Understanding Combustion Reactions and Examples

In this video, we're going to talk about combustion reactions. In the typical chemistry class, the combustion reactions that you'll encounter usually has to do with some sort of hydrocarbon. And sometimes, oxygen might be present. The hydrocarbon always reacts with oxygen in a combustion reaction. And the side products are carbon dioxide and water, if it's a complete combustion reaction. That is, if you react the hydrocarbon with excess oxygen, when there's plenty of oxygen available. Now if you have a limited supply of oxygen, sometimes you might get other products like carbon monoxide in addition to carbon dioxide. But in a complete combustion, you always get CO2 and water, which is typically what you're going to be dealing with. So let's say if we react methane with excess oxygen. What are the products of the reaction and balance it at the same time? Feel free to pause the video. So for any complete combustion reaction, the products are carbon dioxide and water. Now the best way to balance it is to balance the carbon atoms first. We have one carbon on the left, one carbon on the right. So the carbon atoms are balanced. In the next step, balance the hydrogen atoms. Notice that we have 4 on the left, 2 on the right. So therefore, to make it equal, we need to put a 2 in front of H2O. Finally, balance the number of oxygen atoms. On the right side, we have two oxygen atoms in carbon dioxide, and we have two oxygen atoms in the two water molecules. 2 times 1 is 2. So we have a total of four oxygen atoms. So therefore, we've got to put a 2 in front of O2 to balance it. Now let's work on another example. Let's react propane with oxygen gas. Go ahead and try this problem. Predict the products and balance the reaction. So we know the products is going to be CO2 and H2O. Now, just like before, we're going to balance the carbon atoms first. So notice that we have three carbon atoms on the left and only one on the right. Therefore, we've got to put a 3 in front of CO2. Next, balance the hydrogen atoms. So we have eight hydrogen atoms on the left, two on the right. Eight divided by two is four, so we've got to put a 4 in front of H2O. Now we gotta balance the oxygen atoms. What I like to do is count the number of oxygen atoms on the right side, and then figure out what number I need to put in front of O2. So, I have 4 oxygen atoms in the 4 water molecules, because 4 times 1 is 4. And I have 6 here, because 3 times 2 is 6. So basically, you multiply the coefficient by the subscript to get the number of... of oxygen atoms. So I have a total of 4 plus 6, or 10 oxygen atoms on the right side. 10 divided by 2 is 5. So to balance the reaction, I need to put a 5 in front of O2. And so that's That's it for this particular example. Now let's work on some harder examples. Let's react ethane with oxygen gas. This one might be a little bit more trickier than the last two, but feel free to pause the video and try it yourself. So the products are CO2 and H2O. So that part is not going to change. Now let's begin by balancing the carbon atoms first. Notice that we have two on the left, one on the right. So that means we've got to put a two. front of CO2. Next, let's move on to the hydrogen atoms. We have 6 on the left, 2 on the right, 6 divided by 2 is 3. So let's put a 3 in front of H2O. So now we have a total of 6 hydrogen atoms on both sides. Next let's balance the oxygen atoms. So we have 4 in the 2 CO2 molecules because 2 times 2 is 4. And in the 3 water molecules we have 3. 3 times 1 is 3. So we have a total of 7 oxygen atoms on the right side. So what number do we need to put in front of O2? What number times 2 is 7? The opposite of multiplication is division, so you could divide to figure it out. It's going to be 7... divided by 2. So it's a fraction. The reaction is balanced, but we don't want to leave it like this. We want to get rid of the fractions. So what we're going to do is multiply everything by 2, all the coefficients, not the subscripts. So it's 2 ethane molecules. Now, 2 times 7 over 2, the 2's will cancel, giving you 7. So, this is going to yield 7 oxygen molecules. And then, 2 times 2 is 4, so that's going to be... four CO2 