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?
No. Will you please explain how you arrived at those answers.
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.
To determine which reactant is limiting, you need to compare the amounts of reactants available to the stoichiometry of the balanced equation.
First, let's calculate the moles of each reactant available:
Moles of Al = mass of Al / molar mass of Al
Moles of Al = 3.17g / 26.98 g/mol = 0.118 mol
Moles of O2 = mass of O2 / molar mass of O2
Moles of O2 = 2.55g / 32.00 g/mol = 0.080 mol
Next, let's use the stoichiometry of the balanced equation to determine the theoretical number of moles of Al2O3 that can form:
From the balanced equation, the ratio of moles of Al to moles of Al2O3 is 4:2, or simplified, 2:1. This means that for every 2 moles of Al, 1 mole of Al2O3 will be produced.
Let's calculate the theoretical moles of Al2O3 that can form based on the moles of Al:
Theoretical moles of Al2O3 = (0.118 mol Al) * (1 mol Al2O3 / 2 mol Al)
Theoretical moles of Al2O3 = 0.059 mol Al2O3
Similarly, from the balanced equation, the ratio of moles of O2 to moles of Al2O3 is 3:2. This means that for every 3 moles of O2, 2 moles of Al2O3 will be produced.
Let's calculate the theoretical moles of Al2O3 that can form based on the moles of O2:
Theoretical moles of Al2O3 = (0.080 mol O2) * (2 mol Al2O3 / 3 mol O2)
Theoretical moles of Al2O3 = 0.053 mol Al2O3
Comparing the calculated theoretical moles of Al2O3 to the amounts of Al2O3 you provided (1.59 mol and 1.70 mol), it is clear that the smaller value, 0.053 mol, is the limiting reactant.
Thus, the correct answer is that O2 is the limiting reactant, not Al.
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