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Like the concept of the mole,
balancing equations is one of

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those ideas that you learn in
first-year chemistry class.

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It tends to give a lot of
students a hard time, even

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though it is a fairly
straightforward concept.

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I think what makes it difficult
is that there's a

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bit of an art to it.

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So before we talk about
balancing chemical equations,

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what is a chemical equation?

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Well here's some examples right
here, and I have some

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more in the rest
of this video.

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But it essentially just
describes a chemical reaction.

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You've got some aluminum.

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You have some oxygen gas or a
diatomic oxygen molecule.

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And then you end up with
aluminum oxide.

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And you would say, ok, fine,
that's an equation.

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It looks nice.

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I have my reactants, or
the things that react.

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These are the reactants.

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And then I have the products
of this reaction.

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What's left there to do?

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Well you have a problem here.

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The way I've written it right
now, I have one atom of

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aluminum plus two atoms
of oxygen, right?

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They're bonded to each other,
but there's two atoms of

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oxygen here.

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One molecule of diatomic oxygen,
or one molecule of

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oxygen, but they have two
oxygen atoms here.

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And when you add them together,
I have two atoms of

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aluminum and three
atoms of oxygen.

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So I have a different number of
aluminums on both sides of

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this equation.

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On this side, I have
one aluminum.

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On this side, I have two
aluminums. And then I have a

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different number of oxygens.

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On this side, I have
two oxygens.

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And on that side, I have
three oxygens.

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So balancing equations is all
about fixing that problem, so

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that I have the same number of
aluminum on both sides of the

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equation, and the same number
of oxygens on both sides.

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So let's try to do that.

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I'll do it in orange.

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So I said I have one aluminum
here and I have

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two aluminums there.

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So maybe a simple thing is just
to put a two out here.

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So now I have two aluminums on
this side and I have two

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aluminums on this side.

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The aluminums look happy.

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Now let's look at the oxygen.

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Here I have two oxygens on the

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left-hand side of the equation.

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And on the right-hand
side of the equation

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I have three oxygens.

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What can I do here?

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Well if I could kind of have
half atoms, I could just

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multiply this by
one and a half.

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1.5.

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1.5 times 2 is 3.

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So now I have three oxygens on
both sides of this equation

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and I have two aluminums on both
sides of the equation.

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Am I done?

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Well, no, you can't have half
an atom, or one and

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a half of an atom.

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That's not cool.

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So what you do is you just
multiply this so that you end

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up with whole numbers.

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So let's just multiply
both sides of this

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whole equation by two.

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And we have four aluminums plus
three oxygens yields-- we

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multiplied everything
by two-- two

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molecules of aluminum oxide.

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Now, you might have been tempted
at some point in this

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exercise to say, oh, well why
don't I just tweak part of the

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aluminum oxide?

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Why don't I put a 2/3 in
front of this oxygen.

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You cannot do that.

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The equation is as it is.

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The molecule aluminum oxide
is aluminum oxide.

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You can't change the relative
ratios of the aluminum and the

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oxygen within the aluminum
oxide molecule.

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You can just change the number
of aluminum oxide molecules as

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a whole that you have,
in this case, two.

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So what did we do?

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We looked at the aluminum.

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We said, OK, we need two
aluminums to have both sides

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of that be two.

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And then when we looked at
oxygen, we said, well, if I

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multiply this by one and a half,
then that becomes three

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oxygens here, because one and
a half times two oxygens.

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And three oxygens there.

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And then all we said is, oh, I
can't have a one half there,

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so let me multiply both
sides by two.

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And I ended up with four
aluminums plus three oxygen

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molecules, or six oxygen atoms,
yields two molecules of

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aluminum oxide.

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Let's see if we can do
some more of these.

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So here I have methane.

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And that g in parentheses,
I just wanted to

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expose you to that.

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That just means it's a gas.

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So I have methane gas plus
oxygen gas yields carbon

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dioxide gas plus water.

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That's an l there,
so liquid water.

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So what can we do here?