molecules. And if we multiply this number by two, it's going to be six H2O molecules. Now the reaction is balanced. Here's a similar example for you to practice on. Go ahead and react butane with oxygen gas. So take a minute and try this example, because the best way to learn is through practice. And that's how you can master this topic. topic. So the products, as always, will be CO2 and H2O. So let's begin by balancing the carbon atoms. Let's follow the same three steps. Balance the carbon atoms first, then the hydrogen atoms, and then save the oxygen atoms for less because oxygen is a pure element on the left side. If you run into a situation where you get an odd number of oxygen on the right side, go ahead and balance it as a fraction. then simply multiply everything by 2 and it's going to work out. So we have 4 carbon atoms on the left, therefore we need to put a 4 in front of CO2. Now let's move on to the hydrogen atoms. We have 10 on the left, 2 on the right, 10 divided by 2 is 5. So we got to put a 5 in front of H2O. Now let's add up the number of oxygen atoms on the right side. So 4 times 2 is 8. And here we have 5 times 1, which is 5. So we have a total of 13 oxygen atoms on the right side. So what number goes in front of O2? It's going to be 13 divided by 2, or simply 13 over 2. Now the last thing we need to do is multiply everything by 2. So we're going to have 2 butane molecules, C4H10. And 2 times 13 over 2 is just going to be 13. So this is going to be 13 oxygen molecules. And that's going to yield 8 carbon dioxide molecules and 10 water molecules. Here's a different example. So let's say if we have a molecule that contains carbon, hydrogen, and oxygen and we're going to react it with oxygen in a combustion reaction. Go ahead and balance the reaction. So the steps for this one is very similar to the previous examples, with one additional step. Let's balance the carbon atoms first. So we got 2 on the left, which means we need to put a 2 in front of CO2. Next, let's balance the hydrogen atoms. We have a total of 6 hydrogen atoms on the left. 5 plus 1 is 6. And 6 divided by 2 is 3. So we need a 3 in front of H2O. well so that we can have six hydrogen atoms on both sides. Now let's count the oxygen atoms on the right. 2 times 2 is 4 so we have four oxygen atoms and the two CO2 molecules and we have three in the three water molecules. 4 plus 3 is 7. Now we have an odd number but for this example it's not a problem because we already have one oxygen atom in ethanol. And to get to 7, we need 6, because 1 plus 6 is 7, and 6 divided by 2 is 3. So we need a 3 in front of O2. So, notice that we have two carbon atoms on both sides. We have a total of six hydrogen atoms on both sides. 5 plus 1 is 6. And the number of oxygen atoms on both sides is 7. Here we have 4, that's 3. And here, 3 times 2 is 6, plus 1, that's 7. So, everything is balanced. So, this is going to be the last example of this video. So, in this example, we have propanol, and we're going to react it with oxygen. to produce carbon dioxide and water. So let's follow the same examples. We have three carbon atoms on the left, so we're going to put a 3 in front of CO2. Now we have a total of 8 hydrogen atoms. 7 plus 1 is 8, and 8 divided by 2 is 4. put a 4 in front of H2O and then here we have 6 oxygens that's 3 times 2 that's a terrible looking 6 and here we have 4 oxygens in the 4 water molecules now 6 plus 4 is 10 Now we have one oxygen in propanol. So 1 plus 9 is 10. Therefore we need to get nine oxygens in the O2 molecules. So here we're going to get a fraction. 9 divided by 2 is 9 over 2. So what we need to do is multiply everything by 2 to get rid of that fraction. So it's going to be 2 propanol molecules. Plus 2 times 9 over 2, we know it's going to be 2, so 9 oxygen molecules. And that yields 6 CO2 molecules and 8 H2O molecules. So let's double check the work that we have. There are 6 carbon atoms on both sides. Here we have 7 plus 1 times 2. 7 plus 1 is 8, 8 times 2 is 16. So we have 16 hydrogen atoms on both sides. And as for the action items, 6 times 2 is 12, plus 8, that's 20. 9 times 2 is 18, and 1 times 2 is 2. 2 plus 18 equals 12 plus 8. They both equal 20. So everything is balanced in this example.

