1
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2
00:00:00,830 --> 00:00:03,600
Let's say we have some ammonia gas.

3
00:00:03,600 --> 00:00:07,720
That's NH3 and it's a gas, that's why the g is in

4
00:00:07,719 --> 00:00:08,769
parentheses.

5
00:00:08,769 --> 00:00:15,149
And we combine that with some oxygen, molecular oxygen, it's

6
00:00:15,150 --> 00:00:16,359
also a gas.

7
00:00:16,359 --> 00:00:24,379
And that reaction produces some nitrogen monoxide.

8
00:00:24,379 --> 00:00:28,320
NO, monoxide, there's only one oxygen there.

9
00:00:28,320 --> 00:00:31,949
That's also a gas, it's also called nitric oxide, not to be

10
00:00:31,949 --> 00:00:34,689
confused with nitrous oxide.

11
00:00:34,689 --> 00:00:38,229
I'll write nitrous oxide, it's N2O, this is laughing gas.

12
00:00:38,229 --> 00:00:41,449
Anyway, I don't want to divert you too much, we have to focus

13
00:00:41,450 --> 00:00:42,640
on the problem.

14
00:00:42,640 --> 00:00:48,271
So this is nitric oxide or nitrogen monoxide.

15
00:00:48,271 --> 00:00:49,890
It's a pollutant, I think it comes out of

16
00:00:49,890 --> 00:00:50,929
some cigarette smokes.

17
00:00:50,929 --> 00:00:52,450
I think it's used in the body, as well.

18
00:00:52,450 --> 00:00:54,670
You could we could look it up on the internet.

19
00:00:54,670 --> 00:00:55,670
There's your oxide.

20
00:00:55,670 --> 00:00:56,859
Ammonia is an important fertilizer.

21
00:00:56,859 --> 00:00:57,909
You know all about oxygen.

22
00:00:57,909 --> 00:01:00,129
And this also yields some water.

23
00:01:00,130 --> 00:01:04,390
So, plus some H2O.

24
00:01:04,390 --> 00:01:14,269
And we're told that we're given 34 grams of ammonia.

25
00:01:14,269 --> 00:01:16,584
And we're given 32 grams of oxygen.

26
00:01:16,584 --> 00:01:21,119


27
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This is going to be the oxygen molecule, O2.

28
00:01:26,132 --> 00:01:31,320
So the question is, how many grams of nitrogen monoxide, or

29
00:01:31,319 --> 00:01:33,929
the nitric oxide, are going to be produced?

30
00:01:33,930 --> 00:01:41,560
So how much of just the NO is going to be produced in grams?

31
00:01:41,560 --> 00:01:44,650
So this is a stoichiometry problem.

32
00:01:44,650 --> 00:01:47,120
And so the important thing first is to just make sure we

33
00:01:47,120 --> 00:01:49,579
have a balanced equation before we even start anything.

34
00:01:49,579 --> 00:01:52,200
And lo and behold, we don't have a balanced equation.

35
00:01:52,200 --> 00:01:54,740
Let's confirm that it's not balanced.

36
00:01:54,739 --> 00:01:57,530
Let's see, we have one nitrogen here, we have one

37
00:01:57,530 --> 00:01:58,320
nitrogen there.

38
00:01:58,319 --> 00:02:01,709
That looks balanced so far.

39
00:02:01,709 --> 00:02:03,609
And remember, the pattern is start with the complicated

40
00:02:03,609 --> 00:02:07,810
stuff, leave the single atom molecules for last. Because

41
00:02:07,810 --> 00:02:09,889
those you can fix at the end without missing

42
00:02:09,889 --> 00:02:11,060
anything else up.

43
00:02:11,060 --> 00:02:12,569
Hydrogen, we have three hydrogens on

44
00:02:12,569 --> 00:02:14,189
the left-hand side.

45
00:02:14,189 --> 00:02:16,120
On the right-hand side we have two hydrogens.

46
00:02:16,120 --> 00:02:18,789


47
00:02:18,789 --> 00:02:22,030
So let's see, how can we have three hydrogens on the

48
00:02:22,030 --> 00:02:22,610
right-hand side?

