WEBVTT
1
00:00:00.940 --> 00:00:04.379 A:middle L:90%
we can use this picture representing a chemical reaction to
2
00:00:04.379 --> 00:00:07.179 A:middle L:90%
look at both with the overall reaction is as well
3
00:00:07.179 --> 00:00:10.449 A:middle L:90%
as the idea of a limiting an excess free agent
4
00:00:11.039 --> 00:00:13.679 A:middle L:90%
. When we start, we have two different react
5
00:00:13.679 --> 00:00:16.929 A:middle L:90%
. It's on this side. If we call our
6
00:00:16.929 --> 00:00:19.329 A:middle L:90%
green circles be, we can see that there are
7
00:00:19.329 --> 00:00:26.760 A:middle L:90%
molecules of B two. The's react with molecules that
8
00:00:26.760 --> 00:00:29.789 A:middle L:90%
have to bees in an A, which will call
9
00:00:29.800 --> 00:00:36.490 A:middle L:90%
a need to. So our reactions are a B
10
00:00:36.490 --> 00:00:40.530 A:middle L:90%
two plus B two, and then when we look
11
00:00:40.530 --> 00:00:42.229 A:middle L:90%
at the product side, we see would make a
12
00:00:42.229 --> 00:00:48.649 A:middle L:90%
new compound A with three bees. There are also
13
00:00:48.649 --> 00:00:52.729 A:middle L:90%
some B two's still on the side. We can
14
00:00:52.729 --> 00:00:54.850 A:middle L:90%
count how many of each there are. So there's
15
00:00:54.850 --> 00:01:04.219 A:middle L:90%
123456 12345 Btu's. And on this side we have
16
00:01:04.219 --> 00:01:10.840 A:middle L:90%
a total of six A B threes and to be
17
00:01:10.840 --> 00:01:15.159 A:middle L:90%
twos. We can simplify this by combining the B
18
00:01:15.159 --> 00:01:19.189 A:middle L:90%
two since they appear on both sides or the equation
19
00:01:19.189 --> 00:01:27.859 A:middle L:90%
ist six a B two plus three B two make
20
00:01:27.859 --> 00:01:34.750 A:middle L:90%
six a B three, and then finally we can
21
00:01:34.750 --> 00:01:38.510 A:middle L:90%
reduce it to the lowest whole number ratio or dividing
22
00:01:38.510 --> 00:01:46.950 A:middle L:90%
everything by three to a B two plus B two
23
00:01:48.640 --> 00:01:53.099 A:middle L:90%
makes two, maybe three. So this is the
24
00:01:53.099 --> 00:01:57.489 A:middle L:90%
overall equation for the chemical reaction that we see to
25
00:01:57.489 --> 00:02:00.579 A:middle L:90%
identify which react and is limiting. We can also
26
00:02:00.579 --> 00:02:04.480 A:middle L:90%
look at which one we have extra and you can
27
00:02:04.480 --> 00:02:07.510 A:middle L:90%
see that we start off with B two and they're
28
00:02:07.510 --> 00:02:08.650 A:middle L:90%
still be, too, on the other side.
29
00:02:09.039 --> 00:02:13.199 A:middle L:90%
So that means that that's not all reacted where we
30
00:02:13.199 --> 00:02:16.800 A:middle L:90%
say it's an excess. So what that means is
31
00:02:16.800 --> 00:02:23.159 A:middle L:90%
the one determining our amount of product is a be
32
00:02:23.159 --> 00:02:29.080 A:middle L:90%
two or limiting, and we can verify this river
33
00:02:29.080 --> 00:02:32.810 A:middle L:90%
calculation using the mole ratios from the equation. If
34
00:02:32.810 --> 00:02:40.650 A:middle L:90%
I have six moles of a B two, I
35
00:02:40.650 --> 00:02:47.419 A:middle L:90%
should produce according to the equation two moles of a
36
00:02:47.419 --> 00:02:54.699 A:middle L:90%
B three for every two moles of a B two
37
00:02:55.539 --> 00:03:07.759 A:middle L:90%
or six moles of product. If I have five
38
00:03:07.759 --> 00:03:15.009 A:middle L:90%
moles of be to the other reactor, I know
39
00:03:15.009 --> 00:03:22.199 A:middle L:90%
that I'll produce two moles of a B three for
40
00:03:22.199 --> 00:03:28.050 A:middle L:90%
every one mole of B two from the coefficients of
41
00:03:28.050 --> 00:03:31.449 A:middle L:90%
the balanced equation, or I'll produce 10 moles,
42
00:03:34.289 --> 00:03:40.409 A:middle L:90%
maybe three. Since six is less than 10 that
43
00:03:40.409 --> 00:03:45.849 A:middle L:90%
means that a B two is the limiting re agent
44
00:03:52.590 --> 00:03:54.610 A:middle L:90%
. We can also use the information for different quantities
45
00:03:54.610 --> 00:03:58.669 A:middle L:90%
besides those given in the picture. So if,
46
00:03:58.669 --> 00:04:00.849 A:middle L:90%
for example, I have three moles of B two
47
00:04:00.860 --> 00:04:18.959 A:middle L:90%
and five moles of a B two, I can
48
00:04:18.959 --> 00:04:27.980 A:middle L:90%
predict how much product for each one I predict.
49
00:04:27.990 --> 00:04:30.189 A:middle L:90%
I calculate the product like I did above so three
50
00:04:30.189 --> 00:04:35.389 A:middle L:90%
moles of B two, and I know that there
51
00:04:35.399 --> 00:04:40.819 A:middle L:90%
are using the mole ratio two moles of a B
52
00:04:40.819 --> 00:04:48.209 A:middle L:90%
three for every one mole would be to. So
53
00:04:48.209 --> 00:05:00.360 A:middle L:90%
I should produce six malls. Be to if I
54
00:05:00.360 --> 00:05:12.170 A:middle L:90%
have Dr Moles of a B two. According to
55
00:05:12.170 --> 00:05:15.120 A:middle L:90%
my balanced equation, there will be two moles of
56
00:05:15.120 --> 00:05:20.459 A:middle L:90%
a B three for every two moles of a B
57
00:05:20.459 --> 00:05:30.250 A:middle L:90%
two or produce. Five moles have a B three
58
00:05:34.240 --> 00:05:40.259 A:middle L:90%
. Since five is less than six, that's the
59
00:05:40.259 --> 00:05:48.310 A:middle L:90%
amount of product that I'll produce. That means that
60
00:05:48.310 --> 00:05:51.629 A:middle L:90%
I have excess free to. So if I were
61
00:05:51.629 --> 00:06:02.800 A:middle L:90%
to produce five moles of a B three, I
62
00:06:02.800 --> 00:06:16.750 A:middle L:90%
only need half assed much be too or 2.5 malls
63
00:06:17.139 --> 00:06:23.089 A:middle L:90%
be to, but I have three, so three
64
00:06:23.329 --> 00:06:31.199 A:middle L:90%
minus 2.5 indicates that I'll have 0.5 moles be to
65
00:06:31.209 --> 00:06:34.449 A:middle L:90%
in excess were left over