1. Mg(s) + ½O2(g) → MgO(s) DH= _____ -626 kJ
key
Names: ___________________________________
All encompassing activity…A.P. Chemistry Ch. 6
Mg(s) + ½O2(g) → MgO(s) DH= _____
-626 kJ
MgO(s) + 2HCl(aq) → MgCl2(aq) + H2O(l)
DH = _______
-100 kJ
flipped
Add these steps:
Mass of MgO = 0.50 g (0.013 mol)←L.R.
Volume of HCl = 50.0 mL (3.0 M)
DT = 6.3° C (typical temperature change with the amounts used)
qwater = (50.0 g)(4.184 J/g°C)(6.3°C) = 1318 J
qsys = J
DHrxn= J/ 0.013mol = -1.0 x 105 J/mol
MgCl2(aq) + H2O(l) → MgO(s) + 2HCl(aq) DH =
100 kJ
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) DH =
-440 kJ
H2(g) + ½O2(g) → H2O(l) DH=
-286 kJ
Mg(s) + ½O2(g) → MgO(s) DH= _____
-626 kJ
Of Note: the accepted value for the heat of formation of MgO is -602kJ…if you don’t believe me look it it up
Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)
DH = _______
H2(g) + ½O2(g) → H2O(l) DH= _____
-286 kJ
-440 kJ
Mass of Mg = 0.50 g (0.021 mol)←L.R.
Volume of HCl = 50.0 mL (3.0 M)
DT = 44.2° C (typical temperature change with the amount that used)
qwater= (50.0 g)(4.184 J/g°C)(44.2°C) = 9240 J
qsys = -9240
DHrxn= J/ 0.021mol = -440,000 J/mol