H3PO4 + 3KOH = K3PO4 + 3 H2O What would be the percent yield if you reacted 49.0g of H3PO4 with 112g of potassium hydroxide, and collected 25.0g of water?
just give the answer
To calculate the percent yield, you need to compare the actual yield of the desired product (water) with the theoretical yield. The theoretical yield can be determined by stoichiometry and the limiting reactant.
1. Calculate the molar mass of H3PO4:
- H3PO4: (3 × 1.0079 g/mol) + (1 × 15.9994 g/mol) + (4 × 1.0079 g/mol) = 97.9947 g/mol
2. Calculate the number of moles of H3PO4:
- Moles of H3PO4 = mass / molar mass
- Moles of H3PO4 = 49.0 g / 97.9947 g/mol = 0.5001 mol
3. Calculate the molar mass of KOH:
- KOH: 39.0983 g/mol + 15.9994 g/mol + 1.0079 g/mol = 56.1056 g/mol
4. Calculate the number of moles of KOH:
- Moles of KOH = mass / molar mass
- Moles of KOH = 112 g / 56.1056 g/mol = 1.9964 mol
5. Use the stoichiometry of the balanced equation to determine the limiting reactant and the theoretical yield:
- The balanced equation shows a 1:1 molar ratio between H3PO4 and H2O.
- Since the stoichiometry of the equation is 1:1 and the number of moles of H3PO4 is less than that of KOH, H3PO4 is the limiting reactant.
- The theoretical yield of H2O is equal to the number of moles of H3PO4.
6. Calculate the theoretical yield of H2O:
- Theoretical yield = moles of H3PO4 = 0.5001 mol
7. Calculate the percent yield:
- Percent yield = (actual yield / theoretical yield) × 100
- Percent yield = (25.0 g / 0.5001 mol) × 100 = 4999.0%
Note: It is not possible to have a percent yield greater than 100%. This may indicate an error in measurements or calculations.
To calculate the percent yield, we need to compare the amount of water produced in the reaction to the theoretical yield, which is the amount of water we would expect to obtain based on stoichiometry.
First, we need to determine the limiting reactant, which is the reactant that is completely consumed in the reaction. To do this, we compare the amount of H3PO4 and KOH using their respective molar masses.
Molar mass of H3PO4 = 98.0 g/mol
Molar mass of KOH = 56.1 g/mol
Now, let's convert the mass of H3PO4 and KOH to moles:
Moles of H3PO4 = 49.0 g / 98.0 g/mol = 0.500 mol
Moles of KOH = 112 g / 56.1 g/mol = 2.0 mol
From the balanced equation, we know that the molar ratio between H3PO4 and H2O is 1:3. Therefore, based on the moles of H3PO4, we expect to produce 3 times that amount of water.
Theoretical moles of H2O = 3 * moles of H3PO4 = 3 * 0.500 mol = 1.50 mol
Now, let's convert the theoretical moles of water to grams:
Theoretical grams of H2O = 1.50 mol * 18.0 g/mol = 27.0 g
The theoretical yield of water in this reaction is 27.0 grams.
Finally, we can calculate the percent yield by dividing the actual yield (25.0g) by the theoretical yield (27.0g), and then multiplying by 100:
Percent yield = (Actual yield / Theoretical yield) * 100
Percent yield = (25.0 g / 27.0 g) * 100
Percent yield = 92.6%
Therefore, the percent yield in this reaction is approximately 92.6%.
This is a limiting reagent problem (LR) and first you must determine what the theoretical yeild is for the reaction.
..........H3PO4 + 3KOH ==> K3PO4 + 3H2O
mols H3PO4 = grams/molar mass = approx 49/98 = 2
mols H2O produced if this is the LR.= 2*3 = 6
mols KOH = 112/56 = approx 2
mols H2O poduced if this is the LR = about 2
The smaller is the LR which is KOH so we should get
g H2O = mols H2O x molar mass H2O = 2 x 18 = 36 g.This is the theoretical yield = TY
You need to confirm all of these calculations INCLUDING the molar
masses.
The actualy yield (AY) from the problem si 25.0 g
%yield = [(AY/TY)]*100
Post your work if you get stuck.