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?
4 answers
No. Will you please explain how you arrived at those answers.
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
My "no" meant that all of the work leading up to Al being the limiting regent is not right.
Yes, I could not understand what Kylee was doing so did it myself.
1 answer