In this video, I explain the theory of section 4.2 when we write down a chemical reaction, we do it schematically. We use the molecular formulas that we learned about in chapter 3. The first step on the way to one of those chemical equations is a word equation. If you write down a word equation, you write down the substances in words, you use their names, for example, sulfur trioxide, reacts with water to form sulfuric acid. Then, when you have the word equation, you translate the names to their molecular.
Formulas, so3, plus h2o gives h2so4. Then the next step is to balance the reaction. So you have to apply the law of conservation of mass. That means every atom that we have before the arrow. We must also have behind the arrow. I see before the arrow 1s and behind the arrow 1s.
In total, I have 4 o's three from here, plus one from there. So before the arrow in total, there are four oxygen atoms behind the arrow. I also have four o's. Finally, the h I have two here and I have two there. So this one is already.
Balanced, but unfortunately, that's, not the case for every reaction. The next example is a reaction of kmno4, which reacts and forms k2o mno2 and o2. This is a so-called decomposition reaction. We start with one reactant.
Now, if we can again in front of the arrow, I have one k behind the arrow. I have two in front of the arrow. I have one MN, manganese MN is manganese and behind the arrow. Furthermore, I also have one. Furthermore, I have four o atoms and behind the arrow.
One plus two, plus two is five. So this one is not in balance. Yet when you balance a reaction, you may only change the coefficients. So the numbers in front of the substances. So how many times you have a certain molecule? You can't change the formula of the molecule.
I can't change this two. For example, this is the molecule that is formed. I can't change that. So in the case of this example, this becomes the final reaction in the case of this reaction it is easiest to start with this one, because we see in our reactants that we have four o atoms.
So no many how so. No matter how many times I have this molecule. The number of o atoms will always be an even number. There are two o's here and two o's there.
So these are also even, so I must do something with this to make sure that I get an even number here and not an odd number as I had in the beginning. So I start by adding a 2 here just to make it. Even then I have 2 times 2 is 4 k. Each molecule contains 2 k atoms. I have two molecules.
So two times two is four k. So I add a four here, but this four also has.Consequences of for the manganese and the oxygen. Now I have four manganese atoms. I only had one here. So I add a 4 there, 4 MN 4mn. Now I look at my oxygen by writing this 4 here I have 4 times 4 is 16 oxygen atoms.
I have 2 from here. So I have 14 left 4 times 2 is 8. So I have 14 that means now I have 6 left.
So those 6 oxygen atoms I have left must then all form the oxygen molecule. So I put a 3 in front of it. And now the reaction is balanced four case before the arrow, four k's behind four manganese. Four manganese, 16 0 16 0.
There are also reactions where there are brackets. And usually in the beginning, this is thought of as difficult because mainly you don't understand where the brackets come from yet. It helps when you balance it. If you see that what's behind between the brackets as one thing here, in this case, we have o h. And here I have water. It doesn't, look the same, but h2o, if I write it like this h with an o-h attached to it, you do see that o-h come back there.
So the o h is in here. I. Have po4 between brackets here. And I also see that group over here in this case, it's not between brackets, because if it would be then the index behind, it would be one. And the agreement was that if the index or coefficient is one, you don't write it down here in this molecule. It is between brackets with an index 2 to show that this group. This po4 appears twice in this molecule.
If you look at it that way it makes balancing it a bit easier. I start with this molecule. I have three calcium here. Only.
One here because there is no index here. So the calcium in this molecule only appears once, so I put a 3 in front of it. Then my calcium is in balance.
I look at my po4. I have two of those groups here, but only one there. So I use this index and turn it into a coefficient on this side. I put a 2 here.
So now I have 3 times 2 is 6, 0 h groups. And I put a 2 here. So I have 2 times 3 is 6 h's. If I look at my water molecule as this as a h with an o h, I now have 6 0 h groups with 6 additional h's. So I can. Form six complete water molecules. And that is exactly what I do.
Now the reaction is in balance, three, calcium, hydroxide, that's. What this molecule is plus three, phosphoric acid forms, one molecule of calcium phosphate and six water molecules. In addition to balancing the reaction there are two more rules that you do have to keep in mind. And that is that you cannot use a fraction as a coefficient, nor can you use multitudes of the coefficients. So you have to use the lowest possible numbers.
So if I. Regard this reaction this is the combustion of propane octane, I'm, sorry, octane, which is the basis for the gasoline that is put in cars. For example, it reacts with oxygen, and it forms carbon dioxide and water. If I start balancing just from left to right, 8, c's, 8 c's. So I put an 8 there, 18 h's.
So I put a 9 there. 9 times 2 is 18. Then on this side, I have 8 times 2 is 16 plus 9 is 25 0 atoms. Then if I put a 12 and a half here, then I have 12 and a half times 2 is 25. So it is in balance. The only problem. Is, I can't have half a molecule you only have whole molecules.
So what you do is you multiply all the coefficients by 2 to get rid of the fraction you can't just multiply this one because then the rest is not in balance anymore. You have to multiply all the coefficients by the same number in this case, 2. And now it's in balance 2 times 8 is 16 16 c 2. X. 18 is 36. 18 times 2 is 36. 25 times 2 50 o.
16 times 2 is 32 plus 18 is 50. Now in this one, I multiplied these coefficients again by two. So the ratio. Is still intact. The only issue is the rule was they have to be as low as possible. So that means that when you have balance, the reaction you always have to check are the coefficients as low as possible, or can I still divide them by two or three or four or whatever other number.
In this case I can divide them all by two and still get whole numbers. If I were to divide this again by two, I'd get this one. And then I have a fraction, so it's not possible to get lower than this. But it is possible. To get lower than that.
So you must always check for that as well. This was the theory of section 4.2.