Exp.11 Enthalpy of Formation of MgO PRELAB QUESTIONS 1. Read the write up throug
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Exp.11 Enthalpy of Formation of MgO PRELAB QUESTIONS 1. Read the write up through the Data Analysis section. A student carried out the experiment described there. The student used 0.4894 g of Mg in the first reaction and found that the temperature of the solution went from 22.1 to 42.7 C. In the second reaction, 0.803 g of MgO was used and the solution went from 22.7 to 28.7C. Calculate the following. Show all calculations. AH, The enthalpy of the reaction of magnesium with acid Mass of final solution moles of MgExplanation / Answer
Enthalpy of formation of MgO
A) The enthalpy of the reaction of magnesium with acid
Mg + 2HCl ---> MgCl2 + H2
dT = 42.7 - 22.1 = 20.6 oC
mass of final solution = 100 + 0.4894 g = 100.4894 g [HCl = 100 ml or 100 g as density = 1 g/ml]
q = mCpdT = 100.4894 x 4.184 x 20.6 = 8.661 kJ
moles Mg = 0.4894 g/24 g/mol = 0.0204 mol
dHf = -8.661 kJ/0.0204 mol = -424.73 kJ/mol
B) The enthalpy of the reaction of magnesium oxide with acid
MgO + 2HCl --> MgCl2 + H2O
dT = 28.7 - 22.7 = 6.0 oC
mass of final solution = 100 + 0.803 g = 100.803 g [HCl = 100 ml or 100 g as density = 1 g/ml]
q = mCpdT = 100.803 x 4.184 x 6.0 = 2.530 kJ
moles Mg = 0.803 g/40 g/mol = 0.0201 mol
dHf = -2.53 kJ/0.0201 mol = -125.87 kJ/mol
C) Enthalpy of formation of MgO
Mg + 2HCl --> MgCl2 + H2 dHf = -424.73 kJ/mol
MgCl2 + H2O ---> MgO + 2HCl dHf = +125.87 kJ/mol
H2 + 1/2O2 ------> H2O dHf = -285.8 kJ/mol
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Mg + H2O ----> MgO + H2 dHf = -584.7 kJ/mol
So enthalpy of formagion of MgO = -298.86 kJ/mol
% error = (593.3 - 584.7) x 100/593.3 = 14.5%
2. In the reaction between Mg and HCl, the limiting reactant is Mg.
3. In the reaction between MgO and HCl, the limiting reactant is MgO.
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