1 00:00:00,000 --> 00:00:00,710 2 00:00:00,710 --> 00:00:04,200 Like the concept of the mole, balancing equations is one of 3 00:00:04,200 --> 00:00:06,700 those ideas that you learn in first-year chemistry class. 4 00:00:06,700 --> 00:00:09,250 It tends to give a lot of students a hard time, even 5 00:00:09,250 --> 00:00:12,429 though it is a fairly straightforward concept. 6 00:00:12,429 --> 00:00:13,969 I think what makes it difficult is that there's a 7 00:00:13,970 --> 00:00:15,350 bit of an art to it. 8 00:00:15,349 --> 00:00:17,789 So before we talk about balancing chemical equations, 9 00:00:17,789 --> 00:00:18,919 what is a chemical equation? 10 00:00:18,920 --> 00:00:20,660 Well here's some examples right here, and I have some 11 00:00:20,660 --> 00:00:22,179 more in the rest of this video. 12 00:00:22,179 --> 00:00:24,750 But it essentially just describes a chemical reaction. 13 00:00:24,750 --> 00:00:25,719 You've got some aluminum. 14 00:00:25,719 --> 00:00:30,599 You have some oxygen gas or a diatomic oxygen molecule. 15 00:00:30,600 --> 00:00:32,859 And then you end up with aluminum oxide. 16 00:00:32,859 --> 00:00:34,850 And you would say, ok, fine, that's an equation. 17 00:00:34,850 --> 00:00:35,829 It looks nice. 18 00:00:35,829 --> 00:00:38,699 I have my reactants, or the things that react. 19 00:00:38,700 --> 00:00:40,240 These are the reactants. 20 00:00:40,240 --> 00:00:42,510 And then I have the products of this reaction. 21 00:00:42,509 --> 00:00:44,089 What's left there to do? 22 00:00:44,090 --> 00:00:45,400 Well you have a problem here. 23 00:00:45,399 --> 00:00:47,769 The way I've written it right now, I have one atom of 24 00:00:47,770 --> 00:00:51,550 aluminum plus two atoms of oxygen, right? 25 00:00:51,549 --> 00:00:53,699 They're bonded to each other, but there's two atoms of 26 00:00:53,700 --> 00:00:54,250 oxygen here. 27 00:00:54,250 --> 00:00:59,269 One molecule of diatomic oxygen, or one molecule of 28 00:00:59,270 --> 00:01:01,540 oxygen, but they have two oxygen atoms here. 29 00:01:01,539 --> 00:01:05,030 And when you add them together, I have two atoms of 30 00:01:05,030 --> 00:01:07,019 aluminum and three atoms of oxygen. 31 00:01:07,019 --> 00:01:09,549 So I have a different number of aluminums on both sides of 32 00:01:09,549 --> 00:01:10,509 this equation. 33 00:01:10,510 --> 00:01:12,469 On this side, I have one aluminum. 34 00:01:12,469 --> 00:01:14,810 On this side, I have two aluminums. And then I have a 35 00:01:14,810 --> 00:01:15,850 different number of oxygens. 36 00:01:15,849 --> 00:01:17,579 On this side, I have two oxygens. 37 00:01:17,579 --> 00:01:19,359 And on that side, I have three oxygens. 38 00:01:19,359 --> 00:01:22,700 So balancing equations is all about fixing that problem, so 39 00:01:22,700 --> 00:01:24,990 that I have the same number of aluminum on both sides of the 40 00:01:24,989 --> 00:01:28,244 equation, and the same number of oxygens on both sides. 41 00:01:28,245 --> 00:01:30,200 So let's try to do that. 42 00:01:30,200 --> 00:01:31,960 I'll do it in orange. 43 00:01:31,959 --> 00:01:35,099 So I said I have one aluminum here and I have 44 00:01:35,099 --> 00:01:36,599 two aluminums there. 