00:01
Okay, there are several parts to this problem, and we'll be considering indium phosphide.
00:12
Okay.
00:13
I've written the chemical equation here, and notice there's only one coefficient.
00:19
We're given scenario one.
00:21
We have 2 .56 liters of indian.
00:23
I'm not even sure what that is.
00:27
Trimethylindium.
00:28
I don't know what it is.
00:29
At 2 .0 atmospheres allowed to react with 1 .38 liters of phosphorus trihydrate.
00:34
At three atmospheres and were asked what mask of indian 3 phosphide will be produced, assuming an 87 % efficiency.
00:54
Okay, let's go to a blue.
01:00
Probably not my favorite blue.
01:03
So first of all, let's figure out our moles of this one.
01:10
So our moles will equal pv.
01:22
Let me write here, n equals pv over rtv.
01:34
Divided by rt was 900 kelvin.
01:51
And this will give me 0 .0693 moles.
01:59
And my second substance, we're going to calculate this the same way.
02:12
Pressure times our volume.
02:15
And we'll again deval it is up.
02:17
Divide by 0 .0826 and 900 kelvin.
02:33
And this will equal 0 .0561 moles.
02:36
And since the two reactants are a 1 to 1 ratio, the ph 3 is limiting because it's the smaller of the two values.
03:03
So then we can use our value 0 .0561 moles of ph 3 times, and this is also a 1 to 1 .1...