The hydrocarbon always reacts with oxygen in a combustion reaction. And the side products are carbon dioxide and water, if it's a complete combustion reaction. That is, if you react the hydrocarbon with excess oxygen, when there's plenty of oxygen available. Now if you have a limited supply of oxygen, sometimes you might get other products like carbon monoxide in addition to carbon dioxide. But in a complete combustion, you always get CO2 and water, which is typically what you're going to be dealing with.

So let's say if we react methane with excess oxygen. What are the products of the reaction and balance it at the same time? Feel free to pause the video.

So for any complete combustion reaction, the products are carbon dioxide and water. Now the best way to balance it is to balance the carbon atoms first. We have one carbon on the left, one carbon on the right.

So the carbon atoms are balanced. In the next step, balance the hydrogen atoms. Notice that we have 4 on the left, 2 on the right.

So therefore, to make it equal, we need to put a 2 in front of H2O. Finally, balance the number of oxygen atoms. On the right side, we have two oxygen atoms in carbon dioxide, and we have two oxygen atoms in the two water molecules. 2 times 1 is 2. So we have a total of four oxygen atoms. So therefore, we've got to put a 2 in front of O2 to balance it.

Now let's work on another example. Let's react propane with oxygen gas. Go ahead and try this problem.

Predict the products and balance the reaction. So we know the products is going to be CO2 and H2O. Now, just like before, we're going to balance the carbon atoms first. So notice that we have three carbon atoms on the left and only one on the right.

Therefore, we've got to put a 3 in front of CO2. Next, balance the hydrogen atoms. So we have eight hydrogen atoms on the left, two on the right. Eight divided by two is four, so we've got to put a 4 in front of H2O. Now we gotta balance the oxygen atoms.

What I like to do is count the number of oxygen atoms on the right side, and then figure out what number I need to put in front of O2. So, I have 4 oxygen atoms in the 4 water molecules, because 4 times 1 is 4. And I have 6 here, because 3 times 2 is 6. So basically, you multiply the coefficient by the subscript to get the number of... of oxygen atoms. So I have a total of 4 plus 6, or 10 oxygen atoms on the right side.

10 divided by 2 is 5. So to balance the reaction, I need to put a 5 in front of O2. And so that's That's it for this particular example. Now let's work on some harder examples. Let's react ethane with oxygen gas. This one might be a little bit more trickier than the last two, but feel free to pause the video and try it yourself.

So the products are CO2 and H2O. So that part is not going to change. Now let's begin by balancing the carbon atoms first. Notice that we have two on the left, one on the right.

So that means we've got to put a two. front of CO2. Next, let's move on to the hydrogen atoms. We have 6 on the left, 2 on the right, 6 divided by 2 is 3. So let's put a 3 in front of H2O.

So now we have a total of 6 hydrogen atoms on both sides. Next let's balance the oxygen atoms. So we have 4 in the 2 CO2 molecules because 2 times 2 is 4. And in the 3 water molecules we have 3. 3 times 1 is 3. So we have a total of 7 oxygen atoms on the right side. So what number do we need to put in front of O2? What number times 2 is 7?

The opposite of multiplication is division, so you could divide to figure it out. It's going to be 7... divided by 2. So it's a fraction.

The reaction is balanced, but we don't want to leave it like this. We want to get rid of the fractions. So what we're going to do is multiply everything by 2, all the coefficients, not the subscripts.

So it's 2 ethane molecules. Now, 2 times 7 over 2, the 2's will cancel, giving you 7. So, this is going to yield 7 oxygen molecules. And then, 2 times 2 is 4, so that's going to be... four CO2 molecules.

And if we multiply this number by two, it's going to be six H2O molecules. Now the reaction is balanced. Here's a similar example for you to practice on.

Go ahead and react butane with oxygen gas. So take a minute and try this example, because the best way to learn is through practice. And that's how you can master this topic.

topic. So the products, as always, will be CO2 and H2O. So let's begin by balancing the carbon atoms.

