Mg(s) + ½O2(g) → MgO(s) H= _____

Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

H = _______

MgO(s) + 2HCl(aq) → MgCl2(aq) + H2O(l)

H = _______

H2(g) + ½O2(g) → H2O(l) H= _____

All encompassing activity…A.P. Chemistry Ch. 6Names: ___________________________________

MgCl2(aq) + H2O(l) → MgO(s) + 2HCl(aq) H =

Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) H =

H2(g) + ½O2(g) → H2O(l) H=

Mass of MgO = 0.50 g (0.013 mol)←L.R.

Volume of HCl = 50.0 mL (3.0 M)

T = 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 = -1318 J

Hrxn= -1318 J/ 0.013mol = -1.0 x 105 J/mol

-100 kJ

Mass of Mg = 0.50 g (0.021 mol)←L.R.

Volume of HCl = 50.0 mL (3.0 M)

T = 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

Hrxn= -9240 J/ 0.021mol = -440,000 J/mol

-286 kJ

-286 kJ

-440 kJ

100 kJ

flipped

Add these steps:

-626 kJ

Mg(s) + ½O2(g) → MgO(s) H= _____-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

-440 kJ

key