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So the general thing is, do
the complicated molecules

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first, and then at the end you
can worry about the single

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atom molecules.

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Because those are very
easy to play with.

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And the reason why you do that,
whenever you change a

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number here-- let's say we're
trying to engineer how many

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carbons we have on both sides of
this equation-- if I set a

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number here, I'm also changing
the number of hydrogens.

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Then I'll have to play with
the hydrogens there.

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And at the end, I'll have
some number of oxygens.

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And then I have to tweak this
number right there.

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So let's just start
with the carbons.

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It seems complicated, but when
you go step by step and you

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kind of play with things a
little bit, it should proceed

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fairly smoothly.

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So here I have two carbons.

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And here I have, on
the right-hand

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side, only one carbon.

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So ideally I'd want two
carbons on both

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sides of this equation.

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So let me put a two out here.

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So there you go, my
carbons are happy.

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I now have two carbons
and two carbons.

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Now let me move to
the hydrogens.

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Remember, I wanted to do the
oxygens last, because I can

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just set this to whatever I want
it to be without messing

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up any of the other atoms. So on
this side of the equation I

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have four hydrogen atoms. How
many hydrogen atoms do I have

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on this side of the equation?

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So I have four hydrogen
atoms here.

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How many do I have
on this side?

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Well, I have two hydrogen
atoms right there.

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So I want to have four on both
sides, so let me put a two in

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front of the water.

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So now I have four hydrogen
atoms. Cool.

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Finally, oxygen.

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On the left-hand side I have two
oxygen atoms. And on the

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right-hand side,
what do I have?

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I have two times this one
oxygen, right here, so right

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now that's two, right?

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Two waters.

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And in each water I have one
oxygen atom, but I have two

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water molecules, so I
have two oxygens.

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And then in the carbon dioxide
I have two oxygens in each

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carbon dioxide.

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And I have two carbon
dioxides, right?

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When I put that magenta
two out front.

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So how many oxygens do I have?

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Two times two.

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I have four oxygens.

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So on the left-hand side
I have two oxygens.

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On the right-hand side I have
six oxygens, two in the two

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molecules of water and
four in the two

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molecules of carbon dioxide.

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So how do I make this
two into six?

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I want to have six oxygens
on this side.

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I want to them to have
six oxygens on the

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left-hand side as well.

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Well, if I put a three out here,
now I have six oxygens.

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And our equation has
been balanced.

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I have two carbons on this
side, two carbons on that

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side, four hydrogens on this
side, four hydrogens on this

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side, six oxygens on this side,
and then-- four plus

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two-- six oxygens
on that side.

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Next equation to balance.

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This actually becomes quite
fun once you get

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the knack of it.

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So I have ethane plus
oxygen gas yielding

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carbon dioxide and water.

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So this is a combustion
process.

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Let's look at the
carbons first. I

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have two carbons here.

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I have one carbon there.

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So let me put a two here.

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So now I have two carbons.

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Fair enough.

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Remember, I'm going to worry
about the oxygen last. This

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one's actually not
too different

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than the last problem.

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I have six hydrogens here.

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I only have two hydrogens
in the water.

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So then we have three
water molecules.

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And now I've balanced
out the hydrogens.

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I have six hydrogens on both
sides of the equation.

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And now let's deal
with the water.

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Here on the right-hand side--
I'll do it in this orange

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color-- I have two oxygens in
each carbon dioxide molecule.

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And I have two molecules,
so I have four oxygens.

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And I have one oxygen in
each water molecule.

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And I have three molecules, so
I have three oxygens here.

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Is that right?

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Three oxygens on the
right-hand side.

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Yep, three water molecules.

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And then I have four oxygens
in the carbon dioxide.

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Right.

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So I have seven oxygens.

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I have seven oxygens on
this side and I only

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have two on this side.

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So how can I make this
two into seven.

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Well I could multiply it
by three and a half.

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Right?

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Remember, I just want to have
seven on both sides.

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If I have three and a half
of these diatomic oxygen

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molecules-- three and a half
times two is seven-- so now I

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have seven oxygens on both
sides of the equation.