Let's follow the same three steps. Balance the carbon atoms first, then the hydrogen atoms, and then save the oxygen atoms for less because oxygen is a pure element on the left side. If you run into a situation where you get an odd number of oxygen on the right side, go ahead and balance it as a fraction. then simply multiply everything by 2 and it's going to work out.

So we have 4 carbon atoms on the left, therefore we need to put a 4 in front of CO2. Now let's move on to the hydrogen atoms. We have 10 on the left, 2 on the right, 10 divided by 2 is 5. So we got to put a 5 in front of H2O. Now let's add up the number of oxygen atoms on the right side. So 4 times 2 is 8. And here we have 5 times 1, which is 5. So we have a total of 13 oxygen atoms on the right side.

So what number goes in front of O2? It's going to be 13 divided by 2, or simply 13 over 2. Now the last thing we need to do is multiply everything by 2. So we're going to have 2 butane molecules, C4H10. And 2 times 13 over 2 is just going to be 13. So this is going to be 13 oxygen molecules.

And that's going to yield 8 carbon dioxide molecules and 10 water molecules. Here's a different example. So let's say if we have a molecule that contains carbon, hydrogen, and oxygen and we're going to react it with oxygen in a combustion reaction. Go ahead and balance the reaction. So the steps for this one is very similar to the previous examples, with one additional step.

Let's balance the carbon atoms first. So we got 2 on the left, which means we need to put a 2 in front of CO2. Next, let's balance the hydrogen atoms. We have a total of 6 hydrogen atoms on the left. 5 plus 1 is 6. And 6 divided by 2 is 3. So we need a 3 in front of H2O.

well so that we can have six hydrogen atoms on both sides. Now let's count the oxygen atoms on the right. 2 times 2 is 4 so we have four oxygen atoms and the two CO2 molecules and we have three in the three water molecules.

4 plus 3 is 7. Now we have an odd number but for this example it's not a problem because we already have one oxygen atom in ethanol. And to get to 7, we need 6, because 1 plus 6 is 7, and 6 divided by 2 is 3. So we need a 3 in front of O2. So, notice that we have two carbon atoms on both sides. We have a total of six hydrogen atoms on both sides.

5 plus 1 is 6. And the number of oxygen atoms on both sides is 7. Here we have 4, that's 3. And here, 3 times 2 is 6, plus 1, that's 7. So, everything is balanced. So, this is going to be the last example of this video. So, in this example, we have propanol, and we're going to react it with oxygen. to produce carbon dioxide and water.

So let's follow the same examples. We have three carbon atoms on the left, so we're going to put a 3 in front of CO2. Now we have a total of 8 hydrogen atoms. 7 plus 1 is 8, and 8 divided by 2 is 4. put a 4 in front of H2O and then here we have 6 oxygens that's 3 times 2 that's a terrible looking 6 and here we have 4 oxygens in the 4 water molecules now 6 plus 4 is 10 Now we have one oxygen in propanol.

So 1 plus 9 is 10. Therefore we need to get nine oxygens in the O2 molecules. So here we're going to get a fraction. 9 divided by 2 is 9 over 2. So what we need to do is multiply everything by 2 to get rid of that fraction.

So it's going to be 2 propanol molecules. Plus 2 times 9 over 2, we know it's going to be 2, so 9 oxygen molecules. And that yields 6 CO2 molecules and 8 H2O molecules. So let's double check the work that we have. There are 6 carbon atoms on both sides.

Here we have 7 plus 1 times 2. 7 plus 1 is 8, 8 times 2 is 16. So we have 16 hydrogen atoms on both sides. And as for the action items, 6 times 2 is 12, plus 8, that's 20. 9 times 2 is 18, and 1 times 2 is 2. 2 plus 18 equals 12 plus 8. They both equal 20. So everything is balanced in this example.

Transcript for:Understanding Combustion Reactions and Examples

Transcript for:
Understanding Combustion Reactions and Examples