49
00:02:22,610 --> 00:02:28,360
If we multiply this times one and a half, 1.5, now we have

50
00:02:28,360 --> 00:02:30,250
three hydrogens on this side.

51
00:02:30,250 --> 00:02:32,750
1.5 times 2, we have three hydrogens on

52
00:02:32,750 --> 00:02:34,030
the right-hand side.

53
00:02:34,030 --> 00:02:35,280
Things are looking good.

54
00:02:35,280 --> 00:02:37,990


55
00:02:37,990 --> 00:02:41,760
We have two oxygen on the left-hand side.

56
00:02:41,759 --> 00:02:43,299
How many oxygens do we have on the right-hand side?

57
00:02:43,300 --> 00:02:46,900
We have one oxygen here, and we have one oxygen in this

58
00:02:46,900 --> 00:02:49,539
molecule, but we have one and a half of the whole molecule.

59
00:02:49,539 --> 00:02:52,329
So we have one and a half oxygens and then we have one

60
00:02:52,330 --> 00:02:53,040
more oxygen.

61
00:02:53,039 --> 00:02:55,799
So we have two and a half oxygens on the right-hand

62
00:02:55,800 --> 00:02:58,740
side, we only have two in this molecule.

63
00:02:58,740 --> 00:03:00,189
So what do we have to do?

64
00:03:00,189 --> 00:03:02,810
How can we get two and a half oxygens here?

65
00:03:02,810 --> 00:03:10,810
Well if we multiply it by 5/4, or 1.25.

66
00:03:10,810 --> 00:03:14,400
5/4 times 2 is 5/2, which is 2.5.

67
00:03:14,400 --> 00:03:17,050
So now we have 2.5 oxygens.

68
00:03:17,050 --> 00:03:19,070
5/4 times 2 is 2.5.

69
00:03:19,069 --> 00:03:22,810
1 plus 1.5 is also 2.5.

70
00:03:22,810 --> 00:03:24,270
Well, we're not balanced yet.

71
00:03:24,270 --> 00:03:27,850
We can't leave this equation with just these weird decimal

72
00:03:27,849 --> 00:03:28,979
numbers over there.

73
00:03:28,979 --> 00:03:30,229
So let's rewrite it.

74
00:03:30,229 --> 00:03:32,349


75
00:03:32,349 --> 00:03:36,439
If we wanted to get rid of all of these, we can multiply the

76
00:03:36,439 --> 00:03:39,969
entire equation by 2.

77
00:03:39,969 --> 00:03:42,009
No, not 2.

78
00:03:42,009 --> 00:03:44,639
We have to multiply the entire equation by 4 to get rid of

79
00:03:44,639 --> 00:03:46,669
this 4 in the denominator.

80
00:03:46,669 --> 00:03:55,169
So we multiply the entire equation by 4, we have 4

81
00:03:55,169 --> 00:03:59,169
molecules of ammonia.

82
00:03:59,169 --> 00:04:01,269
We can even think of it in terms of moles.

83
00:04:01,270 --> 00:04:03,120
Right now I'm thinking of individual molecules.

84
00:04:03,120 --> 00:04:05,200
But we could say we have 4 moles of ammonia.

85
00:04:05,199 --> 00:04:11,619
4 times 6 times 10 to the 23 molecules of ammonia.

86
00:04:11,620 --> 00:04:13,640
Either way it all works out.

87
00:04:13,639 --> 00:04:15,939
Hopefully, you're starting to see the value of moles.

88
00:04:15,939 --> 00:04:21,009
Plus 5 molecules, I'll just think of it in terms of

89
00:04:21,009 --> 00:04:23,779
individual molecules for now, plus 5 molecules

90
00:04:23,779 --> 00:04:25,315
of molecular oxygen.

91
00:04:25,315 --> 00:04:35,079


92
00:04:35,079 --> 00:04:37,259
We're just multiplying everything by 4, that's what

93
00:04:37,259 --> 00:04:37,990
we're doing.

94
00:04:37,990 --> 00:04:45,550
Yields 4 moles of nitrogen monoxide.

