Exp.11 Enthalpy of Formation of Mg0 PRELAB QUESTIONS Read the write up through t

Question

Exp.11 Enthalpy of Formation of Mg0 PRELAB QUESTIONS 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. In the second reaction, 0.303 g of MgO was used and the solution went from 22.7 to 28.70. Calculate the following. Show all calculations. 1. AKI. The enthalpy of the reaction of magnesium with acid. Mass of final solution moles of Mg

Explanation / Answer

In the first reaction (A), Mg reacts with HCl:
& you are using (0.4894 grams of Mg)/ (24.305 grams per mol Mg) = 0.02014 moles Mg

(A) Mg(s) & 2 HCl(aq) --> MgCl2 (aq) & H2 (g)
we use the specific heat of water as the specific heat of the solution, but
you did not give the grams of solution, or even the ml's of solution.

so I am going to estimate it as 109 grams of solution...
I back calculated that using the dH formation of Mg+2 (aq) being -466.9 kJ/mol, your dT, & spec. ht.
but you will have to re-do the calcs using your mass of solution used.

find Joules for (A)
dH reaction = - (m H2O) (C H2O) (dT H2O)
dH reaction = - (109 g) (4.184 J/g-C) (+ 20.6 C)
dH reaction = - (9395 Joules)
p.s.
which is = (-9395 Joules) / (0.02014 moles Mg) = -466472 Joules
aka dH reaction = - 466 kJ/mol Mg
================================

In the second reaction (B), same situation
you are using (0.803g of MgO) (40.30 grams per mole) = 0.01992 moles of MgO
MgO(s) reacts with HCL:
(B) MgO(s) & 2 HCl(aq) --> MgCl2 (aq) & H2O

again I am going to backtrack, to find the missing grams of solution.

for the net ionic: MgO(s) & 2 H+(aq) --> Mg+2 (aq) & H2O(l)
I find that the joules which should be released, I then use your moles MgO, your dT & water's spec. ht. & find that you likely used about 98 grams of solution.
but you will need to use your own grams of solution.

find Joules for (B)
dH reaction = - (m H2O) (C H2O) (dT H2O)
dH reaction = - (98 g) (4.184 J/g-C) (+6.0 C)
dH reaction = - (2460 Joules)
p.s.
which is = (-2460 Joules) / (0.01992 moles MgO) = -123,504 Joules
aka dH reaction = - 124 kJ/mol MgO


===========================
next you are usually asked to use Hess's law, & the dHf of water:
(C) H2(g) & 1/2 O2(g) --> H2O(l) dHf = -258.8 kJ/mol
to find the heat of combustion of Mg:
Mg(s) & 1/2 O2 --> MgO(s) dH reaction = ?

using these equations:
(A) Mg(s) & 2 HCl(aq) --> MgCl2 (aq) & H2 (g) dH (A) = - 466 kJ/mol Mg
(B) MgO(s) & 2 HCl(aq) --> MgCl2 (aq) & H2O (l) dH (B) = - 124 kJ/mol MgO

(C) H2(g) & 1/2 O2(g) --> H2O(l) dH (C) = -258.8 kJ/mol

in order to get the combustion of Mg
Mg(s) & 1/2O2 (g) --> MgO (s)
we combine the net products & reactants of:
+1 equation (A): Mg(s) & 2 HCl(aq) --> MgCl2 (aq) & H2 (g) dH (A) = - 466 kJ/mol Mg
reverse of (B): MgCl2 (aq) & H2O (l) --> MgO(s) & 2 HCl(aq) dH (-B) = + 124 kJ/mol MgO
+1 equation (C): H2(g) & 1/2 O2(g) --> H2O(l) .... .... .. dH (C) = -258.8 kJ/mol

H2O --> cancels with --> H2O
H2(g) --> cancels with --> H2(g)
2 HCl(aq) --> cancels with --> 2 HCl(aq)
MgCl(aq) --> cancels with --> MgCl(aq)
leaving you only with the net molecular equation: Mg(s) & 1/2 O2 --> MgO(s)

Hess's law says, that if adding 1equation (A), the opposite of (B) with 1 equation (C) gives you the reaction that you wish, then combining the dH's of 1(A) -1(B) +1(C) will give you your dH reaction

dH's of 1(A) & -1(B) & 1(C)
dH's of 1(- 466 kJ) & -1(- 124 kJ) & 1(-258.8kJ)
dH's = - 466 kJ & 124 kJ & -258.8kJ
dH = -601 kJ/ mol Mg burned

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