2CH3OH + 3O2 => 2CO2 + 4H2O dH = -174 kcal/mol and that x 2 = -348 kcal/rxn and I'm taking you at your word that this is -174 kcal/mol and not -174 kcal/reaction for the 2 mols.
-348 kcal x 33.0/64 = ?
methanol CH3OH is used as race car fuel write the balanced equation for the combustion of methanol. change in heat= -174kcal/mol methanol for the process How man kilocal are released u burning 33.0g of methanol
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