00:01
For this question, we are given an amount of cl2 and f2, and we're asked to determine which is the limiting reactant and what is the theoretical yield.
00:10
To do this, we need to convert the information associated with these two gases into moles.
00:17
To do this, we'll use the ideal gas log.
00:21
So we have two liters of each of the gases at 298 kelvin.
00:25
They both have different pressures, though.
00:27
We have 337 millimeters of mercury of chlorine.
00:32
So we'll take the 337 millimeters of mercury divided by 760 to give us atmospheres of chlorine.
00:39
We'll multiply that by the volume in liters, divide by r, and divide by the kelvin temperature, 298.
00:47
This will give us 0 .03627 moles of chlorine.
00:52
We see that it is a 1 to 2 relationship.
00:56
So for every mole of chlorine, we get twice as much clf3...