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Four plus three and
seven here.

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Two carbons.

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And then six hydrogens.

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And I'm almost done, except for
the fact that you can't

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have fractions of molecules.

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So what you do is just multiply
both sides of this

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equation by two or three, or
whatever you need to multiply

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it to get rid of
the fractions.

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So if I multiply everything
by two, I end up with two

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molecules of ethane plus seven
molecules of diatomic oxygen,

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seven O2's, yielding two
molecules of carbon dioxide--

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oh, sorry, I'm multiplying
everything by two, so four

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molecules of carbon dioxide,
plus six molecules of water.

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And just to make sure that all
still works, if you want to,

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you can check.

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How much carbon do we have?

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I have four carbons here.

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I have four carbons here.

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How much hydrogen do I have?

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I have two times six.

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I have 12 hydrogens here.

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I have 12 hydrogens here.

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How much oxygen do I have?

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Let me do a different color.

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I have 14 oxygens here.

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And here I have eight oxygens.

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And here I have-- six times
one-- six oxygens.

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So six plus eight is 14.

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So my equation has
been balanced.

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So that one, a lot of people
might find that to be a hard

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problem, because you have three
and a half, and just

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going straight to this might
seem non-intuitive.

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But if you just work from the
more complicated molecules and

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you go atom by atom, and if you
end up with any fractions

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00:10:07,980 --> 00:10:10,775
you just multiply both sides by
some number to get rid of

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the fraction, of and
then you're done.

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All right.

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Let's do another one.

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So this one looks all hairy.

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I have this iron oxide plus
sulfuric acid yields all this

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00:10:24,190 --> 00:10:25,260
hairy stuff.

250
00:10:25,260 --> 00:10:30,000
But the key here to realize is
that this group right here,

251
00:10:30,000 --> 00:10:35,170
this sulfate group,
stays together.

252
00:10:35,170 --> 00:10:35,480
Right?

253
00:10:35,480 --> 00:10:39,580
You have this SO4 there and
you have this SO4 there.

254
00:10:39,580 --> 00:10:42,740
So to really simplify things for
our head, you can kind of

255
00:10:42,740 --> 00:10:44,010
treat that like an atom.

256
00:10:44,010 --> 00:10:45,410
So let's make a substitution.

257
00:10:45,410 --> 00:10:48,230
We'll substitute that for x.

258
00:10:48,230 --> 00:10:52,230
So if we rewrite this as--
I'll do it in a vibrant

259
00:10:52,230 --> 00:11:03,390
color-- iron oxide plus H2--
there's only one sulfate

260
00:11:03,390 --> 00:11:07,690
group here-- x.

261
00:11:07,690 --> 00:11:15,070
And then that yields
two irons, this

262
00:11:15,070 --> 00:11:15,940
molecule with two irons.

263
00:11:15,940 --> 00:11:18,070
And it has three of these
sulfate groups.

264
00:11:21,140 --> 00:11:22,390
And then plus water.

265
00:11:26,450 --> 00:11:28,870
All I did is replace the
sulfate with an x.

266
00:11:28,870 --> 00:11:32,040
And now we can treat that x like
an atom, and we can just

267
00:11:32,040 --> 00:11:33,830
balance the equation.

268
00:11:33,830 --> 00:11:36,040
Let's see, on the left-hand
side, how

269
00:11:36,040 --> 00:11:37,300
many irons do we have?

270
00:11:37,300 --> 00:11:39,000
We have two irons here.

271
00:11:39,000 --> 00:11:40,820
We have two irons there.

272
00:11:40,820 --> 00:11:45,190
So the irons look balanced,
at first glance.

273
00:11:45,190 --> 00:11:47,630
Let's deal with the oxygens.

274
00:11:47,630 --> 00:11:52,280
So if we have three oxygens
here-- let me do it in a

275
00:11:52,280 --> 00:11:59,660
different color-- and we only
have one oxygen here.

