4.08a Balance the following equation 3

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Del curso dictado por University of Kentucky

Chemistry

545 calificaciones

University of Kentucky

Chemistry

545 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

Write and balance the equation for the combustion of C6H14.

When we have a combustion reaction, what that means is that we're adding oxygen.

And for our hydrocarbons, our only products are going to be CO2 and water.

So, I'm going to start by writing the basics of our reaction.

So, we have C6H14 plus 02

yields C02 plus water.

And because this isn't going to start out balanced, I'm leaving some room for

my coefficients as I go through and balance the equation.

Now, I'm going to create a list of my elements so

that I can basically create a tally.

C, H and O of how many carbons, hydrogens, and

oxygens I have on both sides of the equation.

So, on the left side, I see I have 6 carbons, 14 hydrogens, and 2 oxygens.

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

So, now what I need to do is start putting in coefficients and adjusting my

totals until I get all the coefficients I need to have a balanced chemical equation.

So, I'm going to start by putting a six in front of the CO2,

so that I can balance the carbons.

So, that changes from one carbon to six carbons on the right, but

notice that it also changes the number of oxygens.

Now I have 12 plus one, or 13 oxygens on the right side of the equation.

So, now my carbons are balanced, now I'm going to look at the hydrogens, and

in order to balance the hydrogens, I need a 7 on the right side of the equation.

So, now I have 14 hydrogens, and that also changes the number of oxygens, so,

now I have 12 oxygens plus seven, or 19 oxygens.

Now at this point, both my carbons and my hydrogens are balanced.

It's only the oxygens that are, the not balanced.

And when I look at the left side,

what I notice is that I have oxygen all by itself.

So what I'm going to do, is for a moment, I'm going to pretend.

That the two is not there, and I'm going to say what coefficient would I

need in front of that oxygen in order to get a balanced equation?

And I'm going to say, well if I put a 19 in there.

Then I would have a balanced equation.

But that's assuming that the two is not there.

But I could see that I have the right number of everything.

But I do have the to in there, so

what I have to do is actually change that coefficient to 19 halves, and

return that two into the equation.

So, now I actually do have a balanced equation but I still have

this fractional number in my coefficients which I don't want to leave in there.

So what I can do, is I can multiply everything through by two.

And when I do that I get a 2 here, I get a 19,

in front of the oxygen, I get 12 in

front of CO2 and I get 14 in front of H2O.

Now I can go back and check my count, I see that I have 12 carbons on the left,

I have 28 hydrogens, and I have 38 oxygens.

On the right side of my equation, I have 12 carbons,

I have 24 oxygens in the CO2, plus additional 14 oxygen in the water for

a total of 38 oxygens and I have 28 hydrogens.

Now, I have a balanced chemical equation with integers as coefficients and

I see that that's the lowest possible set of integers because I

cannot divide it by any number and still get whole numbers.

So, it looks a little odd to see values that are this large but

we do see this when we start looking at larger reactants,

particularly when we look at combustion reactions.

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