45 00:01:36,599 --> 00:01:39,129 So maybe a simple thing is just to put a two out here. 46 00:01:39,129 --> 00:01:42,989 So now I have two aluminums on this side and I have two 47 00:01:42,989 --> 00:01:44,289 aluminums on this side. 48 00:01:44,290 --> 00:01:45,400 The aluminums look happy. 49 00:01:45,400 --> 00:01:47,060 Now let's look at the oxygen. 50 00:01:47,060 --> 00:01:49,879 Here I have two oxygens on the 51 00:01:49,879 --> 00:01:51,079 left-hand side of the equation. 52 00:01:51,079 --> 00:01:52,899 And on the right-hand side of the equation 53 00:01:52,900 --> 00:01:54,280 I have three oxygens. 54 00:01:54,280 --> 00:01:55,560 What can I do here? 55 00:01:55,560 --> 00:01:58,960 Well if I could kind of have half atoms, I could just 56 00:01:58,959 --> 00:02:01,839 multiply this by one and a half. 57 00:02:01,840 --> 00:02:03,909 1.5. 58 00:02:03,909 --> 00:02:06,189 1.5 times 2 is 3. 59 00:02:06,189 --> 00:02:09,599 So now I have three oxygens on both sides of this equation 60 00:02:09,599 --> 00:02:12,180 and I have two aluminums on both sides of the equation. 61 00:02:12,180 --> 00:02:13,210 Am I done? 62 00:02:13,210 --> 00:02:15,939 Well, no, you can't have half an atom, or one and 63 00:02:15,939 --> 00:02:17,129 a half of an atom. 64 00:02:17,129 --> 00:02:18,269 That's not cool. 65 00:02:18,270 --> 00:02:20,480 So what you do is you just multiply this so that you end 66 00:02:20,479 --> 00:02:21,144 up with whole numbers. 67 00:02:21,145 --> 00:02:23,230 So let's just multiply both sides of this 68 00:02:23,229 --> 00:02:26,069 whole equation by two. 69 00:02:26,069 --> 00:02:37,699 And we have four aluminums plus three oxygens yields-- we 70 00:02:37,699 --> 00:02:40,209 multiplied everything by two-- two 71 00:02:40,210 --> 00:02:45,090 molecules of aluminum oxide. 72 00:02:45,090 --> 00:02:47,780 Now, you might have been tempted at some point in this 73 00:02:47,780 --> 00:02:50,949 exercise to say, oh, well why don't I just tweak part of the 74 00:02:50,949 --> 00:02:51,859 aluminum oxide? 75 00:02:51,860 --> 00:02:54,940 Why don't I put a 2/3 in front of this oxygen. 76 00:02:54,939 --> 00:02:56,639 You cannot do that. 77 00:02:56,639 --> 00:02:58,379 The equation is as it is. 78 00:02:58,379 --> 00:03:00,719 The molecule aluminum oxide is aluminum oxide. 79 00:03:00,719 --> 00:03:04,629 You can't change the relative ratios of the aluminum and the 80 00:03:04,629 --> 00:03:07,310 oxygen within the aluminum oxide molecule. 81 00:03:07,310 --> 00:03:11,319 You can just change the number of aluminum oxide molecules as 82 00:03:11,319 --> 00:03:13,459 a whole that you have, in this case, two. 83 00:03:13,460 --> 00:03:14,000 So what did we do? 84 00:03:14,000 --> 00:03:14,800 We looked at the aluminum. 85 00:03:14,800 --> 00:03:17,060 We said, OK, we need two aluminums to have both sides 86 00:03:17,060 --> 00:03:18,659 of that be two. 87 00:03:18,659 --> 00:03:20,549 And then when we looked at oxygen, we said, well, if I 88 00:03:20,550 --> 00:03:23,310 multiply this by one and a half, then that becomes three 89 00:03:23,310 --> 00:03:26,340 oxygens here, because one and a half times two oxygens. 90 00:03:26,340 --> 00:03:27,879 And three oxygens there. 91 00:03:27,879 --> 00:03:30,400 And then all we said is, oh, I can't have a one half there, 92 00:03:30,400 --> 00:03:32,539 so let me multiply both sides by two. 93 00:03:32,539 --> 00:03:37,359 And I ended up with four aluminums plus three oxygen 94 00:03:37,360 --> 00:03:41,980 molecules, or six oxygen atoms, yields two molecules of 95 00:03:41,979 --> 00:03:43,659 aluminum oxide. 