4.01b Balance the following equation 2

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来自 University of Kentucky 的课程

Chemistry

488 个评分

University of Kentucky

Chemistry

488 个评分

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.

从本节课中

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

与讲师见面

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

So balance the following equation.

We have calcium, oxygen, and phosphorous involved in this equation.

And so, whenever I want to balance a equation,

the first thing I'm actually going to do is make a list of

all the elements present.

So that I can then tally how many I have on the reactant side and

the product side to aid me in the balancing process.

So I see on the left side I have 1 calcium, I have 11 oxygens,

because I have 1 oxygen here and 10 there for a total of 11.

And I have 4 phosphorous atoms.

On the right side of the equation I have 3 calciums,

I have 2 phosphorus, because I have 2 times 1 for 2.

And for oxygen I have 8 because I have 2 phosphates, and

each phosphate has 4 oxygens for a total of 8.

Now, in order to balance this, I have to make sure that

everything has the same number of atoms on the, both sides of the equation.

One thing I notice is that oxygen is present in two of my reactants.

And so I'm actually going to save that one for last in hopes that

balancing the calcium and the phosphorus will actually balance out the oxygen.

So the first thing I'm going to do is add in my coefficient for

calcium because I want to balance that.

So I need a three in front of the calcium oxide,

when I do that changes the calcium from one on the left to three on the left.

It also changes my oxygens.

Now I have 3 oxygens and 10 oxygens, for a total of 13 oxygens.

So now my calcium is balanced, my oxygen is still not balanced, and

my phosphorus is not balanced.

But I see that I have 4 phosphorus on the left and

2 phosphorous on the right so I'm going to have to add in the coefficient.

I'm going to add in a 2 on the right side.

Now, I have 6 calciums, because I have 2 times 3.

Then I look and say, I've got, well, 2 phosphorous in 1 unit of calcium

phosphate, but I have 2 units so that's going to give me 4 phosphorus atoms.

And it's also going to give 16 oxygen atoms because I

have 2 times 4 is 8 times 2 is 16.

So, in balancing the phosphorus, I actually got the calcium out of balance.

So now I need to go back and change that coefficient to a 6.

And when I do, now I have 6 calcium, and I can recount my oxygens.

I have 6 oxygens of the calcium oxide, and

10 oxygens in the P4O10, which gives me a total of 16.

So it appears that everything is balanced, but

I'm going to go back and check everything just to make sure.

So on the left, I have 6 calciums, on the right I have 2 times 3 so 6 calciums.

On the left I have 6 plus 10 oxygens, and

on the right I have 2 times 4 is 8 times 2 is 16 oxygens.

And then I have 4 phosphorous and 2 times 1 times 2 is 4 phosphorous on the right.

So it appears that,

with these coefficients, I have a balanced chemical equation.

These are the lowest set of coefficients I can have because my coefficients all

can't be divided by any number and still get whole integers for the coefficients.

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