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Course: Chemistry library > Unit 3
Lesson 1: Balancing chemical equations- Chemical reactions introduction
- 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 1
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Balancing chemical equation with substitution
Using the substitution method to balance a reaction where a reactant and product have a polyatomic ion in common.
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- Wouldn't it be easier to simply recognize that there are 3 sulphur atoms on the right, so you have to have a 3 coefficient in front of the H2S04 on the left? Then, that gives you 6 hydrogen atoms on the left, so you need to have a 3 coefficient in front of the H20 on the right..."and we are done".(10 votes)
- That basically is what Sal did. He just did it much slower, and very step-by-step for anyone who doesn't have the intuition yet.(38 votes)
- Erm, sorry... Valencies are what? Part of the electrons, right?(0 votes)
- The valence electrons make up the outer shell of the atom. this is what causes chemical reactions to occur.(28 votes)
- I thought Fe represents iron why does sal cal it ferric(3 votes)
- Fe is indeed the chemical symbol for iron. There iron here is bonded to oxygen though in an iron oxide whose common name is ferric oxide which is what Sal referred to it as. So Fe is iron, but Fe2O3 is ferric oxide. Ferric oxide is the common name meaning it has some sort of historical significance as to why its named as such, but it also has a more modern systematic name called iron(III) oxide. This time the iron is simply named iron, the roman numeral represent the 3+ charge on the iron atom, and the oxide means it's bonded to oxygens.
Hope that helps.(9 votes)
- Is there a video explaining how you get the products from the reactants? I get the balancing part, but I don't understand how you got Iron(III) sulfate and water from the iron oxide and sulfuric acid.(5 votes)
- So, this is 5 years late, but better late than never in case other people had this question.
My understanding is that this is a special kind of reaction called a neutralization reaction. When you have an acid and a base reacting together (eg. sulfuric acid + iron (iii) oxide), it has been observed in the lab that you will get a "salt" (iron (iii) sulfate) and water.
if you're not sure why iron (iii) oxide is a base, or really why all metal oxides are basic, have no worries. that's stuff you will learn later in Gen Chem II.
edit: a google search tells me that Fe2O3 is capable of being amphoteric. in other words, it can act either as an acid or base depending on what it reacts with. in this context, H2SO4 is a strong acid/has a bigger Ka, so it makes Fe2O3 act basic in this case.(6 votes)
- Can you make the chemical equation from words eg. Sodium oxide + sodium nitrate(2 votes)
- Of course, but it's standard practice to use the chemical symbols instead: Na₂O + NaN0₃(8 votes)
- sir, seemingly unrelated to this video, but how can we know the valency of elements that we do not know? At school, only the first 30 elements are taught. I found it slightly difficult to answer the first quiz in this course.(3 votes)
- I mean, you just have to learn the others too. In the same way you previously didn’t know the valency of the first 30 elements, you had to learn those.(4 votes)
- why was the original equation multiplied by 2? why not another number? why 2?(3 votes)
- We don't like the idea of half a molecule in our equations, because it could reduced all the way down to saying how many molecules react. Half a molecule of water doesn't make too much sense for example.(3 votes)
- What happens to the 4 in front of SO? Does it get substituted with the SO?(2 votes)
- The 4 represents 4 oxygen so it's like in math when you multiply both by the number outside. Since Sulfer had a subscript of 1 it becomes an S3 and since there was 4 oxygen it becomes O12.(1 vote)
- Wait Why multiply by 2? would we always multiply by 2?(2 votes)
- You would only multiply by two if you have a coefficient that isn't a whole number, since typically the coefficients should be whole numbers.
Since the coefficient in question ends with.5
, we can multiply the.5
by two in order to turn the.5
into a1
, which is a whole number. In order not to completely change the value, though, you have to multiply all the coefficients by two.
