnesium - 366.1 � 0.4 - 322.2 � 0.4 Calcium oxide N/A - 141.7 � 1.0 Calcium - 192.6 � 0.5 - 173.6 � 0.5 Notes: * The heats of formation were calculated using the formula ?H = Q/nlimiting reactant Ex: Magnesium oxide Trial 1: n(MgO) = (1.08 g � 0.01 g)/40.304 g/mol = 0.026796 mol n(HCl) = (100.0 g � 0.5 g)/36.461 g/mol = 2.7427 mol MgO 1 mol 0.026796 mol ����= ����� = ������������ H2O 1 mol 0.026796 mol HCl 2 mol 2.7427 mol ��� = ����� = ���������� H2O 1 mol 1.3714 mol Therefore MgO is the limiting reactant. Assuming HCl(aq) has the same density as water, the 100.0 mL of HCl(aq) has a mass of 100.0 g. Qsurr = mc?T = (100.0 g)(4.184 J/(g��C)(25.5�C - 19.0�C) = (100.0 g)(4.184 J/g��C)(6.5�C) ...read more.

Middle

+ 2HCl(aq) --> H2(aq) + MgCl2(aq) - 322.2 � 0.4 H2(g) + 1/2O2(g) --> H20(l) - 285.8 Table XI: Enthalpy of Formation of MgO Original Equation Enthalpy of Formation (kJ/mol) Mg(s) + 1/2O2(g) --> MgO(s) - 526.2 � 0.9 Table XII: Enthalpies of Reaction After Manipulating The Given Equations for CaO Manipulated Equation Enthalpy of Reaction (kJ/mol) CaCl2(aq) + H2O(l) --> CaO(s) + 2HCl(aq) 141.7 � 1.0 Ca(s) + 2HCl(aq) --> H2(aq) + CaCl2(aq) - 173.6 � 0.5 H2(g) + 1/2O2(g) --> H20(l) - 285.8 Table XIII: Enthalpy of Formation of CaO Original Equation Enthalpy of Formation (kJ/mol) Ca(s) + 1/2O2(g) --> CaO(s) - 317.7 � 1.5 Notes: * The enthalpy of formation was calculated by adding together the enthalpies of reaction Ex: Combustion of magnesium Trial 1: ?Hf = (101.5 - 366.1 - 285.8)kJ/mol = - 550.4 kJ/mol � 0.9 kJ/mol Error: 0.5 + 0.4 = 0.9 Table XIV: Enthalpy of Formation Averages Original Equation Average Enthalpy of Formation (kJ/mol) Mg(s) ...read more.

Conclusion

was determined experimentally. There were many sources of error that could have accounted for this. When the enthalpy of reaction for magnesium was calculated, the magnesium was first sanded with sandpaper to remove and magnesium oxide that had formed on it. Similarly, calcium oxide forms on calcium, however no attempt was made to remove the calcium oxide. This would mean that more moles of calcium were accounted for than were actually present, and that calcium oxide was also reacting but this was not recorded and the amount of calcium oxide was not measured. As is visible in Table VIII, the enthalpy of reaction of calcium is greater than that of calcium oxide. Since it was not all calcium and some calcium oxide reacting in the trial that was meant to be only calcium, the calcium oxide could be responsible for the enthalpy of reaction being lower than the accepted value. This difference between the magnesium and calcium experiments could be one source of error accounting for the large difference in percent error between the two. The experiment could be improved by sanding the calcium as well. ...read more.

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