96 00:03:43,659 --> 00:03:46,069 Let's see if we can do some more of these. 97 00:03:46,069 --> 00:03:48,269 So here I have methane. 98 00:03:48,270 --> 00:03:50,200 And that g in parentheses, I just wanted to 99 00:03:50,199 --> 00:03:50,929 expose you to that. 100 00:03:50,930 --> 00:03:53,390 That just means it's a gas. 101 00:03:53,389 --> 00:03:58,339 So I have methane gas plus oxygen gas yields carbon 102 00:03:58,340 --> 00:04:00,090 dioxide gas plus water. 103 00:04:00,090 --> 00:04:02,409 That's an l there, so liquid water. 104 00:04:02,409 --> 00:04:04,639 So what can we do here? 105 00:04:04,639 --> 00:04:07,439 So the general thing is, do the complicated molecules 106 00:04:07,439 --> 00:04:11,139 first, and then at the end you can worry about the single 107 00:04:11,139 --> 00:04:13,000 atom molecules. 108 00:04:13,000 --> 00:04:15,180 Because those are very easy to play with. 109 00:04:15,180 --> 00:04:17,139 And the reason why you do that, whenever you change a 110 00:04:17,139 --> 00:04:19,680 number here-- let's say we're trying to engineer how many 111 00:04:19,680 --> 00:04:22,680 carbons we have on both sides of this equation-- if I set a 112 00:04:22,680 --> 00:04:24,680 number here, I'm also changing the number of hydrogens. 113 00:04:24,680 --> 00:04:26,139 Then I'll have to play with the hydrogens there. 114 00:04:26,139 --> 00:04:28,719 And at the end, I'll have some number of oxygens. 115 00:04:28,720 --> 00:04:30,630 And then I have to tweak this number right there. 116 00:04:30,629 --> 00:04:33,769 So let's just start with the carbons. 117 00:04:33,769 --> 00:04:36,049 It seems complicated, but when you go step by step and you 118 00:04:36,050 --> 00:04:38,100 kind of play with things a little bit, it should proceed 119 00:04:38,100 --> 00:04:39,550 fairly smoothly. 120 00:04:39,550 --> 00:04:41,069 So here I have two carbons. 121 00:04:41,069 --> 00:04:43,680 122 00:04:43,680 --> 00:04:45,790 And here I have, on the right-hand 123 00:04:45,790 --> 00:04:47,810 side, only one carbon. 124 00:04:47,810 --> 00:04:49,790 So ideally I'd want two carbons on both 125 00:04:49,790 --> 00:04:50,780 sides of this equation. 126 00:04:50,779 --> 00:04:53,929 So let me put a two out here. 127 00:04:53,930 --> 00:04:55,699 So there you go, my carbons are happy. 128 00:04:55,699 --> 00:04:58,459 I now have two carbons and two carbons. 129 00:04:58,459 --> 00:04:59,849 Now let me move to the hydrogens. 130 00:04:59,850 --> 00:05:02,010 Remember, I wanted to do the oxygens last, because I can 131 00:05:02,009 --> 00:05:05,000 just set this to whatever I want it to be without messing 132 00:05:05,000 --> 00:05:08,069 up any of the other atoms. So on this side of the equation I 133 00:05:08,069 --> 00:05:11,370 have four hydrogen atoms. How many hydrogen atoms do I have 134 00:05:11,370 --> 00:05:12,689 on this side of the equation? 135 00:05:12,689 --> 00:05:16,319 So I have four hydrogen atoms here. 136 00:05:16,319 --> 00:05:18,149 How many do I have on this side? 137 00:05:18,149 --> 00:05:21,069 Well, I have two hydrogen atoms right there. 138 00:05:21,069 --> 00:05:24,099 So I want to have four on both sides, so let me put a two in 139 00:05:24,100 --> 00:05:25,510 front of the water. 