- Balancing chemical equations 1
- Balancing chemical equations 2
- Balancing chemical equations
- Balancing more complex chemical equations
- Visually understanding balancing chemical equations
- Balancing another combustion reaction
- Balancing chemical equation with substitution
- Balancing Chemical Equations Intuition
Example of balancing the combustion reaction of ethylene, C₂H₄. Some tips on how to balance more complicated reactions.
Want to join the conversation?
- When referring to C2H4, Sal kept saying that it was ethylene. Would the name of that molecule not just be "Ethene". We are in the middle of studying Organic Chemistry in school and I am under the impression that is how it is pronounced. Any help with this?(48 votes)
- Good question.
Actually ethene is an IUPAC name and ethylene is a common name . It was used by scientists earlier but now they have switched to ethene.
*Hope it helps :)*(40 votes)
- Why must C2H4+O2 makes CO2 and H2O, why not H2O3 and C?(10 votes)
- That is because trioxidane (H₂O₃) is exceedingly unstable and reactive, especially in the presence of water. Even if trioxidane were formed, it would decompose into water and O almost immediately. Then, because single atoms of O are one of the most reactive known substances, the O would react with C or O₂ to form either ozone, CO or CO₂.
However, if there is not enough oxygen available, the combustion of ethane will also produce CO (g) and C(s) instead of just CO₂ and water.(29 votes)
- to balance the equation don't you just ad a 2 at h2o and a 2 at the 02(13 votes)
- Depends on the formulas but if you had Fe(NO3)3+H2O that won't work u have to break it down element by element and remember their charges(8 votes)
- When you multiplied 2 times the hydrogen atoms, why would you in turn multiply 2 times the oxygen atoms? Are you multiplying two times the entire subset? Or do you not have to multiply the whole equation by two? Also, is this equation equal to start? Or is this equation technically not equal. Because if it is equal, shouldn't we be doing the reverse, and doing it to both sides? Or does that not work in chemistry?(11 votes)
- When balancing chemical equations, it is not necassary to perform operation on the whole equation like you would do in certain math fields e.g. Matrices. Therefore, it would be a sheer concidence that you would be balancing by 2. The equal of atoms in the equation is not equal and that would be the reason you would have to balance it. Hope that helps.(4 votes)
- I'm a little confused one of my Chemistry problems. Any help would be greatly appreciated.
Give a balanced chemical equation for the reaction between phosphoric acid (H₃PO₄) and CsOH.
The balanced chemical equation:
H₃PO₄ + 3CsOH → Cs₃PO₄ + 3H₂O
I understand it for the most part, but why does Cs have a subscript of 3 on the product's side? If someone could explain this to me, it would be awesome. Thank you!(6 votes)
- It's helpful to understand why H₃PO₄ has this formula. This is because it is made up of H⁺ and the phosphate polyatomic ion PO₄³⁻. In order to balance out these charges, you need three hydrogens. This will result in the charges being 3+ and 3- which when added give 0 (no charge overall).
In this reaction, the caesium displaces the hydrogen as you can see in the balanced equation H₃PO₄ + 3CsOH → Cs₃PO₄ + 3H₂O. Caesium is in group 1 as well so the same principle applies as above. You need three Cs atoms (total charge 3+) bonding with the phosphate ion (charge 3-) so that the charges balance out to zero.
Hope that helps!(8 votes)
- At0:19the video says there is a correction from ethylene to ethane, but isnt C2H4 ethylene ? I mean Sal was correct right ?(4 votes)
- Isn't it actually ethene because of the double bond? Ethane would only have a single bond.
Either way, some of the other commenters are correct; there may be different names for a certain structure but they're all the same.(2 votes)
- I understand why it is necessary to balance chemical reaction equations as it needs to represent reality by being consistent with the Conversion of Matter. But why would anyone ever right an unbalanced chemical reaction equation in the first place? Where do these unbalanced equations come from? I would guess it comes from real world chemistry applications maybe? I'm not sure, its not that obvious to me where these unbalanced equations come from.(3 votes)
- When you react hydrogen gas (H2) and oxygen gas (O2) together you produce water, H2O
We can turn this into an equation:
H2 + O2 -> H2O
But this equation isn’t balanced, yet it is correctly describing the chemical reaction that takes place.(2 votes)
- Is there a certain order I should try balancing certain elements?
E.g. should I balance H first instead of O as a general rule?(2 votes)
- You can actually balance them in any order. Choose the order which does not confuse you.
