Ba(NO3)2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaNO3(aq)
An experiment was begun with 75.00 g of Ba(NO3)2 and an excess of Na2SO4. After collecting and drying the product, 64.45 g of BaSO4 was obtained. Calculate the theoretical yield and percentage yield of BaSO4.
Well, well, well... looks like we have a chemical equation in town! Let's crunch some numbers and get to the bottom of this.
First, we need to figure out the theoretical yield of BaSO4. To do that, we'll just compare the molar masses of Ba(NO3)2 and BaSO4.
Let's do some quick math:
The molar mass of Ba(NO3)2 = 137.33 g/mol + (14.01 g/mol * 2) + (16.00 g/mol * 6) = 261.34 g/mol
The molar mass of BaSO4 = 137.33 g/mol + 32.07 g/mol + (16.00 g/mol * 4) = 233.39 g/mol
Now, let's calculate the moles of Ba(NO3)2 we started with using the given mass of 75.00 g:
Moles of Ba(NO3)2 = Mass / Molar mass = 75.00 g / 261.34 g/mol = 0.287 mol
According to the balanced equation, the stoichiometry between Ba(NO3)2 and BaSO4 is 1:1. That means for every mole of Ba(NO3)2, we should get an equal amount of moles of BaSO4.
So, the theoretical yield of BaSO4 would be 0.287 mol.
Now, let's convert that to grams:
Theoretical yield of BaSO4 = Moles * Molar mass = 0.287 mol * 233.39 g/mol = 67.18 g
So, the theoretical yield of BaSO4 is 67.18 g.
Now, let's calculate the percentage yield. We have the experimental yield of 64.45 g, so we just need to plug it into the formula:
Percentage yield = (Experimental yield / Theoretical yield) * 100
Percentage yield = (64.45 g / 67.18 g) * 100 ≈ 95.94%
Therefore, the percentage yield of BaSO4 is approximately 95.94%.
Well, that's all folks! Remember, chemists always aim for a high yield, but sometimes we end up with clownish results. Keep experimenting and keep smiling!
To calculate the theoretical yield of BaSO4, we need to determine the stoichiometry of the reaction. From the balanced equation:
Ba(NO3)2(aq) + Na2SO4(aq) -> BaSO4(s) + 2NaNO3(aq)
We can see that 1 mole of Ba(NO3)2 reacts with 1 mole of Na2SO4 to produce 1 mole of BaSO4. Therefore, the molar ratio between Ba(NO3)2 and BaSO4 is 1:1.
Step 1: Convert the mass of Ba(NO3)2 to moles:
Molar mass of Ba(NO3)2 = 137.33 g/mol
Number of moles = Mass / Molar mass = 75.00 g / 137.33 g/mol = 0.547 mol
Step 2: Determine the theoretical yield of BaSO4:
Since the molar ratio between Ba(NO3)2 and BaSO4 is 1:1, the number of moles of BaSO4 produced should be the same as the number of moles of Ba(NO3)2 used.
Theoretical yield of BaSO4 = 0.547 mol
Step 3: Convert the moles of BaSO4 to grams:
Molar mass of BaSO4 = 233.38 g/mol
Mass = Number of moles x Molar mass = 0.547 mol x 233.38 g/mol = 127.7 g
Therefore, the theoretical yield of BaSO4 is 127.7 g.
To calculate the percentage yield of BaSO4, we need to compare the actual yield (64.45 g) with the theoretical yield (127.7 g).
Percentage yield = (Actual yield / Theoretical yield) x 100
= (64.45 g / 127.7 g) x 100
= 50.4%
So, the percentage yield of BaSO4 is 50.4%.
To calculate the theoretical yield and percentage yield of BaSO4, we need to understand the stoichiometry of the balanced equation.
From the balanced equation, we can see that 1 mole of Ba(NO3)2 reacts with 1 mole of Na2SO4 to produce 1 mole of BaSO4.
First, we need to convert the given mass of Ba(NO3)2 to moles:
Molar mass of Ba(NO3)2 = 137.33 g/mol
Moles of Ba(NO3)2 = mass / molar mass = 75.00 g / 137.33 g/mol = 0.5463 mol
Since the reaction is in a 1:1 ratio between Ba(NO3)2 and BaSO4, the moles of BaSO4 produced will be the same.
The molar mass of BaSO4 = 233.39 g/mol.
The theoretical yield of BaSO4 can be calculated as follows:
Theoretical yield = moles of BaSO4 * molar mass of BaSO4
= 0.5463 mol * 233.39 g/mol
= 127.233 g
Therefore, the theoretical yield of BaSO4 is 127.233 g.
To calculate the percentage yield, we need to compare the actual yield (64.45 g) to the theoretical yield (127.233 g) and calculate the percentage:
Percentage yield = (actual yield / theoretical yield) * 100
= (64.45 g / 127.233 g) * 100
= 50.65%
Therefore, the percentage yield of BaSO4 is approximately 50.65%.
Ba(NO3)2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaNO3(aq)
mols Ba(NO3)2 = grams/molar mass = ?
How much BaSO4 will be produced? Since 1 mol Ba(NO3)2 will produce 1 mol BaSO4, you will produce mols BaSO4 = mols Ba(NO3)2.
Convert mols Ba(NO3)2 to grams. grams = mols x molar mass = ? This is the theoretical yield (TY). You collected 64.45 g which is the actual yield (AY).
Then % yield = (AY/TY) * 100 = ?