Asked by Lena
I did a lab where I put Copper (II) Chloride in water with Alumnim.
2Al + 3CuCl2 -> 3Cu + 2AlCl3
The aluminum on the reactant side weighs 0.5 g and the CuClw weighs 1.69 grams.
It said to find the theoricital yield. Did I do this right?
2Al + 3CuCl2 -> 3Cu + 2AlCl3
# of mols Al = 0.50 g / 26.98 g/mol = 0.018532246 mol.
# of mols CuCl2 = 1.69 g / [63.55+(35.45 x 2)] g/mol = 0.012569728 mol
(0.012569728 mol CuCl2) * n mol AlCl3 / 0.012569728 mol CuCl2 = (0.012569728 mol CuCl2) * 2 mol AlCl3 / 3 mol AlCl3
n mol AlCl3 = 0.008379818
(0.0185322468 mol Al) * n mol AlCl3 / 0.018532246 mol Al = (0.018532246 mol Al) * 2 mol AlCl3 / 2 mol Al
n mol AlCl3 = 0.018532246
Therefore, the limiting reagent is the Copper (II) Chloride
Theoretical Yield of Aluminium Chloride:
Mass of AlCl3 = (0.008379818 mol) * (133.33 grams/mol)
= 1.1173 grams
2Al + 3CuCl2 -> 3Cu + 2AlCl3
The aluminum on the reactant side weighs 0.5 g and the CuClw weighs 1.69 grams.
It said to find the theoricital yield. Did I do this right?
2Al + 3CuCl2 -> 3Cu + 2AlCl3
# of mols Al = 0.50 g / 26.98 g/mol = 0.018532246 mol.
# of mols CuCl2 = 1.69 g / [63.55+(35.45 x 2)] g/mol = 0.012569728 mol
(0.012569728 mol CuCl2) * n mol AlCl3 / 0.012569728 mol CuCl2 = (0.012569728 mol CuCl2) * 2 mol AlCl3 / 3 mol AlCl3
n mol AlCl3 = 0.008379818
(0.0185322468 mol Al) * n mol AlCl3 / 0.018532246 mol Al = (0.018532246 mol Al) * 2 mol AlCl3 / 2 mol Al
n mol AlCl3 = 0.018532246
Therefore, the limiting reagent is the Copper (II) Chloride
Theoretical Yield of Aluminium Chloride:
Mass of AlCl3 = (0.008379818 mol) * (133.33 grams/mol)
= 1.1173 grams
Answers
Answered by
DrBob222
Some comments.
First, you don't say theoretical yield of what? You seem to have calculated theoretical yield of AlCl3 and I don't see anything wrong with your calculations if that was the idea. And you are correct that CuCl2 is the limiting reagent.
Second, you have used far too many places (significant figures). It is useless (and a waste of time) to copy all of those numbers that appear in the window of your calculator. Technically, you have only one s.f. since the 0.5 g Al was used (although that may have been 0.500 and you just didn't type in the extra zeros.). Otherwise, you have 3 from the 1.69. I usually carry one more place than is "allowed" then I round at the end. So I would have rounded the 1.1173 g AlCl3 to 1.12 g.
First, you don't say theoretical yield of what? You seem to have calculated theoretical yield of AlCl3 and I don't see anything wrong with your calculations if that was the idea. And you are correct that CuCl2 is the limiting reagent.
Second, you have used far too many places (significant figures). It is useless (and a waste of time) to copy all of those numbers that appear in the window of your calculator. Technically, you have only one s.f. since the 0.5 g Al was used (although that may have been 0.500 and you just didn't type in the extra zeros.). Otherwise, you have 3 from the 1.69. I usually carry one more place than is "allowed" then I round at the end. So I would have rounded the 1.1173 g AlCl3 to 1.12 g.
Answered by
Lena
If it was broken into 2 different questions. number one being find the limiting reagant and number 2 being find the theoritical yield of AlCl2 would it still be 1 significant digit?
Answered by
DrBob222
The part dealing with 0.5 g would be 1 place (unless as I said before you actually had 0.500 g and just neglected to type in the zeros). The part dealing with 1.69 g would be 3 places. Since the 1.69 part is the limiting reagent, then that is the only calculation that matters in the final number; therefore, you are still allowed three places. For the final percent yield that you calculated above, since the 0.27 has two, then you are allowed two in the final answer which is what you rounded it to; i.e., 24%.
Answered by
carson
HALP
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