0:00
All right.
00:01
So in this question, we have a reaction of chlorine and fluorine gas to form chlorine trifluoride.
00:10
And we're asked to determine what is the limiting reactant and also what's the theoretical yield.
00:15
So starting, first thing to do is figure out the number of moles that we have of both these things.
00:22
Using the ideal gas law, we can just say that n is equal to pv over rt.
00:30
So from there we calculate n, let's say, of cl2, which is going to be equal first to the pressure, which it tells us the problem is 337 millimeters mercury, and the volume which tells us a 2 .0 liter vessel.
00:47
Over r, 62 .36 is going to be the value we use, since it's just in millimeters of mercury.
00:54
Units are going to be millimeters of mercury times liters over mole times kelvin.
01:03
And lastly, a temperature of 298 kelvin.
01:08
All right, so all of our units check out, and we should get moles from this.
01:11
If we do that, we're going to have 0 .0362 moles.
01:17
We can do the exact same thing for fluorine, right? we can say moles of f2.
01:21
And this time i'm going to leave off the units just for sake of time, since we already made sure all the units check out.
01:27
So we're going to have 729 for the pressure, 2 again for the volume, 62.
01:36
0 .36 again for r and 298 again for temperature, giving us 0 .05 to 3 moles...