Molar Enthalpy of Reaction for Mg(s)

Question

Using 50.0 mL of 1.0 mol/L hydrochloric acid (HCl) and the specific heat of water is 4.19J/g°C

Calculate the molar enthalpy of reaction for Mg(s) for each of the 3 trials.

Trial 1: Mass of Mg(s)= 0.15g┃Change in Temp. 13.6°C

Trial 2: Mass of Mg(s)= 0.25g┃Change in Temp. 22.7°C

Trial 3: Mass of Mg(s)= 0.35g┃Change in Temp. 31.8°C

Answer in kJ. Thank you!

Answer 1

Reaction: Mg(s) + 2HCl(aq) ==> MgCl2(aq) + H2(g) .. balanced equation

moles HCl present = 50.0 ml x 1 L / 1000 ml x 1.0 mol / L = 0.05 mols HCl

moles Mg present =

Trial 1: 0.15 g Mg x 1 mol / 24.3 g = 0.00617 mols

Trial 2: 0.25 g Mg x 1 mol / 24.3 g = 0.0103 mols

Trial 3: 0.35 g Mg x 1 mol / 24.3 g = 0.0144 mols

Using q = mC∆T to find heat (q), and assuming that the density of the solution is the same as that for water, i.e. 1 g / ml and assuming a specific heat that is the same as that of water, we can proceed.

Trial 1:

q = mC∆T = (50.0 g)(4.19 J/gº)(13.6º) = 2849 J

q = 2849 J / 0.00617 mols Mg x 1 kJ / 1000 J = 462 kJ / mol

Trial 2:

q = mC∆T = (50.0 g)(4.19 J/gº)(22.7º) = 4756 J

q = 4756 J / 0.0103 mols Mg x 1 kJ / 1000 J = 462 kJ / mol

Trial 3:

q = mC∆T = (50.0g)(4.19 J/gº)(31.8º) = 6662 J

q = 6662 J / 0.0144 mols Mg x 1 kJ / 1000 J = 463 kJ / mol