00:01
We are asked to consider the production of methane at high temperature and pressure from carbon monoxide.
00:06
We are given 4 .5 x 10 to the second milliliters of carbon monoxide and 8 .25 x 10 to the second milliliters of hydrogen.
00:25
And the first thing we're asked to solve for is which reactant is in excess.
00:40
I'm going to read ahead to see what i'm solving next and how much of the excess remains and what volume is produced.
00:46
Okay.
00:51
So let's see if we can figure out the moles.
00:59
If i have 4 .5 x 10 to the second milliliters of co, that will be a 1 to 1 mole ratio or a liter ratio at constant temperature and pressure.
01:15
So i'm going to go with milliliters to make it lovelier.
01:24
And that will equal 4 .50 x 10.
01:28
This should have been 4 .50.
01:30
I'm bad.
01:38
There we go.
01:40
10 to the second milliliters of methanol.
01:45
And if i have 8 .25 x 10 to the second.
01:53
Boy, i just copied lots of things down here.
01:55
8 .25 x 10 to the second milliliters of hydrogen.
02:03
That will have a 2 to 1 milliliter ratio for hydrogen and methanol...