00:01
Okay, chapter 7, section problem number 46, asks us to find the limiting reactant for each initial amount of reactants.
00:12
And it gives us the balanced equation for, you know, four aluminum plus three oxygen goes to two aluminum oxide.
00:22
And we're asked in each case to say what the limiting react agent is.
00:27
Okay.
00:28
So the most surefire way to answer a question.
00:32
Like this and get it right every single time is to convert the amount of moles of the starting material into the product and see which one gives you less product.
00:43
In terms of limiting reagents, you always want to just convert it all the way through the product and see which one gives you the least amount of product and that's how you know.
00:53
So if you were to make a table for this question, maybe using excel or something like that or on your page.
01:03
Oh, this is not going to do what i wanted to do.
01:15
If you were to make a table, i'll try and recover from this, that had what we're doing right here.
01:24
So we've got aluminum, oxygen, and then really, this is kind of broken.
01:40
I kind of break this up into like two to more cells being product from the aluminum.
01:50
And then the product that you can form from the oxygen.
02:04
Right.
02:04
Okay.
02:05
Because it gives you all these in moles.
02:08
So we can kind of just go from there.
02:10
Okay, so we've got one, one, and let's start here.
02:16
So if we convert one mole of aluminum, how many moles of the al2 -03 are we going to get? okay, well, one mole of aluminum.
02:34
And we look at our balance equation.
02:36
It's four moles of aluminum.
02:38
Required to produce two moles of al2 -03.
02:45
So we can see that we'll form half of one here, 0 .5.
02:50
Now, what about one mole of 02? if we consider the stoichiometry of our equation, the coefficients of our equation, is 3 -0 -2's moles for, again, 2 moles of a -l -203.
03:10
So it's 0 .6, right? 0 .6 or 0 .7, whatever we're talking about.
03:17
Okay, so here, there is, there's actually, we can form more of our product, the al2, 03, with the 02, then we can with the al2.
03:27
So this is our limiting reagent right here.
03:35
And that's the process for all of these...