140 00:05:25,509 --> 00:05:29,310 So now I have four hydrogen atoms. Cool. 141 00:05:29,310 --> 00:05:30,850 Finally, oxygen. 142 00:05:30,850 --> 00:05:35,800 On the left-hand side I have two oxygen atoms. And on the 143 00:05:35,800 --> 00:05:37,780 right-hand side, what do I have? 144 00:05:37,779 --> 00:05:41,969 I have two times this one oxygen, right here, so right 145 00:05:41,970 --> 00:05:44,290 now that's two, right? 146 00:05:44,290 --> 00:05:44,980 Two waters. 147 00:05:44,980 --> 00:05:49,450 And in each water I have one oxygen atom, but I have two 148 00:05:49,449 --> 00:05:52,170 water molecules, so I have two oxygens. 149 00:05:52,170 --> 00:05:56,270 And then in the carbon dioxide I have two oxygens in each 150 00:05:56,269 --> 00:05:56,844 carbon dioxide. 151 00:05:56,845 --> 00:05:58,950 And I have two carbon dioxides, right? 152 00:05:58,949 --> 00:06:01,289 When I put that magenta two out front. 153 00:06:01,290 --> 00:06:02,560 So how many oxygens do I have? 154 00:06:02,560 --> 00:06:03,420 Two times two. 155 00:06:03,420 --> 00:06:05,330 I have four oxygens. 156 00:06:05,329 --> 00:06:07,819 So on the left-hand side I have two oxygens. 157 00:06:07,819 --> 00:06:11,019 On the right-hand side I have six oxygens, two in the two 158 00:06:11,019 --> 00:06:13,430 molecules of water and four in the two 159 00:06:13,430 --> 00:06:15,769 molecules of carbon dioxide. 160 00:06:15,769 --> 00:06:18,459 So how do I make this two into six? 161 00:06:18,459 --> 00:06:20,409 I want to have six oxygens on this side. 162 00:06:20,410 --> 00:06:21,689 I want to them to have six oxygens on the 163 00:06:21,689 --> 00:06:22,855 left-hand side as well. 164 00:06:22,855 --> 00:06:30,110 Well, if I put a three out here, now I have six oxygens. 165 00:06:30,110 --> 00:06:31,810 And our equation has been balanced. 166 00:06:31,810 --> 00:06:35,209 I have two carbons on this side, two carbons on that 167 00:06:35,209 --> 00:06:38,779 side, four hydrogens on this side, four hydrogens on this 168 00:06:38,779 --> 00:06:42,039 side, six oxygens on this side, and then-- four plus 169 00:06:42,040 --> 00:06:44,810 two-- six oxygens on that side. 170 00:06:44,810 --> 00:06:46,839 Next equation to balance. 171 00:06:46,839 --> 00:06:49,269 This actually becomes quite fun once you get 172 00:06:49,269 --> 00:06:51,319 the knack of it. 173 00:06:51,319 --> 00:06:58,509 So I have ethane plus oxygen gas yielding 174 00:06:58,509 --> 00:07:00,574 carbon dioxide and water. 175 00:07:00,574 --> 00:07:03,229 So this is a combustion process. 176 00:07:03,230 --> 00:07:05,280 Let's look at the carbons first. I 177 00:07:05,279 --> 00:07:06,809 have two carbons here. 178 00:07:06,810 --> 00:07:09,009 I have one carbon there. 179 00:07:09,009 --> 00:07:11,050 So let me put a two here. 180 00:07:11,050 --> 00:07:13,240 So now I have two carbons. 181 00:07:13,240 --> 00:07:14,090 Fair enough. 182 00:07:14,089 --> 00:07:15,899 Remember, I'm going to worry about the oxygen last. This 183 00:07:15,899 --> 00:07:16,909 one's actually not too different 184 00:07:16,910 --> 00:07:18,420 than the last problem. 185 00:07:18,420 --> 00:07:21,129 I have six hydrogens here. 186 00:07:21,129 --> 00:07:23,399 I only have two hydrogens in the water. 187 00:07:23,399 --> 00:07:28,089 So then we have three water molecules. 