For the reaction shown, find the limiting reactant for each of the following initial amounts of reactants.
4Al(s)+3O2 (g)-->2Al2O3 (s)
4 mol Al
2.6 mol O2
The equation tells you that 4 mol Al will require 3 mol O2. You have only 2.6 mol O2; therefore, O2 must be the limiting reagent.
To find the limiting reactant, we need to compare the stoichiometric ratios of the reactants in the balanced chemical equation to the given amounts.
The balanced equation shows that the stoichiometric ratio between Al and O2 is 4:3, meaning that for every 4 moles of Al, we need 3 moles of O2.
Let's calculate the moles of Al required for the given amount of O2:
Using the stoichiometric ratio, we can calculate the moles of Al required to react with 2.6 mol O2 as follows:
2.6 mol O2 * (4 mol Al / 3 mol O2) = 3.4667 mol Al
Since we have 4 mol of Al, which is greater than the calculated 3.4667 mol Al required for the given amount of O2, Al is in excess.
Now, let's calculate the moles of O2 required for the given amount of Al:
Using the stoichiometric ratio, we can calculate the moles of O2 required to react with 4 mol Al as follows:
4 mol Al * (3 mol O2 / 4 mol Al) = 3 mol O2
Since we have 2.6 mol of O2, which is less than the calculated 3 mol O2 required for the given amount of Al, O2 is the limiting reactant.
Therefore, for the given initial amounts of reactants:
- Al is in excess
- O2 is the limiting reactant.
To determine the limiting reactant, you need to compare the mole ratios of the reactants in the balanced chemical equation to the given amounts of the reactants. The reactant that produces fewer moles of the desired product will be the limiting reactant. Let's calculate the moles of Al2O3 produced from each reactant quantity:
For Al:
Using the balanced chemical equation, 4 mol Al yields 2 mol Al2O3.
Therefore, 4 mol Al will produce (2 mol Al2O3 / 4 mol Al) = 1 mol Al2O3.
For O2:
Using the balanced chemical equation, 3 mol O2 yields 2 mol Al2O3.
Therefore, 2.6 mol O2 will produce (2 mol Al2O3 / 3 mol O2) = 1.73 mol Al2O3.
Comparing the moles of Al2O3 produced from each reactant:
From 4 mol Al, you get 1 mol Al2O3.
From 2.6 mol O2, you get 1.73 mol Al2O3.
Since the reaction cannot produce fractional moles of substances, we need to consider whole numbers.
Therefore, 4 mol Al will produce more Al2O3 than 2.6 mol O2, making Al the limiting reactant.