A student carried out the experiment described there. the student used 0.4894 g

Question

A student carried out the experiment described there. the student used 0.4894 g of Mg the first reaction and found that the temperature of the solution went from 22.1 to 42.7 C. In the second reaction, 0.803g of MgO was used and then solution went from drom 22.7 to 28.7C. Calculate the following. Show all calculations.

?T=_____________

Mass of final solution=______________

q=________________________

moles of Mg=_______________________

?H1=___________________

?H4 The enthalpy of the reaction of magnesium oxide with acid.

Mass of final solution=______________________

q=______________________

moles MgO=____________________

?H4=________________

?Hr_____________________________

%Error_____________________

2. In tghe reaction between Mg and HCl, which is the limiting reactant?

3. In the reaction between MgO and HCl, what is the reactant?

Explanation / 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

----------------------------------------------------------------------------

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%

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