00:01
Hi, guys.
00:02
So in this problem, we're being asked to find the percent of n2h4 or hydrazine in the initial mixture of ammonia and hydrogen gas.
00:16
So since we're changing three variables across these reactions, we're changing the pressure, the temperature, as well as the moles of gas.
00:29
This is going to be a combined gas lot problem.
00:33
So first let's write out what the combined gas lock is.
00:36
It's p1 v1 over n1, t1 equals p2 v2 over n2, t2.
00:52
Since this problem specifies that we're talking about a sealed container, we can assume that the initial and final volumes are going to be the same.
01:00
So we can cross them out in this equation.
01:04
So that leaves us with just p1 over n1t1 equals p2 over n2t2.
01:10
So let's think about what our final goal with this equation is.
01:17
What we really want from here is the ratio of the initial number of molds to the final number of moles.
01:25
Because if we're trying to find the percent of n2h4 in the original mixture, we're going to need to find the ratio of the moles of n2h4 to the moles of ammonia.
01:38
And finding our mole ratio here is going to lead us to that answer later on.
01:46
So let's do that first.
01:49
So our pressure initially is 0 .5 atm.
01:55
N1 is going to be left as a variable.
01:59
T1 is 300 kelvin's.
02:05
P2 is going to be 4 .5 atm.
02:10
N2 again is going to be left as a variable.
02:14
And t2 is 1 ,200 kelvin.
02:19
So when we rearrange this equation to give us n2 in terms of n1, we get n2 is equal to 2 .25n1.
02:39
So what this is telling us is that our final number of moles of gas is going to be 2 .25 times the initial number of moles of gas.
02:51
Okay, now let's look at the reactions that are actually taking place.
02:57
So the first one is the decomposition of ammonia, which is nh3.
03:03
So two nh3 molecules are decomposing to use.
03:11
1 and 2 and 3 hydrogen molecules.
03:22
So let's talk about the initial and final number of moles of gas here.
03:30
So initial is going to be the reactants.
03:33
Final is going to be with the products.
03:39
So initially we have two moles of gas and you can get that just by looking at the co -coval in front of nh3.
03:52
On the product side, however, we have 1 plus 3, so 4 moles of gas.
04:03
That means over the course of the reaction, we are multiplying the number of modes of gas by 2.
04:13
So if we assume that we have an initial number of x moles of gas, our final value is going to be 2x moles of gas.
04:37
Okay, now let's repeat this for the other equation...