Teachers Notes:

Problem 1:
Heat (q) = m x c x ΔΤ. ΔΤ = 76.0 °C - 4.0 °C = 72.0 °C.
Q = 300.0 g x 4.184 J/g ·°C x 72.0 °C = 90,400 J of heat or 90.4 kJ of heat is released by the heater pad or gained by the water.

Problem 2:
Mg (s) + 2 H₂O (l) → Mg(OH)₂ (s) + H₂ (g)
ΔH_rxn = Σ ΔH_products -ΣΔH_reactants
ΔH_rxn = (1 mol x -924.54 kJ/mol) - (2 mol x -285.830 kJ/mol) = -352.88 kJ

Problem 3:
Mg (s) + 2 H₂O (l) → Mg(OH)₂ (s) + H₂ (g)
ΔS_rxn = Σ S_products -Σ S_reactants
ΔS_rxn = (63.18 J/K· mol +130.684 J/K· mol) - (32.68 J/K· mol + 2 x 69.91 J/K· mol) =
21.364 J. The reaction in more disordered because the ΔS_rxn > 0

Problem 4:
ΔG_rxn = ΔH_rxn - T ΔS_rxn
ΔG_rxn = -352.88 kJ - (298 K x 0.021364kJ) = -359.25 kJ
The reaction is spontaneous because the value for ΔG_rxn < 0