For the reaction shown, find the limiting reactant for each of the following initial amounts of reactants.
4Al(s) + 3O2(g) -> 2Al2O3(s)
4mol Al, 2.6mol O2?
a) 4mol Al,
b) 2.6mol O2.
I seriously have no idea what they are asking! What do you do with the 4mol Al & 2.6mol O2...to the equation given?!! Do you just multiply something?
My question doesn't have to be answered. I figured out what they were asking and got it correct.
To find the limiting reactant, you need to compare the number of moles of each reactant to the stoichiometric ratio in the balanced chemical equation.
Let's start with option a) 4 mol Al.
From the balanced chemical equation, we know that 4 mol Al reacts with 3 mol O2 to produce 2 mol Al2O3.
So, we need to determine how many moles of Al2O3 can be formed with 4 mol Al.
Using stoichiometry, we set up a ratio based on the coefficients in the balanced equation:
4 mol Al : 3 mol O2 = 2 mol Al2O3
The stoichiometric ratio tells us that for every 4 moles of Al, we need 3 moles of O2 to produce 2 moles of Al2O3.
Now let's move on to option b) 2.6 mol O2.
To find out how much Al2O3 can be formed with 2.6 mol O2, we again use stoichiometry:
4 mol Al : 3 mol O2 = 2 mol Al2O3
Since the stoichiometric ratio is in terms of Al to O2, we need to convert the given amount of O2 to Al by using the same ratio:
2.6 mol O2 * (4 mol Al / 3 mol O2) = 3.47 mol Al
We find that to react with 2.6 mol O2, we need 3.47 mol Al.
Now, compare the initial amounts of reactants to the calculated amounts:
a) 4 mol Al (initial amount) - 4 mol Al (required amount) = 0 mol Al remaining
b) 2.6 mol O2 (initial amount) - 3.47 mol O2 (required amount) = -0.87 mol O2 remaining
Based on the calculations, we can conclude that the limiting reactant is option a), 4 mol Al. This means that all the Al will react, and some O2 will be left over.
I hope this explanation helps you understand how to determine the limiting reactant using stoichiometry.