When 9.0g of Al were treated with an excess of chlorine, 20.0g of Al2Cl6 were collected.
2AL + 3Cl6 -------> Al2Cl6
What was the percentage yield?
This is what i did so far
# moles in Al = 0.333
# moles in Cl2 = 0.282
The limiting reactant = Cl2
27g of Al forms 71g of Cl2
9g of Al should form 9/27 * 71 = 23.7g
The ratio between Cl2 & Al2Cl6 is 3 : 1
I'm stuck There
need Help here tnx
When 9.0g of Al were treated with an excess of chlorine, 20.0g of Al2Cl6 were collected.
2AL + 3Cl6 -------> Al2Cl6
What was the percentage yield?
This is what i did so far
# moles in Al = 0.333
# moles in Cl2 = 0.282
The limiting reactant = Cl2 <b< absolutely not. There was an excess of Cl2, according to the problem. There was no limiting reactant.
27g of Al forms 71g of Cl2 Huh? Aluminum does not form Cl2
9g of Al should form 9/27 * 71 = 23.7g
The ratio between Cl2 & Al2Cl6 is 3 : 1
I'm stuck There
You are going the wrong way, not stuck. For every mole of aluminum, you should get 1/2 mole of Al2Cl6. You used .333mole Al (I did not check that calc), so you should get .333/2 moles of Al2Cl6.
Precentage yield= grams got of product/grams should have got * 100>/b>
You started just great but faltered after the first step. The # mols Al is correct at 0.333.
BUT, the problem says you have an EXCESS of Cl2 so you have all you need and there is no limiting reagent (well, of course the Al is the limiting reagent but this is a regular stoichiometry problem).
Step 2. Convert mols Al to mols Al2Cl6 (is that the problem or is that what you think is formed?).
Step 3. Convert mols Al2Cl6 to grams Al2Cl6. This is the theoretical yield and what you would get at 100%.
Step 4.
% yield = [20.0/theoretical]x 100 = ??
I think you can work it from here but post your work if you get stuck.
Step 2. Convert mols Al to mols Al2Cl6 (is that the problem or is that what you think is formed?).
Lost there
OO
i got it
the yield is 45%.. the ans. is in the book
It says in the question that Cl is in excess so the limiting reactant is Al so
9.0*1/27g/mol = 0.333mol of Al
Use the mole ratio to find the mol of Al2Cl6 2:1
0.333*1/2 =0.1665mol of Al2Cl6
0.1665mol*267/1 =44.5g of Al2Cl6
Then 20.0g/44.5 *100% = 45%
Great job! I'm glad you figured it out. Sometimes chemistry problems can be a bit confusing, but you managed to solve it. Keep up the good work! And hey, 45% isn't too bad. So close to 50%, you're almost there!
To calculate the percentage yield, you need to compare the actual yield (20.0g of Al2Cl6 collected) to the theoretical yield.
First, you need to calculate the theoretical yield of Al2Cl6 using stoichiometry. From the balanced equation, you know that the molar ratio between Al and Al2Cl6 is 2:1. So, you can convert the moles of Al to moles of Al2Cl6.
Moles of Al = mass of Al / molar mass of Al
Moles of Al = 9.0g / 27g/mol = 0.333 mol Al
Since the molar ratio between Al and Al2Cl6 is 2:1, the moles of Al2Cl6 would be half of the moles of Al:
Moles of Al2Cl6 = 0.333 mol Al / 2 = 0.1665 mol Al2Cl6
Now, you can calculate the theoretical yield of Al2Cl6 using the moles of Al2Cl6 and the molar mass of Al2Cl6:
Theoretical yield = moles of Al2Cl6 * molar mass of Al2Cl6
Theoretical yield = 0.1665 mol * (27 + (2*35.5) g/mol) = 15.0055 g Al2Cl6
Now you can calculate the percentage yield:
Percentage yield = (actual yield / theoretical yield) * 100
Percentage yield = (20.0g / 15.0055g) * 100 = 133.29%
Therefore, the percentage yield is approximately 133.29%.
Glad to hear that you figured it out! The answer is indeed 45% as you calculated. It seems that there was a mistake in the calculations in your initial approach.
To clarify the steps:
Step 1: Calculate the moles of Al used:
Mass of Al = 9.0 g
Molar mass of Al = 26.98 g/mol
Moles of Al = Mass of Al / Molar mass of Al
= 9.0 g / 26.98 g/mol
= 0.333 mol (approximately)
Step 2: Calculate the moles of Al2Cl6 formed:
According to the balanced equation, for every 2 moles of Al, 1 mole of Al2Cl6 is produced:
Moles of Al2Cl6 = Moles of Al / 2
= 0.333 mol / 2
= 0.1665 mol (approximately)
Step 3: Calculate the theoretical yield of Al2Cl6:
Molar mass of Al2Cl6 = 26.98 g/mol + (2 x 35.45 g/mol) = 133.33 g/mol
Theoretical yield of Al2Cl6 = Moles of Al2Cl6 x Molar mass of Al2Cl6
= 0.1665 mol x 133.33 g/mol
= 22.21 g (approximately)
Step 4: Calculate the percentage yield:
Percentage yield = (Actual yield / Theoretical yield) x 100
Actual yield = 20.0 g
Percentage yield = (20.0 g / 22.21 g) x 100
= 0.901 x 100
= 90.1% (approximately)
Therefore, the percentage yield is approximately 90.1%.