For reference, see this video (starting at3:33): https://www.khanacademy.org/science/chemistry/chemical-reactions-stoichiome/balancing-chemical-equations/v/balancing-another-combustion-reaction
Hope this helps!(3 votes)
- Hi guys, I just wanted to ask, for these types of chemical equations, like ones with brackets, is substitution the only way of solving it? Are there any other ways and if there are can you show me? Thanks :)(2 votes)
- Well the way Sal did it is the way to balance chemical reactions with the least math needed. I'm assuming when you say the ones with brackets you're referring to the SO4, the sulfate. A good tip I give to people balancing these is to focus on the 'most complicated' chemical first and balance the more rare elements before the more common ones. What I mean by this is focus on the iron sulfate, the Fe2(SO4)3, and balance everything other than that. The sulfate is known as a polyatomic ion and stays in tact when going from reactants to products so you can almost think of like it is its own element. Meaning if you have 3 on the products side, you also need 3 on the reactants side which is why you would put a 3 in front of the sulfuric acid. The irons are already balanced because there are two on both sides of the reactions so we can move onto the more common elements of hydrogen and oxygen. For the hydrogens, since we added a 3 in front of the sulfuric acid to balance the sulfates, we also increased the number of hydrogens on the left to 6. Which means we need 6 hydrogens on the right, hence we need 3 waters to give 6 hydrogens. This also automatically balanced the oxygen giving us 3 on either side. This will often happen that the more common elements get balanced automatically in trying to balance the more rare elements. So a happy byproduct. This kind of thinking works for other reactions with brackets too.
Now if you are interested and are decent in math, there is a way to balance reactions using matrices. I won't go into the entire process because it is quite long, but I will link a very good video explaining the process. https://www.youtube.com/watch?v=Pdtdeqod4Aw (make sure you use the whole url) Hope this helps.(3 votes)
Video transcript
- [Voiceover] All right,
let's see what's going on in this chemical reaction. On the reactant side, we have an iron oxide. This is ferric oxide right over here, reacting with sulfuric acid, and it's producing, this is ferric sulfate and water. We want to balance this chemical equation. I encourage you to pause this
video and try to balance this. I'm assuming you've had a go at it, and you might have been able
to successfully balance it. If you didn't, if you weren't able to successfully balance it, one theory of why you
weren't able to is because this sulfate group made
things really confusing. You have four oxygens, three oxygens here, you have seven on this side, then you have four oxygens
in the sulfate group here, but you have three sulfate groups. This is 12 oxygens here, and then you have another oxygen here. This seems really, really,
really, really confusing, especially with all this
sulfate group business. The key here is to appreciate
that the sulfate group is kind of staying together. You can kind of treat it ... Instead of just saying,
"Hey, let's try to balance "all of the oxygens," you can say, "Let's balance the oxygens "that are outside of the
sulfate groups separately, "and let's balance the
sulfate groups separately." To help our brains grapple with that, I'm going to rewrite
this chemical equation with a substitution. I'm going to say, let's say that x is equal to a sulfate, a sulfate group. Let me rewrite all of this business. I have ... It's nice to have, I guess
this is close to a rust color, which seems appropriate. Let's say I have some ferric oxide. I'm just rewriting what I
have above right over here. Ferric oxide plus some sulfuric acid. But instead of writing H2, and then writing the sulfate group, I'm going to write H2 and then x. H2 and then x is going to yield ... Is going to yield ferric sulfate. Ferric sulfate has three sulfate groups. So x is a sulfate group. It's going to have three of them. Ferric sulfate plus molecular, plus molecular water. Now, even though x
represents an entire group, let's treat it like an
element and just ... make sure we have the same number of x's, or the same number of
sulfate groups on both sides. Let's balance this chemical equation. Let's start with the iron. Over here, I have two irons. And over here, on the right-hand
side, I have two irons. It doesn't seem like I have
to tweak the irons at all. Now let's move on to the oxygens. I have three oxygens here, on the left-hand side. On the right-hand side,
I only have one oxygen. But I can change that by saying, "Let's have three water molecules." Then this is going to be three, right over here. Now let's focus on the hydrogens. Focus on the hydrogens, I
have two hydrogens here, and I have six hydrogens right over here. If I have six hydrogens right over there, how do I get six hydrogens here? Well, I'll have three
molecules of sulfuric acid. Each of them have two hydrogen atoms. Now I have six hydrogens. My hydrogens, my irons, and my oxygens that are not part of the sulfate group are all balanced. Now let's see if we can balance the sulfate groups. On the left-hand side, I
have three sulfate groups. Let me do that in that magenta color. I have three sulfate groups. On the right-hand side, I also
have three sulfate groups. I'm all balanced. And if we want to un-substitute, we just go back up here. Okay, we didn't change the coefficient on this molecule, on the ferric oxide. We did change the coefficient
on the sulfuric acid. We say we have, for every
molecule of ferric oxide, we have three molecules of sulfuric acid. We didn't change this. They're going to yield one for every one
molecule of ferric oxide, and three molecules of sulfuric acid. It's going to yield, the product, are going to be one
molecule of ferric sulfate and three molecules of water. And we are done.