95
00:04:45,550 --> 00:04:50,689
So 4NO plus, we multiplied both sides by

96
00:04:50,689 --> 00:04:53,629
4, so plus 6 waters.

97
00:04:53,629 --> 00:04:55,939
6 H2O.

98
00:04:55,939 --> 00:04:57,500
So there you go, we got some good

99
00:04:57,500 --> 00:04:59,199
practice balancing equations.

100
00:04:59,199 --> 00:05:01,779
So let's go back to the original problem.

101
00:05:01,779 --> 00:05:04,715
We're given 34 grams of ammonia.

102
00:05:04,716 --> 00:05:08,040


103
00:05:08,040 --> 00:05:11,020
So what we need to figure out is how many moles of ammonia

104
00:05:11,019 --> 00:05:11,750
were we given?

105
00:05:11,750 --> 00:05:15,000
So what's the atomic mass of ammonia?

106
00:05:15,000 --> 00:05:17,300
Not the atomic mass, what's the molecular mass of ammonia?

107
00:05:17,300 --> 00:05:21,509


108
00:05:21,509 --> 00:05:23,789
We're dealing with nitrogen 14, so it has a

109
00:05:23,790 --> 00:05:27,340
mass number of 14.

110
00:05:27,339 --> 00:05:30,560
Hydrogen has a mass number of 1.

111
00:05:30,560 --> 00:05:36,649
So each nitrogen has a mass of 14.

112
00:05:36,649 --> 00:05:39,289
And then the hydrogens each have a mass of 1.

113
00:05:39,290 --> 00:05:40,819
Remember, hydrogen is kind of strange.

114
00:05:40,819 --> 00:05:42,889
It doesn't have neutrons, or at least in its most

115
00:05:42,889 --> 00:05:43,779
traditional form.

116
00:05:43,779 --> 00:05:45,129
So it has just a mass number of 1.

117
00:05:45,129 --> 00:05:47,790
It's just a proton and an electron if it's neutral.

118
00:05:47,790 --> 00:05:50,110
So this is 3 times 1.

119
00:05:50,110 --> 00:05:53,569
We have three hydrogen atoms. So the mass, the atomic mass,

120
00:05:53,569 --> 00:05:57,719
of one molecule of ammonia is 14 plus 3 is

121
00:05:57,720 --> 00:06:00,770
17 atomic mass units.

122
00:06:00,769 --> 00:06:06,669
Or, another way to write that, if one molecule of ammonia is

123
00:06:06,670 --> 00:06:14,990
17 atomic mass units, then 1 mole of

124
00:06:14,990 --> 00:06:19,300
ammonia is how many grams?

125
00:06:19,300 --> 00:06:23,590
It's going to be 17 grams.

126
00:06:23,589 --> 00:06:25,439
So how many moles of ammonia are we given?

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00:06:25,439 --> 00:06:29,990
We're given 34 grams of ammonia, 1 mole is 17 grams.

128
00:06:29,990 --> 00:06:31,470
So we're given 2 moles.

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00:06:31,470 --> 00:06:33,020
34 is 2 times 17.

130
00:06:33,019 --> 00:06:35,069
So this is 2 moles.

131
00:06:35,069 --> 00:06:39,480
We're given 2 moles of ammonia.

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00:06:39,480 --> 00:06:41,400
Let's see how much oxygen we're given.

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00:06:41,399 --> 00:06:44,489
Or how much of the molecular oxygen we've been

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00:06:44,490 --> 00:06:46,160
given in this case.

135
00:06:46,160 --> 00:06:52,660
So let's see, the mass of just oxygen by itself is 16.

136
00:06:52,660 --> 00:06:56,590
The mass of just the atomic oxygen, you have to be a

137
00:06:56,589 --> 00:06:59,049
little bit careful here because sometimes people say

138
00:06:59,050 --> 00:07:01,340
we have 32 grams of oxygen when they're really talking

139
00:07:01,339 --> 00:07:04,009
about the molecular oxygen.

140
00:07:04,009 --> 00:07:06,539
Well, I guess it doesn't matter either way.

