4. Suppose there are 72.0 g of O2 available, along with the 8.08 grams of H2.
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?
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.
hello just for the heck of it could you show me your way to do the limiting regent problem? this if for DRBOB
How come O2 is the limiting reagent ? If o2 is 2.25 and h2 is 4.04
To determine the limiting reagent and find the theoretical yield, you will need to compare the amount of each reactant to the stoichiometry of the balanced equation.
The balanced equation for the reaction is:
2H2 + 1O2 → 2H2O
First, convert the given mass of O2 and H2 into moles by using their molar masses. The molar mass of O2 is 32.0 g/mol, and the molar mass of H2 is 2.016 g/mol.
1. Moles of O2:
molsO2 = massO2 / molar massO2
molsO2 = 72.0 g / 32.0 g/mol
molsO2 = 2.25 mol
2. Moles of H2:
molsH2 = massH2 / molar massH2
molsH2 = 8.08 g / 2.016 g/mol
molsH2 = 4.01 mol
To determine the limiting reagent, you need to compare the mole ratios of the reactants according to the balanced equation. The ratio is 2 moles of H2 to 1 mole of O2.
3. Compare the mole ratios:
H2:molsH2 = 4.01 mol:1 --> H2 is in excess
O2:molsO2 = 2.25 mol:1 --> O2 is the limiting reagent
Since O2 is the limiting reagent, the theoretical yield of this reaction is based on the amount of O2 used.
Now, to find the theoretical yield in moles and grams:
4. Theoretical yield in moles:
According to the balanced equation, 1 mole of O2 reacts to form 2 moles of H2O.
So, the theoretical yield of H2O is twice the amount of O2 used.
molesH2O = 2 × molesO2
molesH2O = 2 × 2.25 mol
molesH2O = 4.50 mol
5. Theoretical yield in grams:
To convert the moles of H2O to grams, use the molar mass of H2O, which is 18.015 g/mol.
massH2O = molesH2O × molar massH2O
massH2O = 4.50 mol × 18.015 g/mol
massH2O = 81.07 g
Therefore, the theoretical yield of H2O in grams is 81.07 g.