188 00:07:28,089 --> 00:07:30,469 And now I've balanced out the hydrogens. 189 00:07:30,470 --> 00:07:33,520 I have six hydrogens on both sides of the equation. 190 00:07:33,519 --> 00:07:36,060 And now let's deal with the water. 191 00:07:36,060 --> 00:07:40,329 Here on the right-hand side-- I'll do it in this orange 192 00:07:40,329 --> 00:07:45,589 color-- I have two oxygens in each carbon dioxide molecule. 193 00:07:45,589 --> 00:07:48,779 And I have two molecules, so I have four oxygens. 194 00:07:48,779 --> 00:07:52,229 And I have one oxygen in each water molecule. 195 00:07:52,230 --> 00:07:56,240 And I have three molecules, so I have three oxygens here. 196 00:07:56,240 --> 00:07:57,579 Is that right? 197 00:07:57,579 --> 00:08:00,189 Three oxygens on the right-hand side. 198 00:08:00,189 --> 00:08:02,060 Yep, three water molecules. 199 00:08:02,060 --> 00:08:07,199 And then I have four oxygens in the carbon dioxide. 200 00:08:07,199 --> 00:08:07,860 Right. 201 00:08:07,860 --> 00:08:09,790 So I have seven oxygens. 202 00:08:09,790 --> 00:08:11,590 I have seven oxygens on this side and I only 203 00:08:11,589 --> 00:08:13,129 have two on this side. 204 00:08:13,129 --> 00:08:15,574 So how can I make this two into seven. 205 00:08:15,574 --> 00:08:19,360 Well I could multiply it by three and a half. 206 00:08:19,360 --> 00:08:19,689 Right? 207 00:08:19,689 --> 00:08:21,620 Remember, I just want to have seven on both sides. 208 00:08:21,620 --> 00:08:25,290 If I have three and a half of these diatomic oxygen 209 00:08:25,290 --> 00:08:27,640 molecules-- three and a half times two is seven-- so now I 210 00:08:27,639 --> 00:08:30,829 have seven oxygens on both sides of the equation. 211 00:08:30,829 --> 00:08:32,870 Four plus three and seven here. 212 00:08:32,870 --> 00:08:34,058 Two carbons. 213 00:08:34,058 --> 00:08:35,819 And then six hydrogens. 214 00:08:35,820 --> 00:08:38,429 And I'm almost done, except for the fact that you can't 215 00:08:38,428 --> 00:08:40,558 have fractions of molecules. 216 00:08:40,558 --> 00:08:42,408 So what you do is just multiply both sides of this 217 00:08:42,408 --> 00:08:45,759 equation by two or three, or whatever you need to multiply 218 00:08:45,759 --> 00:08:46,960 it to get rid of the fractions. 219 00:08:46,960 --> 00:08:50,800 So if I multiply everything by two, I end up with two 220 00:08:50,799 --> 00:09:00,579 molecules of ethane plus seven molecules of diatomic oxygen, 221 00:09:00,580 --> 00:09:07,520 seven O2's, yielding two molecules of carbon dioxide-- 222 00:09:07,519 --> 00:09:10,199 oh, sorry, I'm multiplying everything by two, so four 223 00:09:10,200 --> 00:09:18,920 molecules of carbon dioxide, plus six molecules of water. 224 00:09:18,919 --> 00:09:21,599 And just to make sure that all still works, if you want to, 225 00:09:21,600 --> 00:09:22,629 you can check. 226 00:09:22,629 --> 00:09:23,830 How much carbon do we have? 227 00:09:23,830 --> 00:09:25,980 I have four carbons here. 228 00:09:25,980 --> 00:09:28,180 I have four carbons here. 229 00:09:28,179 --> 00:09:29,519 How much hydrogen do I have? 230 00:09:29,519 --> 00:09:31,419 I have two times six. 231 00:09:31,419 --> 00:09:33,629 I have 12 hydrogens here. 232 00:09:33,629 --> 00:09:36,100 I have 12 hydrogens here. 233 00:09:36,100 --> 00:09:37,639 How much oxygen do I have? 234 00:09:37,639 --> 00:09:40,149 235 00:09:40,149 --> 00:09:41,419 Let me do a different color. 