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?

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.

To calculate the percent yield, we need to compare the actual amount of the desired product obtained (in this case, dihydrogen monoxide) with the theoretical amount that would be obtained if the reaction went to completion.

First, let's determine the theoretical amount of dihydrogen monoxide that should be produced. We can use the balanced chemical equation for the reaction between oxygen and hydrogen:

2H2 + O2 -> 2H2O

From the equation, we can see that 1 mole of O2 reacts with 2 moles of H2 to produce 2 moles of H2O.

1 mole of O2 has a molar mass of 32.00 g/mol, so 6.0 g of O2 is equivalent to:
6.0 g / 32.00 g/mol = 0.1875 mol of O2

2 moles of H2 corresponds to:
2 moles H2 x 2.02 g/mol = 4.04 g of H2

Therefore, if the reaction goes to completion, we would expect the formation of 4.04 g of dihydrogen monoxide.

Now, let's calculate the percent yield using the actual amount of dihydrogen monoxide obtained. Given that 5.0 g of dihydrogen monoxide was produced, we can calculate the percent yield as follows:

Percent yield = (actual yield / theoretical yield) x 100

Percent yield = (5.0 g / 4.04 g) x 100 = 123.76%

Therefore, the percent yield is 123.76%.