Using the Hess Law and the thermochemical equations below,
MgO(s)
+ + 2HCl(aq) ------>
MgCl2(g) +
H2O(l)
ΔHrxn = -111.7
kJ/mol
Mg(s) + 2 HCl(aq) ------> MgCl2(g) + H2(g) ΔHrxn = -548.3 kJ/mol
H2(g)
+ 1/2 O2(g)
------>
H2O(l)
ΔHrxn = -142.9 kJ/mol
find the heat of reaction of the following: (0.25 pt.)
Mg(s) + 1/2
O2(g)
------>
MgO(g) ΔHrxn
= ?
b) If the theoretical enthalpy of this reaction is -602 kJ/mol,
calculate the percent error (0.05 pt.)
the data given is
MgO(s)
+ + 2HCl(aq) ------>
MgCl2(g) +
H2O(l)
ΔHrxn = -111.7
kJ/mol (1)
reversing the reaction-1, MgCl2(g)+ H2O(l)------>MgO(s)+ 2HCl(aq), ΔHrxn= 111.7 KJ/mole (1A)
Mg(s) + 2 HCl(aq) ------> MgCl2(g) + H2(g) ΔHrxn = -548.3 kJ/mol (2)
Eq.1A+ Eq.2 gives H2O(l)+Mg(s)---->MgO(s)+ H2(g), ΔHrxn =111.7-548.3 =-436.6 KJ/mole (2A)
H2(g) + 1/2 O2(g) ------> H2O(l) ΔHrxn = -142.9 kJ/mol (3)
Addition of Eq.2A and 3 gives Mg(s)+0.5O2(g)------>MgO(s), ΔHrxn =-142.9-436.6 =-579.5 Kj/mole
given Mg(s)+0.5O2(g)------>MgO(s), ΔHrxn = -602. 2 Kj/mole
% error= 100*{-602.5-(-579.5)/-602.5}=3.82%
Using the Hess Law and the thermochemical equations below, MgO(s) + + 2HCl(aq) ------> MgCl2(g) ...
a) Using the Hess Law and the thermochemical equations below, MgO(s) + + 2HCl(aq) ------> MgCl2(g) + H2O(l) ΔHrxn = -111.7 kJ/mol Mg(s) + 2 HCl(aq) ------> MgCl2(g) + H2(g) ΔHrxn = -548.3 kJ/mol H2(g) + 1/2 O2(g) ------> H2O(l) ΔHrxn = -142.9 kJ/mol find the heat of reaction of the following: Mg(s) + 1/2 O2(g) ------> MgO(g) ΔHrxn = ? b) If the theoretical enthalpy of this reaction is -602 kJ/mol, calculate the percent error
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