141
00:07:06,540 --> 00:07:09,640
But, sometimes when they talk about oxygen you have to make

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00:07:09,639 --> 00:07:11,889
sure whether it is molecular or atomic oxygen.

143
00:07:11,889 --> 00:07:14,584
But the atomic mass number of oxygen is 16.

144
00:07:14,584 --> 00:07:17,159
I can confirm that by looking at the

145
00:07:17,160 --> 00:07:18,670
periodic table down here.

146
00:07:18,670 --> 00:07:21,420


147
00:07:21,420 --> 00:07:26,710
So what's the molecular mass of the diatomic molecule O2?

148
00:07:26,709 --> 00:07:31,109
Well it has 2 oxygen, so it's going to be 2 times 16 equals

149
00:07:31,110 --> 00:07:33,879
32 atomic mass units.

150
00:07:33,879 --> 00:07:37,129
One molecule of O2 is 32 atomic mass units.

151
00:07:37,129 --> 00:07:44,730
Or 1 mole of O2 is how many grams?

152
00:07:44,730 --> 00:07:49,689
Well if one molecule is 32 atomic mass units, then 6

153
00:07:49,689 --> 00:07:53,449
times 10 to the 23 of that molecule are going to be that

154
00:07:53,449 --> 00:07:58,969
many grams. 32 grams. So how many moles of oxygen have we

155
00:07:58,970 --> 00:08:00,070
been given?

156
00:08:00,069 --> 00:08:02,990
We've been given exactly 32 grams of oxygen, which is

157
00:08:02,990 --> 00:08:04,500
exactly 1 mole.

158
00:08:04,500 --> 00:08:08,139
So we've been given 34 grams of ammonia, which is 2 moles.

159
00:08:08,139 --> 00:08:10,879
Let me write that in a nice vibrant color.

160
00:08:10,879 --> 00:08:16,759


161
00:08:16,759 --> 00:08:23,750
And we've been given 1 mole of the oxygen molecule.

162
00:08:23,750 --> 00:08:28,779
Now, when we look at this reaction, for every 4 moles of

163
00:08:28,779 --> 00:08:32,178
ammonia, we need 5 moles of oxygen.

164
00:08:32,178 --> 00:08:36,158
Or for every 5 moles of oxygen we need 4 moles of ammonia.

165
00:08:36,158 --> 00:08:38,080
So something doesn't gel here.

166
00:08:38,080 --> 00:08:40,020
Normally, we need more moles of oxygen

167
00:08:40,019 --> 00:08:41,350
than we have ammonia.

168
00:08:41,350 --> 00:08:44,930
In the example that we're working through, we've been

169
00:08:44,929 --> 00:08:47,349
given less moles of oxygen than ammonia.

170
00:08:47,350 --> 00:08:50,290
We've been given less oxygen than we need

171
00:08:50,289 --> 00:08:53,000
for all of this ammonia.

172
00:08:53,000 --> 00:08:56,850
In an ideal world, if we had 2 moles of ammonia, we would

173
00:08:56,850 --> 00:08:59,259
need 2.5 moles of oxygen.

174
00:08:59,259 --> 00:09:00,819
The ratio of ammonia to oxygen, let

175
00:09:00,820 --> 00:09:02,170
me write that down.

176
00:09:02,169 --> 00:09:04,439
In a different color, this is getting boring.

177
00:09:04,440 --> 00:09:12,000
The ratio of ammonia, NH3 to oxygen in our balanced

178
00:09:12,000 --> 00:09:14,779
equation, let me put that in a square so you know this is the

179
00:09:14,779 --> 00:09:17,259
most important part of what I'm writing.

180
00:09:17,259 --> 00:09:20,679
The ratio in this reaction is 4:5.

181
00:09:20,679 --> 00:09:24,209


182
00:09:24,210 --> 00:09:29,680
So if I'm given 2 moles of ammonia,

183
00:09:29,679 --> 00:09:31,149
this is equal to what?

184
00:09:31,149 --> 00:09:34,039
If I'm given 2 moles of ammonia, how many moles of

185
00:09:34,039 --> 00:09:34,969
oxygen do I need?

