How much methanol, in grams, must be burned to produce 683 kJ of heat?

Question

When methanol, CH3OH, is burned in the presence of oxygen gas, O2, a large amount of heat energy is released. For this reason, it is often used as a fuel in high performance racing cars. The combustion of methanol has the balanced, thermochemical equation

CH3OH(g)+32O2(g)⟶CO2(g)+2H2O(l) Δ𝐻=−764 kJ

mass: ?g

Answer 1

Be careful when you copy or cut/paste because the 32O2 in your equation should be 3/2 (1.5.) In addition, you never asked a question. For example, did you mean to ask "how many grams of methanol are required to produce 879 kJ of energy? If so, we proceed as follows:

CH₃OH(g) + 3/2 O₂(g) ==> CO₂ (g) + H₂O(l) .. Δ𝐻=−764 kJ

879 kJ x 1 mol CH3OH / 764 kJ = 1.15 mols

1.15 mols CH3OH x 32.0 g / mol = 36.8 g