Balancing chemical equations
Balancing chemical equation with substitution
- [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.