00:01
So we have three different reactions, and the question is asking us, if we had 20 grams of each reactant in all three reactions, what is our limiting reagent? and then it tells us which product we're looking for in particular in parentheses.
00:18
So when we look at limiting reagents, a reaction can only continue as long as you have both products or excuse me, both reactants available.
00:30
Once you run out of a single reactant, you are no longer able to make any more product.
00:36
So the analogy that i like to think about is, if i had meat, cheese, and bread to make sandwiches, i can make as many sandwiches as long as i have enough of all three ingredients.
00:49
Once i run out of bread, it doesn't matter how much meat and cheese i have, i can't make any more.
00:54
I can't make any more.
00:56
So then my reaction, my sandwich -making ability, stops.
01:00
Same thing with chemical reactions.
01:02
As long as i have both reactants, but once i run out of one, i'm stopped.
01:08
That's all i can make.
01:10
And so that whatever one has the least amount is our limiting reagent.
01:15
So in order to find this, what you need to use is dimensional analysis and stoichiometry and calculate the amount of product you can make with both reactants.
01:24
And whatever one is the least amount, that's your limiting region.
01:28
So this first one, a, we have aluminum in chlorine, and they're going to form aluminum chloride.
01:44
So we need to figure out if i have 20 grams of my aluminum and 20 grams of my chlorine, which one of these is going to be my limiting reagent? so and then it also tells us we're looking for our grams of my product are aluminum chloride.
02:09
So we're just going to go ahead.
02:10
We're going to convert from 20 grams of each reactant to grams of my product.
02:15
And whatever one makes the least amount, that's going to be our limiting reagent.
02:19
So we'll start with aluminum first.
02:22
So if i have 20 grams of aluminum and i'm going to convert from grams to moles.
02:30
And so when i look up on my periodic table of elements, i go and i see that for every one mole of aluminum, it tells me that i have 26 .982 grams.
02:46
And then i know that for every one mole of aluminum, or excuse me, for every two moles of aluminum, i get two moles of my aluminum chloride.
02:58
And then to figure out my moles of, or excuse me, my molar mass of aluminum chloride, all i need to do is i need to add up all the individual elements and their molar masses.
03:14
And then that's that's the molar mass for my total compound.
03:18
So let's take aluminum.
03:21
We know that it's got 26 .92.
03:24
And my chlorine is a 35 .45.
03:29
But in my aluminum chloride, i have three chlorines.
03:34
And so i'm going to take this number times three, and then i'm going to add it to my aluminum.
03:41
So when i get 26 .92 plus 35 .45 times three, my total molar mass from my aluminum chloride is going to be a 133 .3 .3 .3 .3 to rand for mold.
04:02
So i'm going to use this number to convert back to grams.
04:07
So i know that for every one mole i just calculated of aluminum chloride, i have 133 .332 grams.
04:20
And so all i'm going to do now is i'm going to multiply across the top and then divide by the numbers on the bottom.
04:26
So i'm going to take 20 times 1 times 2 times 133.
04:31
So i take all those numbers and multiply it at the top, 5 ,333 .28.
04:40
And i'll multiply all these numbers across the bottom.
04:43
So 1 times 26 .92 times 2 times 1.
04:49
And so when i multiply those numbers, i get it 53 .964.
04:55
So i'm going to divide the top by the bottom.
04:58
And so the amount of aluminum chloride i can make with 20 grams of aluminum is 98 .8 grams.
05:10
So that's how much i can make of aluminum chloride.
05:14
So now what i need to do is figure out, okay, if i have 20 grams of chlorine, how much aluminum chloride can i make? so we're going to do the same process.
05:27
So my 20 grams of chlorine, we're going to convert that to mold.
05:33
So i know that there's 35 .45 grams of chlorine for every one mold.
05:40
And then i know that my mold -mole ratios are slightly different on this one.
05:47
So if i look up at my balanced equation here, i have three moles for every two moles of my product.
06:00
Let's take that ratio down here.
06:02
So i have three moles of chlorine for every two moles of aluminum.
06:08
Chloride and then for every one mole of aluminum chloride we just solved that i get 133 .33332 grams so same process i multiply across the top multiply all these numbers across the bottom so when i do that i get a total of 25 .1 grams of aluminum chloride and so looking at this this is my smaller number.
06:41
Because this is my smaller number, that means this is all the product i'm allowed to make.
06:47
And that my chlorine, if i have 20 grams of it and 20 grams of aluminum, my chlorine is going to be my limiting reagent.
06:55
So once all that chlorine is used up, after that time, i've only made 25 .1 grams of aluminum chloride.
07:03
So this is all i can make.
07:06
And then chlorine is going to be my limiting reagent.
07:11
So we're just going to use that same process for the other two reactions.
07:16
So on the second reaction, i got pneumonia and oxygen, forming nitrogen and oxide in water.
07:37
And i have 20 grams of each reacted again.
07:39
So i have 20 grams of my ammonia.
07:43
I have 20 grams of my oxygen.
07:47
In and the parentheses for this one was for water so we're trying to figure out how much water can i make and then which one of these is going to be my limiting region so same process now these this one's not an element this one's so these are compounds and so i need to do the same process i did earlier to find out my molar mass so in order to find out my molar mass i'm just going to add up the molar masses of all the individual elements that make up my compound so let's start with ammonia.
08:22
So nitrogen is going to be 14 .007 and hydrogen is going to be 1 .008.
08:31
But i have three of them in this one compound.
08:34
So i'm going to take that times three.
08:36
So when i add this up, i get 17 .031 grams per mole.
08:44
So here is my molar mass of my ammonia.
08:51
And that'll come in handy later.
08:54
My molar mass of oxygen, so remember oxygen is diatomic.
09:01
And so i'm gonna take my 15 .998 times two.
09:06
And that's gonna give me my molar mass of oxygen because there's two of them.
09:12
So that's gonna be 31 .998.
09:20
And then my water, molar mass of water.
09:27
So i have hydrogen, which is 1 .008.
09:30
I have two hydrogens.
09:32
I'm going to take that times two.
09:34
Oxygen, which is said, it was a 15 .999.
09:38
And we'll add those together...