4.08 Combustion Reactions

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Del curso dictado por University of Kentucky

Chemistry

506 calificaciones

University of Kentucky

Chemistry

506 calificaciones

This course is designed to cover subjects in advanced high school chemistry courses, correlating to the standard topics as established by the American Chemical Society. This course is a precursor to the Advanced Chemistry Coursera course. Areas that are covered include atomic structure, periodic trends, compounds, reactions and stoichiometry, bonding, and thermochemistry.

De la lección

Week 4

We will explore how compounds react with one another to form new substances and then write balanced chemical equations to represent what is happening in a reaction. We will explore several different types of reactions including precipitation, acid-base, oxidation-reduction, and combustion reaction.

4.01 Writing Balanced Chemical Equations 12:10

4.01a Balance the following equation 1 2:57

4.01b Balance the following equation 2 3:32

4.02 Aqueous solutions 12:52

4.03 Solubility of ionic compounds 6:23

4.04 Precipitation Reactions6:34

4.05 Molecular, Ionic and Net ionic Equations 6:14

4.05a Net ionic equation 5:46

4.06 Acid-base reactions 15:16

4.07 Oxidation-reduction reactions 16:49

4.07a Oxidation Numbers 2:25

4.07b Oxidized and Reduced species identification 2:37

4.08 Combustion Reactions 5:35

4.08a Balance the following equation 3 4:12

Conoce a los instructores

Dr. Allison Soult
Lecturer
Chemistry
Dr. Kim Woodrum
Sr. Lecturer
Chemistry

In this module, we're going to look at Combustion Reactions.

By the end of this module, you should be able to write and

balance the equation for the combustion of a hydrocarbon compound.

Just a reminder of what a hydrocarbon is,

this is a type of organic compound, it contains hydrogen and carbon.

We see that we can have branched chains, we can have straight chained compounds,

ring structures, single, double or triple bonds are all possible,

lots of different ways to attach these atoms to one another.

And we're going to worry about what the combustion looks like for

these substances.

So, when we have a combustion reaction,

a hydrocarbon reacts with oxygen to produce CO2 and water.

Now the actual balancing the equation,

we'll go back to what we learned before about balancing.

We know that combustion always means adding oxygen and forming CO2 and water.

So, lets look at an example of how we would write and

balance an equation, for the combustion of C3H8, which is also known as propane.

So, the first thing we want to do is we want to write our formula for

our reactant, C3H8.

We know that because it's a combustion reaction, it's going to react with oxygen,

and we're going to produce CO2 and water.

And whenever we're dealing with combustions of a hydrocarbon,

this is always the basic reactant and

products that we're going to add to our hydrocarbon.

Now, we have to look at is to balance the equation.

So, just as I did before,

I'm going to make a list of the elements, of each element in our compounds and

our reactions, and I'm going to tally up the number that I have.

So, here, I have 3 carbons, 8 hydrogens, and 2 oxygen atoms, on the reactant side.

On the product side, I have 1 carbon, I have 2 hydrogens, and

I have 3 oxygens, notice that I've got oxygens in two different places.

So, I have a 3 there for my oxygens.

I'm going to start with my carbon because I have 3 carbons on the left and

only 1 on the right.

So, I need to add a coefficient of 3 in front of C02.

When I do that, that changes to having 3 carbons but

it also changes the number of oxygen atoms.

So, now I have 6 plus 1 or seven oxygen atoms.

Next, I'm going to look at my the hydrogen atoms to balance those, and

I'm going to do that by adding a coefficient in front of water.

So, I have eight hydrogens on the left and only two on the right.

So, I'm going to add a coefficient of 4 in front of water.

When I do that,

that gives me 8 of my hydrogens, but it also changes the number of oxygens.

So I have 6 from my CO2,

and I have 4 from the water, so that makes a total of 10 oxygens.

And notice how each time I add a coefficient, I update my tally here, so

that I can keep track of what I have of each element.

Now I see that my carbon is balanced, my hydrogens are balanced.

The only thing that's unbalanced is my oxygens.

I have ten oxygen atoms on the right.

I need ten oxygen atoms on the left, since I have O2,

I can add a five as a coefficient, so now I have 10 oxygen atoms on the left.

Before I do anything else, I'm going to go back and double check, three carbons,

three carbons, eight hydrogens, eight hydrogens, ten oxygens, and

then six plus four for a total of ten oxygens.

So whenever you balance and equation, always make sure that you go back and

check at the end to make sure everything is truly balanced.

So, let's look at an example where you determine the coefficient for O2,

in the balanced equation for the combustion of C2H4.

Remember that when I have a combustion, I'm always adding oxygen and producing CO2

and water, it's no different, because I have a compound here with a double bond.

So, I have my C2H4,

plus oxygen yields C02 plus H20.

Just as I did before, I'm going to make a list of my elements,

which frequently in combustion reactions is simply carbon, hydrogen and oxygen.

I have 2 carbons on the left, 4 hydrogens and 2 oxygens on the left.

On the right, I have 1 carbon, I have 2 hydrogens and 3 oxygens.

So, I see that none of my elements are balanced.

I'm going to start with the carbon.

I'm going to add a coefficient of 2 there in front of C02.

That changes the number of carbons from one to 2, but it also

changes the number of oxygens, I had four oxygens plus the one more makes 5 oxygens.

I now need to balance the number of hydrogens, so

I'm going to add a 2 infront of the water molecule, so

that now I have 4 hydrogens, but that also changes the number of oxygens.

So, I have four oxygens here and two more there for a total of 6.

Now, I see that my carbons and

hydrogens are both balanced, the only thing that's unbalanced is the oxygens.

And so, if I have 2 oxygen on the left and 6 on the right, I simply need to add

a 3 as a coefficient on the left side, and that gives me 6 oxygen atoms on the left.

Just as I've always done with balanced equations,

I want to go back at the end and check everything.

I see 2 carbons on the left, 2 carbons on the right, 4 hydrogens on the left,

4 hydrogens on the right, 6 oxygens on the left, and 4 plus 2 oxygens on the right.

So now I know, I have a balanced chemical equation for the combustion of C2H4.

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