236 00:09:41,419 --> 00:09:44,189 I have 14 oxygens here. 237 00:09:44,190 --> 00:09:48,040 And here I have eight oxygens. 238 00:09:48,039 --> 00:09:50,860 And here I have-- six times one-- six oxygens. 239 00:09:50,860 --> 00:09:53,009 So six plus eight is 14. 240 00:09:53,009 --> 00:09:56,000 So my equation has been balanced. 241 00:09:56,000 --> 00:09:58,230 So that one, a lot of people might find that to be a hard 242 00:09:58,230 --> 00:10:00,259 problem, because you have three and a half, and just 243 00:10:00,259 --> 00:10:02,129 going straight to this might seem non-intuitive. 244 00:10:02,129 --> 00:10:05,659 But if you just work from the more complicated molecules and 245 00:10:05,659 --> 00:10:07,980 you go atom by atom, and if you end up with any fractions 246 00:10:07,980 --> 00:10:10,774 you just multiply both sides by some number to get rid of 247 00:10:10,774 --> 00:10:13,029 the fraction, of and then you're done. 248 00:10:13,029 --> 00:10:13,589 All right. 249 00:10:13,590 --> 00:10:15,850 Let's do another one. 250 00:10:15,850 --> 00:10:17,950 So this one looks all hairy. 251 00:10:17,950 --> 00:10:24,190 I have this iron oxide plus sulfuric acid yields all this 252 00:10:24,190 --> 00:10:25,260 hairy stuff. 253 00:10:25,259 --> 00:10:30,000 But the key here to realize is that this group right here, 254 00:10:30,000 --> 00:10:35,169 this sulfate group, stays together. 255 00:10:35,169 --> 00:10:35,479 Right? 256 00:10:35,480 --> 00:10:39,580 You have this SO4 there and you have this SO4 there. 257 00:10:39,580 --> 00:10:42,740 So to really simplify things for our head, you can kind of 258 00:10:42,740 --> 00:10:44,009 treat that like an atom. 259 00:10:44,009 --> 00:10:45,409 So let's make a substitution. 260 00:10:45,409 --> 00:10:48,230 We'll substitute that for x. 261 00:10:48,230 --> 00:10:52,230 So if we rewrite this as-- I'll do it in a vibrant 262 00:10:52,230 --> 00:11:03,389 color-- iron oxide plus H2-- there's only one sulfate 263 00:11:03,389 --> 00:11:07,689 group here-- x. 264 00:11:07,690 --> 00:11:15,070 And then that yields two irons, this 265 00:11:15,070 --> 00:11:15,940 molecule with two irons. 266 00:11:15,940 --> 00:11:18,070 And it has three of these sulfate groups. 267 00:11:18,070 --> 00:11:21,140 268 00:11:21,139 --> 00:11:22,389 And then plus water. 269 00:11:22,389 --> 00:11:26,449 270 00:11:26,450 --> 00:11:28,870 All I did is replace the sulfate with an x. 271 00:11:28,870 --> 00:11:32,039 And now we can treat that x like an atom, and we can just 272 00:11:32,039 --> 00:11:33,829 balance the equation. 273 00:11:33,830 --> 00:11:36,040 Let's see, on the left-hand side, how 274 00:11:36,039 --> 00:11:37,299 many irons do we have? 275 00:11:37,299 --> 00:11:39,000 We have two irons here. 276 00:11:39,000 --> 00:11:40,820 We have two irons there. 277 00:11:40,820 --> 00:11:45,190 So the irons look balanced, at first glance. 278 00:11:45,190 --> 00:11:47,630 Let's deal with the oxygens. 279 00:11:47,629 --> 00:11:52,279 So if we have three oxygens here-- let me do it in a 280 00:11:52,279 --> 00:11:59,659 different color-- and we only have one oxygen here. 281 00:11:59,659 --> 00:12:03,480 Remember, there were some oxygens in this x group, but 282 00:12:03,480 --> 00:12:05,590 the sulfate group stayed together, so we can just treat 283 00:12:05,590 --> 00:12:08,700 those separately. 