00:01
Reading this, i don't know if i should have done this one.
00:05
Ammonia gas reacts with chlorine gas, chlorine trifluoride, to produce nitrogen, chlorine, and hydrogen fluoride gases.
00:14
So we are given that we have 30 grams of ammonia, and it's reacting with 175 grams of chlorine gas.
00:30
We're asked to find the mass of each product.
00:38
So we're going to figure out which one is limiting.
00:40
I've got another one to do, too.
00:44
Ok, i'm going to have to go quick on this one.
00:49
Ok, so let's start with 30 .0 grams of nh3 and figure out 17 .04 grams of ammonia per mole.
01:09
Figure out how many grams of my first product are formed.
01:14
I have a 2 to 1 mole ratio for n2 and nh3.
01:21
That's for my balanced chemical equation.
01:23
And there are 28 .02 grams of nitrogen per mole of nitrogen.
01:31
This will give me, for my first possible theoretical yield, 30 times 28 .02 divided by 17 .04 divided by 2.
01:49
This is 24 .67.
01:51
I'll round later.
01:58
Now i'm going to do the similar calculation for 175 grams of clf3.
02:08
Clf3.
02:17
Clf3 is 92 .45 grams per mole.
02:34
Here i have a 2 to 1 mole ratio again.
02:45
And finally, again, 28 .02 grams per mole.
02:52
And this will equal 175 times 28 .02 divided by 92 .45 divided by 2.
03:10
This is 26 .52 grams of n2.
03:18
So let me get back to my problem here.
03:21
What is the mass of each product? my mass of n2 will equal 24 .7 grams.
03:38
Then i'm going to find the mass of my other two using my 30 grams of nh3.
04:04
On each of these, we will convert grams to moles using the molar mass.
04:24
For chlorine and for hydrogen fluoride gas, we have this.
04:39
And then for this, we have two.
04:52
Then for both of these, we're going to multiply by our molar mass of the product in question.
05:20
Oopsie.
05:25
And here's my second one.
05:35
Now i can put these numbers in my calculator.
05:37
30 divided by 17 .04...