Heat energy released (thermochemistry)

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) + 3/2 O2(g) —> CO2(g) + 2H2O(l)

^H=-764 kJ

Based on the balanced thermochemical reaction, how much heat is produced when 40.4g of methanol reacts with 62.6g of oxygen?

Which reactant is limiting?

Answer 1

CH₃OH(g) + 3/2 O₂(g) ==> CO₂(g) + 2H₂O(l) ... balanced equation, but I prefer to write it as...

2CH₃OH + 3O₂ ==> 2CO₂ + 4H₂O ∆H = -764 kJ

Since we are given the mass of both reactants, we must find which, if either, is limiting.

40.4 g CH₃OH x 1 mol/32.04 g = 1.26 moles CH₃OH x 2 mol CO₂/2 mol CH₃OH = 1.26 moles CO₂

62.6 g O₂ x 1 mol/32 g = 1.96 moles O₂ x 2 mol CO₂/3 mol O₂ = 1.31 moles CO₂

LIMITING REACTANT = CH₃OH

In case the question asks for ∆H, it would be as follows:

1.26 mol CH₃OH x -764 kJ/2 mol = 481 kJ