If 4.02g of hydrogen and 25.0g of oxygen (at standard conditions) are reacted to form water vapor, what quantity of heat is evolved from this reaction given that standard enthalpy of formation of H2O is -242kJ/mol?

Answer 1

A combination limiting reagent problem with a heat reaction problem.

2H2 + O2 ==> 2H2O

First let's determine the limiting reagent and the amount of H2O vapor formed.
mols H2 = g/molar mass = 4.02/2.01 = 2 mols.
mols O2 = 25/16 = 1.56

Now convert mols of each to mols H2O produced. For H2, it is
2.0 mols H2 x (2 mols H2O/2 mol H2) = 2 mols H2O vapor.
For O2 it is
1.56 mols O2 x (2 mols H2O/1 mol O2) = 3.12 mols H2O produced.
These answers are not the same; obviously one of them must be wrong. In limiting reagent problems the correct answer is ALWAYS the smaller value and the reagent producing that value is the limiting reagent. So H2 is the limiting reagent and we will produce 2 mols H2.

Since the heat formation for H2O is 242 kJ/mol and we have two mols, the heat reaction will be 242 x 2 kJ.

Answer 2

Thank you Dr. Bob. Could you explain why you used double the molecular weight for hydrogen and not for oxygen? I thought that since both are diatomic that I would double it for both molecules.

Answer 3

Kay, I screwed up. Too much of a hurry I guess. You are absolutely correct. mols O2 = 25/32 = 0.781
That may change the limiting reagent so you need to go through the problem with that correction. Thanks for bringing this to my attention.