My question: 4a1+302--2A1203
Which reactant is limiting if 0.32 mol A1and 0.26mol oxygen is available.
My answer is 0.228 is missing
Let's make this a little simpler. Why not call a1 just a. Then
4a + 3O2 ==> 2A2O3. On second thought, I'll bet that is Al (for aluminum so that's an ell and not a number 1) so
4Al + 3O2 ==> 2Al2O3.
What I do for limiting regent(LR) problems is to convert Al to mols Al2O3. Do the same with O2 to mols Al2O3. The one producing the smaller number of mols for Al2O3 is the LR.
0.32 mols Al x (2 mols Al2O3/4 mols Al) = 0.32 x 2/4 = ?
0.26 mols O2 x (2 mols Al2O3/3 mols O2) = 0.26 x 2/3 = ?
The reagent that produces the smaller number of mols Al2O3 is the LR.
I don't know what 0.228 missing means for your answer.
To determine which reactant is limiting, we need to compare the moles of the reactants to their stoichiometric coefficients in the balanced equation.
The balanced equation for the reaction is:
4Al + 3O2 -> 2Al2O3
According to the equation, it takes 4 moles of Al to react with 3 moles of O2 to produce 2 moles of Al2O3.
Given:
- Moles of Al available = 0.32 mol
- Moles of O2 available = 0.26 mol
To determine the limiting reactant, we can compare the moles of Al and O2 to their stoichiometric coefficients in the balanced equation.
For Al:
0.32 mol Al / 4 mol Al = 0.08
For O2:
0.26 mol O2 / 3 mol O2 = 0.0867
From the calculations, we can see that the moles of Al available (0.08) are greater than the moles of O2 available (0.0867). Therefore, Al is in excess and O2 is the limiting reactant.
To determine the limiting reactant, you need to compare the number of moles of each reactant available with the stoichiometric ratio of the reaction. Let's break down the given reaction and calculate the number of moles of each reactant:
4A1 + 3O2 -> 2A1203
Given:
0.32 mol A1
0.26 mol O2
Now, let's calculate the number of moles of A1 and O2 needed to react completely:
From the balanced chemical equation, we can see that for every 4 moles of A1, we need 3 moles of O2 to react completely.
Moles of A1 needed = (0.32 mol A1) / (4 mol A1) = 0.08 mol A1
Moles of O2 needed = (0.08 mol A1) * (3 mol O2 / 4 mol A1) = 0.06 mol O2
As we can see, we only have 0.06 mol of O2, which is less than what is required (0.08 mol). Therefore, O2 is the limiting reactant in this case.
The missing value you mentioned (0.228) might not be directly related to this question. If you provide more context or further information, I would be happy to assist you.