If 6.0 g of Oxygen (O2) reacts with 12.0 g of hydrogen (H2) to give 5.0 g of dihydrogen monoxide. What's the % yield?

Answer 1

O2 + 2H2 ==> 2H2O
This is a limiting reagent (LR) problem. You know that when amounts are given for BOTH reactants.

mols O2 = grams/molar mass = approx 0.18 but you need a more accurate answer for this and all of the calculations that follow.
mols H2 = 12.0/2 = 6

Using the coefficients in the balanced equation, convert mols O2 to mols H2O if we had all of the H2 we needed. That's 0.18 x (2 mol H2O/1 mol O2) = 0.36 mols H2O
Do the same to convert mols H2 to mols H2O. That's 6 mols H2 x (2 mols H2O/2 mols H2) = 6 x 1/1 = 6 mols H2O.
You see the values don't agree which is typical of LR problems and the correct answer is ALWAYS the smaller value. The reagent responsible for the smaller value is the LR.
Now using the smaller value convert that to grams H2O produced. 0.36 mols H2O x mols mass H2O = approx 6.5g. This is the theoretical yield (TY). The actual yield (AY) is 5.0 g. from the problem.
% yield = (AY/TY)*100 =
(5.0/6.5)*100 = ?
REMEMBER, you need to go through and get more accurate answers for each step.