how much heat is produced when 22.9 g of methanol reacts with 35.5 g of oxygen?

Question

When methanol, CH3OH, is burned in the presence of oxygen gas, O2, a large amount of heat energy is released, as shown in the combustion reaction.

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

Based on the balanced thermochemical reaction, how much heat is produced when 22.9 g of methanol reacts with 35.5 gof oxygen?

heat: ? kJ

Answer 1

CH₃OH(g) + 3/2 O₂ ==> CO₂(g) + 2H₂O(g) .. ∆H= -764 kJ

Since we are given the amounts of BOTH reactants, we must first find which one is limiting. One way to do this is to simply divide moles of each by the corresponding coefficient in the balanced equation. Whichever value comes out less represents the limiting reactant.

For CH3OH: 22.9 g x 1 mol / 32.0 g = 0.7156 moles (÷1->0.716)

For O2: 35.5 g x 1 mol / 32 g = 1.11 moles (÷3/2->0.74)

Since 0.716 is less than 0.74, CH3OH is the limiting reactant and we use moles of CH3OH (0.716) to calculate energy generated.

0.716 mols CH3OH x 764 kJ / 1 mole = 547 kJ of energy released

NOTE: the value would normally have a negative sign, but since we state that energy is released, the negative sign isn't needed.