Hope this helps :)(3 votes)
- When balancing equations is there always a solution to solving these types of problems and if so are they always "whole numbers"?(2 votes)
- There is always a solution if the reaction is correctly written out in the form of its elements.
They don't need to be whole numbers: the key is to have a ratio. So if the ratio is 2.5 : 1 then the numbers are 5 and 2.(3 votes)
- At4:02, he says that there are 4 oxygens, 2 O2s. Then, at4:09, he says that there are 2 oxygens, despite that there are 2 O2s. ? This makes no sense...(1 vote)
- Both things are the same. 2 oxygens means that there are 2 oxygen molecules, while 2 O2s means the same. The difference is how Sal said it. the O2 is the subscript and means that there are two oxygen atoms in one molecule of oxygen. In this case there are two molecules of oxygen, which is why he said 2 oxygens. Hope This Helps! Please add upvote.(4 votes)
- [Voiceover] Let's now see if we can balance a chemical equation with slightly more complex molecules. So, here we have a chemical equation, describing a chemical reaction. This is actually a combustion reaction. You have some ethylene right over here, in the presence of oxygen, and you need to get a little bit of energy to get this going, but then you're going to have this reaction that's actually going to release energy as well, but we're not accounting for the energy, at least the way we've written it. Right over here, you have some ethylene, and this little g in parentheses, says it's in the gas form or gaseous form, so gaseous ethylene plus some dioxygen molecule, which is the most prevalent form of oxygen molecule that you would find in the atmosphere. And so, that's also in the gas form. Put them together, you end up with some carbon dioxide gas and some liquid water. This is the classic combustion reaction. But now let's think about, how do we balance this thing? Let's make sure we have the same number of each atom on both sides. And when you see something more complicated like this, where, you know, here I have an oxygen and two different molecules over here, and a lot of these molecules have multiple elements in it. It might be very daunting. Where do I start? And this is where the art of balancing chemical equations starts to come into play. The general idea is, Try to balance the... try to balance the molecules that have multiple elements in them first, and leave the... molecules that only have one element in them for last. And the idea there is, is that these are harder. They're going to have all sorts of implications, and then, at the end of the day, you can just set a number here for the number of dioxygens. If you saved, say the ethylene for last, then every time, and you're trying to balance the carbons, you try to change the number of carbons, you're going to change the number of hydrogens, which is going to change the... You're going to have to balance over and over, and you're going to go into this really really really confusing circle. So, the best thing to do, try to balance the complex molecules first, and then save the single element molecules for last. So let's do that. So, let's start with the carbons. So, over here, I have two carbons. Over here, I only have one carbon. I only have one carbon. So, it seems like the best way to balance it is, I should have two molecules of carbon dioxide, and I haven't even thought about the oxygens yet. By putting that two there, that's going to change the number of oxygens I have on the righthand side. But at least it balances my carbons. I now have two carbons on the lefthand side, and I have two carbons on the righthand side. I’m no longer magically destroying a carbon atom, all right. Now, let's move on to the hydrogens, and remember, what I said is, let's wait to do the oxygens last, because we have a molecule that only contains oxygen right over here, so we'll save oxygen for last. So, let's do hydrogen next. So, hydrogen, right over here, we have four hydrogens. And on the righthand side, we have two hydrogens. So, it seems like the easiest thing to do to balance the hydrogens is to have two of these water molecules. Now I have four hydrogens here, and I have four hydrogens there. Now, let's do the oxygen. Now, let's do the oxygen. I've balanced the carbons and the hydrogens. And the reason why oxygen's going to be interesting, I can just count the amount of oxygen I now have here, after changing the amount of molecules I have. And then I can adjust this accordingly, because this is only going to affect the number of oxygens that I have on the lefthand side. Right now, on the lefthand side, I have two oxygens, and on the righthand side, let me count this, I have two O two's, really. So, this is going to be four oxygens here, and then I have, each of these water molecules has one oxygen, but I have two water molecules, so this is going to be two oxygens, two oxygens here. So, on the righthand side, I have four plus two oxygens. So, I have six oxygens on the righthand side. I need six oxygens on the lefthand side. I need this number to be six. So, how do I do that? Well, I just need three of these molecules. If I have three molecules, each of them have two oxygens, I'm going to have a total of six oxygens. And just like that, we have balanced this combustion reaction, this chemical equation.