Volume of s.00 M HCL 5.0 ML toom1L-a pt. cach) 2.409 (I pt.cach) L189 5.0 Mass o
ID: 1043400 • Letter: V
Question
Volume of s.00 M HCL 5.0 ML toom1L-a pt. cach) 2.409 (I pt.cach) L189 5.0 Mass of MgO bcalculated) (calculated) Mass of 3.00 MHCI Total Mass Mole product if HCI limiting reactant Mole product if (calculated) SME realculated) (calculated) ig 145i 7.338 3/ns 76 (calculated) calculated) (1 pt. each) (lpt, each) MgO limiting reactant Limiting reactant ?19,1"L Initial Temperature Final Temperature 39.0 AHm.rn 2 43,1°0(I pt. each) k (4 pts. cach) Write the balanced equation for the reaction of magnesium metal with hydrochloric acid. Write the balanced equation for the reaction of magnesium oxide with hydrochloric acid. ? Combine the above two equationk With that for the formation of liquid water (Eq. 6) and use Hess's Law to clearly show how you arrive at the equation for the formation of magnesium oxide (Eq. 5). In this same space use the average values for ??mr?n 1, ??.nnn 2 and AH no to determine AHMgo. Show work. (5 pts for correctly shown work.) Calculated AH Mgo (4 pts) Accepted AH: Mgo (1 t)Explanation / Answer
Hess's law
Reaction 1,
Mg + 2HCl --> MgCl2 + H2
Reaction 2,
MgO + 2HCl ---> MgCl2 + H2O
Reaction 3- water,
H2 + 1/2O2 ---> H2O
Add reaction 1 and 3,
Mg + 2HCl --> MgCl2 + H2
H2 + 1/2O2 ---> H2O
-----------------------------------------
Mg + 2HCl + 1/2O2 ---> MgCl2 + H2O ------(4)
Now subtract Reaction 3 from (4),
Mg + 2HCl + 1/2O2 ---> MgCl2 + H2O
MgCl2 + H2O ---> MgO + 2HCl
-----------------------------------------------------
Mg + 1/2O2 --> MgO
Is the reaction for formation of MgO
From above data,
limiting reactant = MgO
dHrxn,2 = (7 kJ x 40.3 g/mol/1.788 g + 8.3 kJ x 40.3 g/mol/2.466 g)/2 = 147.6 kJ/mol
So,
dHof for formation of MgO = 1.53 - 286 - 147.6 = -432.07 kJ/mol (calculated)
[pl. note, dHrxn,1 is incorrect above, you need to convert kJ to kJ/mol from limiting reactant moles, then redo the dHof for MgO calculation to get correct value]
dHof accepted = -601.8 kJ/mol
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