186
00:09:34,970 --> 00:09:38,550
I need 2.5 moles of oxygen.

187
00:09:38,549 --> 00:09:41,240
Whatever this is, is that right?

188
00:09:41,240 --> 00:09:42,750
Yeah, 4/5.

189
00:09:42,750 --> 00:09:44,809
1.25.

190
00:09:44,809 --> 00:09:48,239
If you doubled both of these numbers, you get 4/5.

191
00:09:48,240 --> 00:09:49,529
So I need 2.5 moles.

192
00:09:49,529 --> 00:09:51,490
But I don't have 2.5 moles of oxygen.

193
00:09:51,490 --> 00:09:53,460
I only have 1 mole of oxygen.

194
00:09:53,460 --> 00:09:55,580
So oxygen is going to be the limiting

195
00:09:55,580 --> 00:09:58,860
reagent in this reaction.

196
00:09:58,860 --> 00:10:00,409
I don't have enough oxygen.

197
00:10:00,409 --> 00:10:03,059
I have plenty of ammonia, but I don't have enough oxygen to

198
00:10:03,059 --> 00:10:03,679
react with it.

199
00:10:03,679 --> 00:10:04,929
So this is the limiting reagent.

200
00:10:04,929 --> 00:10:08,669


201
00:10:08,669 --> 00:10:12,079
And I said before, the word reagent and reactant are used

202
00:10:12,080 --> 00:10:12,435
interchangeably.

203
00:10:12,434 --> 00:10:14,619
But when people talk about the limiting ones, they tend to

204
00:10:14,620 --> 00:10:16,190
call it the reagent.

205
00:10:16,190 --> 00:10:17,800
So oxygen is the limiting reagent.

206
00:10:17,799 --> 00:10:19,189
So we have extra ammonia.

207
00:10:19,190 --> 00:10:23,280
So given that we have 1 mole of oxygen, how many moles of

208
00:10:23,279 --> 00:10:25,259
ammonia can I react with that?

209
00:10:25,259 --> 00:10:27,830


210
00:10:27,830 --> 00:10:33,259
So this reaction is going to look something like this.

211
00:10:33,259 --> 00:10:35,169
I only have 1 mole of oxygen.

212
00:10:35,169 --> 00:10:40,149
So instead of 5O2, I have to write 1O2.

213
00:10:40,149 --> 00:10:42,904
Let me make sure that's not of 10.

214
00:10:42,904 --> 00:10:46,559
Let me do it in a different color.

215
00:10:46,559 --> 00:10:49,119
I only have 1O2 instead of 5O2.

216
00:10:49,120 --> 00:10:53,039
So how many ammonia are going to react with that?

217
00:10:53,039 --> 00:10:54,759
Well the ratio is 4:5.

218
00:10:54,759 --> 00:10:57,889
So I'm going to have 0.8 ammonia.

219
00:10:57,889 --> 00:11:01,049
4 is to 5 as 0.8 is to 1.

220
00:11:01,049 --> 00:11:05,240
And so essentially, if I take this whole equation up here

221
00:11:05,240 --> 00:11:07,740
and divide it by 5, I'll get what's

222
00:11:07,740 --> 00:11:09,230
actually going to happen.

223
00:11:09,230 --> 00:11:13,639
So this divided by 5 is 0.8, nitric

224
00:11:13,639 --> 00:11:16,429
oxide or nitrogen monoxide.

225
00:11:16,429 --> 00:11:21,079
Plus 6/5 moles of H2O.

226
00:11:21,080 --> 00:11:24,960
And so the original question in the beginning is, how many

227
00:11:24,960 --> 00:11:26,870
grams of nitric oxide are we going to

228
00:11:26,870 --> 00:11:30,399
produce, or nitrogen monoxide?

229
00:11:30,399 --> 00:11:36,889
So we have 1 mole of oxygen, 0.8 moles of ammonia, and

230
00:11:36,889 --> 00:11:45,500
we're going to produce 0.8 moles of nitrogen monoxide.

