In: Chemistry
Find heat absorbed by water (kJ) and ΔrH (kJ mol-1) of the following reaction
Mg + 2 HCl (aq) MgCl2 + H2
Mass of Mg: 20.8mg
Moles of Mg: calculate using formula m/M
Ti= 19.9 C
Tf= 33.4 C
density of water: 1g/mol
5M of 1.0 M HCl
V of water is 5g
Given reaction is Mg + 2 HCl (aq) MgCl2 + H2
The change in enthalpy H = ?
q(solution) = q(reaction) = m(solution) C(solution)T
msolution = mass of the solution = 5 g
CSolution = Specific heat capacity of water = 4.2 J/g oC
T = Tfinal - TInitial = 33.4 - 19.9 = 13.5 oC
q(reaction) = m(solution) C(solution)T
= (5 g) x (4.2 J/g oC) x (13.5 oC)
= 283.5 J/g
= 0.283 KJ/g
No of moles of Magnesium = mass/Molar mass
Mass of Magnesium = 20.8 mg = 0.0208 g
Molar mass of Magnesium = 24.31 g/mol
moles of Mg = 0.0208 g / (24.31 g/mol) = 0.0008556 moles
The heat evolved for this reaction per mole of Magnesium metal = (Heat evolved in this reaction) / (Moles of Magnesium metal)
= 0.283 KJ / (0.0008556 moles) (Note it is the heat per 24.31 g magnesium)
= 330.762 KJ/mole
The heat energy produced in this reaction per mole of Magnesium metal = 330.762 KJ/mol