Asked by Kylee
Given the following:
4Al+ 3O2 ==> 2Al2O3
If 3.17g of Al and 2.55g of O2 are available, which reactant is limiting?
My answer is: 1.59 mol Al2O3 and 1.70 mol Al2O3. Al is limiting.
Is my answer and work correct?
4Al+ 3O2 ==> 2Al2O3
If 3.17g of Al and 2.55g of O2 are available, which reactant is limiting?
My answer is: 1.59 mol Al2O3 and 1.70 mol Al2O3. Al is limiting.
Is my answer and work correct?
Answers
Answered by
DrBob222
No. Will you please explain how you arrived at those answers.
Answered by
Damon
checking my way
Al = 27 g/mol
3.17 g = .1174 mols of Al
I need 3 mol O2 for 4 mol Al
mol O2/mol Al = 3/4 = x/.1174
x = .0881 mols O2 required
O2 = 32 grams/mol
.0081*32 = 2.82 grams of O2 needed
BUT I only have 2.55 grams O2
SO Al is limiting
Al = 27 g/mol
3.17 g = .1174 mols of Al
I need 3 mol O2 for 4 mol Al
mol O2/mol Al = 3/4 = x/.1174
x = .0881 mols O2 required
O2 = 32 grams/mol
.0081*32 = 2.82 grams of O2 needed
BUT I only have 2.55 grams O2
SO Al is limiting
Answered by
DrBob222
My "no" meant that all of the work leading up to Al being the limiting regent is not right.
Answered by
Damon
Yes, I could not understand what Kylee was doing so did it myself.
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