231
00:11:45,500 --> 00:11:47,000
Because we only have one oxygen.

232
00:11:47,000 --> 00:11:55,009
So 0.8 moles of nitric oxide or nitrogen monoxide.

233
00:11:55,009 --> 00:11:57,054
So what's the atomic mass, the molecular

234
00:11:57,054 --> 00:12:00,250
mass, of nitrogen monoxide?

235
00:12:00,250 --> 00:12:05,039
Nitrogen has 14 atomic mass units.

236
00:12:05,039 --> 00:12:08,879
Oxygen is 16.

237
00:12:08,879 --> 00:12:09,689
We've done that before.

238
00:12:09,690 --> 00:12:13,260
But you can confirm, your oxygen is 16, nitrogen is 14.

239
00:12:13,259 --> 00:12:26,460
So one molecule of nitrogen monoxide is equal to 30 atomic

240
00:12:26,460 --> 00:12:28,150
mass units.

241
00:12:28,149 --> 00:12:37,360
1 mole or 6.02 times 10 to the 23 molecules of nitrogen

242
00:12:37,360 --> 00:12:42,284
monoxide, therefore, will be 30 grams. And how many moles

243
00:12:42,284 --> 00:12:44,069
are we producing in this reaction?

244
00:12:44,070 --> 00:12:47,100
Because oxygen was the limiting reagent, we only had

245
00:12:47,100 --> 00:12:49,250
1 mole of oxygen here.

246
00:12:49,250 --> 00:12:51,269
Because of that, we can only produce 0.8

247
00:12:51,269 --> 00:12:53,710
moles of nitrogen monoxide.

248
00:12:53,710 --> 00:12:58,190
So 0.8 moles of nitrogen monoxide, 1 mole is 30 grams.

249
00:12:58,190 --> 00:13:07,050
So 0.8 moles of NO is going to be equal to 0.8 times 30,

250
00:13:07,049 --> 00:13:16,139
which is equal to 24 grams. So we're going to be able to

251
00:13:16,139 --> 00:13:19,240
produce 24 grams of nitrogen monoxide.

252
00:13:19,240 --> 00:13:25,720
And so you might ask a question, we're only using 0.8

253
00:13:25,720 --> 00:13:26,950
moles of ammonia.

254
00:13:26,950 --> 00:13:33,210
We were given, in the original problem, 2 moles of ammonia,

255
00:13:33,210 --> 00:13:35,790
so what happens with all the leftover ammonia?

256
00:13:35,789 --> 00:13:37,740
Assuming we mix it really good, we're going to literally

257
00:13:37,740 --> 00:13:41,250
end up with 1.2 moles of ammonia just doing

258
00:13:41,250 --> 00:13:43,370
nothing at the end.

259
00:13:43,370 --> 00:13:47,570
So we're going to have 24 grams of nitric oxide.

260
00:13:47,570 --> 00:13:55,370
And then we used 0.8 moles of ammonia.

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And we're going to have left 1.2 moles of ammonia.

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And if you want, you can figure out how many of the

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original grams of ammonia that is.

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You just figure out how many grams 1 mole of ammonia is,

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it's 17 grams. And then just multiply that times 1.2.

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Anyway, I hope you found that interesting.

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If you found this a little bit confusing, just watch the

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video again and pause it and try to solve it yourself.

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But I think it should start making a little bit of

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intuitive sense.

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I think the hard part is just the conversions between moles

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and grams and getting that part right.

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And then making sure that you understand the ratios.

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You understand this is a 4:5 ratio, there's always going to

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be less ammonia than oxygen.

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And if I only have 1 mole of oxygen, I'm going to have to

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have less than 1 mole of ammonia, even

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though I was given 2.

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So oxygen, in this example, was the rate limiting factor.

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Let's say I had 10 moles of oxygen and 2 moles of ammonia.

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In that case, ammonia would have been the

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rate limiting factor.

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Because then I would have had more than enough oxygen.

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Because for 2 moles of ammonia, I only need

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2.5 moles of oxygen.

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So 10 moles would have been overkill.

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Anyway, see you in the next video.

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