284 00:12:08,700 --> 00:12:10,820 We want to have three oxygens on the right-hand side, as 285 00:12:10,820 --> 00:12:13,250 well, so let's stick a three here. 286 00:12:13,250 --> 00:12:14,500 Oops. 287 00:12:14,500 --> 00:12:19,710 288 00:12:19,710 --> 00:12:21,040 So let me put that three there. 289 00:12:21,039 --> 00:12:23,959 So now we have three oxygens, as well. 290 00:12:23,960 --> 00:12:28,733 And then finally, let's look at the hydrogens. 291 00:12:28,732 --> 00:12:35,949 292 00:12:35,950 --> 00:12:37,160 Let's see, how many hydrogens do we have here. 293 00:12:37,159 --> 00:12:41,269 We have six hydrogens now, on the right-hand side. 294 00:12:41,269 --> 00:12:42,840 Six hydrogens here. 295 00:12:42,840 --> 00:12:43,725 Three times two. 296 00:12:43,725 --> 00:12:46,769 And we want to have six hydrogens here, so we have to 297 00:12:46,769 --> 00:12:48,909 have three of these molecules. 298 00:12:48,909 --> 00:12:52,199 And then finally, let's look at the sulfate group, that x. 299 00:12:52,200 --> 00:12:53,790 We have three x's here. 300 00:12:53,789 --> 00:12:57,089 301 00:12:57,090 --> 00:13:01,350 And lucky for us, we have three x's over there. 302 00:13:01,350 --> 00:13:03,420 So our equation has been balanced. 303 00:13:03,419 --> 00:13:05,759 And if we want to write it back in terms of the sulfate 304 00:13:05,759 --> 00:13:09,490 terms, we can just un-substitute the x, and we're 305 00:13:09,490 --> 00:13:16,450 left with one molecule of iron oxide plus three molecules of 306 00:13:16,450 --> 00:13:21,540 sulfuric acid, H2SO4 -- I just un-substituted the x with SO4 307 00:13:21,539 --> 00:13:28,569 -- yields one of these molecules, SO4 3. 308 00:13:28,570 --> 00:13:32,270 309 00:13:32,269 --> 00:13:39,539 I just un-substituted the x plus three molecules of water. 310 00:13:39,539 --> 00:13:40,669 Let's do one more. 311 00:13:40,669 --> 00:13:43,799 And then I think we'll be all balanced out. 312 00:13:43,799 --> 00:13:47,750 Carbon dioxide plus hydrogen gas yields methane plus water. 313 00:13:47,750 --> 00:13:51,700 So let's deal with the carbon first. I have one carbon here, 314 00:13:51,700 --> 00:13:52,540 one carbon there. 315 00:13:52,539 --> 00:13:53,939 The carbons look happy. 316 00:13:53,940 --> 00:13:54,890 Let's look at the oxygen. 317 00:13:54,889 --> 00:13:56,919 I have two oxygens here. 318 00:13:56,919 --> 00:13:58,990 I have one oxygen there. 319 00:13:58,990 --> 00:14:02,350 So I want two oxygens here so let me stick a two there. 320 00:14:02,350 --> 00:14:04,300 And then let's deal with the hydrogens last, because this 321 00:14:04,299 --> 00:14:06,649 is the easiest one to play with because it doesn't affect 322 00:14:06,649 --> 00:14:11,029 any of the other atoms. So if I have two here and I have 323 00:14:11,029 --> 00:14:14,059 four, plus four here, right? 324 00:14:14,059 --> 00:14:15,864 I have four hydrogens there. 325 00:14:15,865 --> 00:14:18,159 And I have four hydrogens there. 326 00:14:18,159 --> 00:14:20,789 So I need eight hydrogens on the left-hand side. 327 00:14:20,789 --> 00:14:22,439 So I just put a four there. 328 00:14:22,440 --> 00:14:24,350 We're all done balancing. 329 00:14:24,350 --> 00:14:27,259 Anyway, hopefully you found that useful.