The heat produced by a combustion of 125g of methanol under standard state of the condition is 2836.29 kJ. The combustion reaction of methanol is 2CH₃OH+3O₂→2CO₂+4H₂O
The value for standard enthalpy of the combustion of a methanol is ΔH°=-726.1kJ/mol. It means that 1 mole of methanol on combustion releases 726.1 kJ of energy. First, we need to find molar mass of CH₃OH.
The molar mass of CH₃OH is
Molar mass=12+(4×1)+16
=12+4+16
=32g/mol
And the number of moles for CH₃OH is
moles=125g×(1 mol/32g)
moles=3.9062 mol
For one mole, we have 726.1 kJ of energy. Hence, energy for 3.9062 mol is
(726.1 kJ/mol)×3.9062 mol
=2836.29 kJ
Therefore, the heat produced by 125g of methanol is 2836.29 kJ.
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