Calculating molar heat of combustion?

Density of water = 1.00 g/mL
Mass of water heated = (75.0 mL) × (1.00 g/mL) = 75.0 g

Mass of methanol burned = (113.54 - 112.97) g = 0.57 g
No. of moles of methanol burned = (0.57 g) / (32.05 g/mol) = 0.01778 mol

Heat released in burning methanol
= (Heat absorbed by Al) + (Heat absorbed by H₂O)
= m(Al) × c(Al) × ΔT + m(H₂O) × c(H₂O) × ΔT
= (44.5 g) × (0.900 J/g°C) × (23.9 - 16.5)°C + (75.0 g) × (4.18 J/g°C) × (23.9 - 16.5)°C
= 2616 J

Molar heat of the combustion of methanol = (2616 J) / (0.01778 mol) = 147000 J/mol = 147 kJ/mol

here's the idea
.. heat gained by water + heat gained by aluminum = heat released by combustion of methanol
.. the water and aluminum started at the same temp and ended at the same temp

and since there were no phase changes,
.. heat (Q) = m * Cp * dT

so..
.. (m * Cp * dT) H2O + (m * Cp * dT) Al = Q from combustion of MeOH (methanol)

and since dT was the same for both
.. [(m * Cp) H2O + (m * Cp) Al] * dT = Q from combustion of MeOH (methanol)

plugging in the data and reversing the equation
.. Q comb MeOH = [(75.0mL * 1g/1mL * 4.18J/g°C) + (44.5g * 0.900J/g°C)] * (23.9°C-16.5°C) = 2.62x10³ J

and that was for the alcohol burned and we need to convert that to kJ/mol

.. .... .. .... 2.62x10³J... ... .. 1kJ... ... 32.05g
.. Q = ------ ----- ----- ----- x ---- ---- x ---- ----- = 147 kJ/mol
.. ... .. 113.54g - 112.97g.. 1000J... .. 1 mol

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that's not a very good estimate... from here
https://en.wikipedia.org/wiki/Heat_of_combustion
Q combustion methanol = 638 kJ/mol... LHV