276
00:11:59,660 --> 00:12:03,480
Remember, there were some
oxygens in this x group, but

277
00:12:03,480 --> 00:12:05,590
the sulfate group stayed
together, so we can just treat

278
00:12:05,590 --> 00:12:08,700
those separately.

279
00:12:08,700 --> 00:12:10,820
We want to have three oxygens
on the right-hand side, as

280
00:12:10,820 --> 00:12:13,250
well, so let's stick
a three here.

281
00:12:13,250 --> 00:12:14,500
Oops.

282
00:12:19,710 --> 00:12:21,040
So let me put that
three there.

283
00:12:21,040 --> 00:12:23,960
So now we have three
oxygens, as well.

284
00:12:23,960 --> 00:12:28,733
And then finally, let's
look at the hydrogens.

285
00:12:35,950 --> 00:12:37,160
Let's see, how many hydrogens
do we have here.

286
00:12:37,160 --> 00:12:41,270
We have six hydrogens now,
on the right-hand side.

287
00:12:41,270 --> 00:12:42,840
Six hydrogens here.

288
00:12:42,840 --> 00:12:43,725
Three times two.

289
00:12:43,725 --> 00:12:46,770
And we want to have six
hydrogens here, so we have to

290
00:12:46,770 --> 00:12:48,910
have three of these molecules.

291
00:12:48,910 --> 00:12:52,200
And then finally, let's look at
the sulfate group, that x.

292
00:12:52,200 --> 00:12:53,790
We have three x's here.

293
00:12:57,090 --> 00:13:01,350
And lucky for us, we have
three x's over there.

294
00:13:01,350 --> 00:13:03,420
So our equation has
been balanced.

295
00:13:03,420 --> 00:13:05,760
And if we want to write it back
in terms of the sulfate

296
00:13:05,760 --> 00:13:09,490
terms, we can just un-substitute
the x, and we're

297
00:13:09,490 --> 00:13:16,450
left with one molecule of iron
oxide plus three molecules of

298
00:13:16,450 --> 00:13:21,540
sulfuric acid, H2SO4 -- I just
un-substituted the x with SO4

299
00:13:21,540 --> 00:13:28,570
-- yields one of these
molecules, SO4 3.

300
00:13:32,270 --> 00:13:39,540
I just un-substituted the x plus
three molecules of water.

301
00:13:39,540 --> 00:13:40,670
Let's do one more.

302
00:13:40,670 --> 00:13:43,800
And then I think we'll
be all balanced out.

303
00:13:43,800 --> 00:13:47,750
Carbon dioxide plus hydrogen gas
yields methane plus water.

304
00:13:47,750 --> 00:13:51,700
So let's deal with the carbon
first. I have one carbon here,

305
00:13:51,700 --> 00:13:52,540
one carbon there.

306
00:13:52,540 --> 00:13:53,940
The carbons look happy.

307
00:13:53,940 --> 00:13:54,890
Let's look at the oxygen.

308
00:13:54,890 --> 00:13:56,920
I have two oxygens here.

309
00:13:56,920 --> 00:13:58,990
I have one oxygen there.

310
00:13:58,990 --> 00:14:02,350
So I want two oxygens here so
let me stick a two there.

311
00:14:02,350 --> 00:14:04,300
And then let's deal with the
hydrogens last, because this

312
00:14:04,300 --> 00:14:06,650
is the easiest one to play with
because it doesn't affect

313
00:14:06,650 --> 00:14:11,030
any of the other atoms. So if
I have two here and I have

314
00:14:11,030 --> 00:14:14,060
four, plus four here, right?

315
00:14:14,060 --> 00:14:15,865
I have four hydrogens there.

316
00:14:15,865 --> 00:14:18,160
And I have four hydrogens
there.

317
00:14:18,160 --> 00:14:20,790
So I need eight hydrogens
on the left-hand side.

318
00:14:20,790 --> 00:14:22,440
So I just put a four there.

319
00:14:22,440 --> 00:14:24,350
We're all done balancing.

320
00:14:24,350 --> 00:14:27,260
Anyway, hopefully you
found that useful.