00:01
All right, so in this question, we are giving this potential reaction to produce hydrogen gas, and we're asked to calculate the percent yield of this reaction.
00:10
So the first thing we need to do is first calculate which one of these, whether it's methane or water vapor, it's going to be the limiting reagent.
00:18
How we do that is we're going to calculate the number of moles each.
00:22
Notice that these both have a one -to -one molar ratio.
00:26
So whichever one has less moles should just be the limiting reactant.
00:31
All right, so to start, we're going to calculate the moles of methane.
00:38
So we're going to have n of ch4 as equal to, we know for this whole problem, that n is equal to pv over rt.
00:50
So with that being said, we can calculate the number of moles of methane.
00:55
Using the pressure it gives us, which is 700 or 732 tours, times the volume, which it gives us is 25 .5 liters over and then rt.
01:08
R is going to be 62 .36, and the units for that is going to be tor times liter over moles times kelvin.
01:20
And then lastly, it gives us a temperature of 25 degrees celsius, which should be equivalent to 298 kelvin.
01:33
So if we run that calculation, you're going to find that we're going to get 1 .00 moles.
01:42
All right.
01:43
So now we get to do the moles of water vapor.
01:46
We're essentially going to go through the exact same process.
01:50
We can find that the moles of h2o equal to the pressure gives us, which is 702, tor, the volume, which is 22 .8 liters over.
02:07
We're going to use the exact same r value.
02:20
And this time it gives us a temperature of 398 degrees kelvin.
02:29
All right, so with all that being said, if you run this calculation, you'll find that the moles of water is 0 .64.
02:37
So with that being said, water is going to be the limiting reagent of this reaction.
02:42
So now we're going to calculate our theoretical yield.
02:46
So i'm just going to write theory here...