Final answer:
The heat produced by combustion per liter of methanol can be calculated by determining the enthalpy of combustion of methanol and converting it into heat per liter. The enthalpy of combustion of 1 mole of methanol is -726 kJ/mol. By taking into account the density of methanol and converting the enthalpy of combustion from kJ/mol to J/L, the heat produced by combustion per liter of methanol is -917 J/L.
Step-by-step explanation:
The heat produced by combustion per liter of methanol can be calculated by determining the enthalpy of combustion of methanol and converting it into heat per liter. First, we need to calculate the enthalpy of combustion of methanol per mole using the balanced equation:
2H₂(g) + CO(g) → CH3OH(g)
From this equation, we can see that one mole of methanol is formed during the combustion of H₂ and CO. Using the enthalpies of formation from Appendix G, we can calculate the enthalpy of combustion of 1 mole of methanol to be -726 kJ/mol. Next, we need to convert the enthalpy of combustion from kJ/mol to kJ/liter by taking into account the density of methanol. Methanol has a density of 0.791 g/mL, which is equivalent to 791 g/L. So the enthalpy of combustion per liter of methanol can be calculated by dividing the enthalpy of combustion per mole (-726 kJ/mol) by the number of grams per liter (791 g/L), and then multiplying by 1000 to convert from kJ to J:
Enthalpy of combustion per liter = (-726 kJ/mol) / (791 g/L) * 1000 J/kJ = -917 J/L
Therefore, the heat produced by combustion per liter of methanol is -917 J/L.