4. Suppose there are 72.0 g of O2 available, along with the 8.08 grams of H2.

Question

Use ratios of stoichiometric coefficients to find the limiting reagent for the reaction. Then, find the theoretical yield in both moles and grams.

2H2 + 1O2 à 2H2O

I got 2.25 grams for o2
Then 4.48 grams for h2

I divide 8.08 by 2?

Answer 1

I prefer to do limiting reagent problems (LR) another way but I'll stick to this way for this problem.
You have 8.08/2 = 4.04 mols H2 and that will use 4.04 x 1/2 = 2.02 mols O2. Do you have that many mols O2? Yes, you have 72/32 = 2.25 so H2 is the LR and O2 is the excessive reagent.
How much H2O will be produced by 4.04 mol H2. You will produce 4.04 mols H2 x (2 mols H2O/2 mols H2O = 4.04 x 2.2 = 4.04 mols H2O and this the theoretical yield (in mols).
Theoretical yield in grams is g = mols x molar mass = 4.04 x 18 = about 73 grams.

You may want to go through the math and calculate all of the numbers more accurately.

Note: You 2.25 mols O2 is correct and it will take 4.5 mols H2 to react with that; however, you don't have 4.5 mols H2 (you have 8.08/2 = 4.04 mols) so H2 is the limiting regent and O2 is the excess reagent.

Answer 2

hello just for the heck of it could you show me your way to do the limiting regent problem? this if for DRBOB

Answer 3

How come O2 is the limiting reagent ? If o2 is 2.25 and h2 is 4.04