WEBVTT

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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,

00:09:21.600 --> 00:09:22.630
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.

00:09:56.000 --> 00:09:58.230
So that one, a lot of people
might find that to be a hard

00:09:58.230 --> 00:10:00.260
problem, because you have three
and a half, and just

00:10:00.260 --> 00:10:02.130
going straight to this might
seem non-intuitive.

00:10:02.130 --> 00:10:05.660
But if you just work from the
more complicated molecules and

00:10:05.660 --> 00:10:07.980
you go atom by atom, and if you
end up with any fractions

00:10:07.980 --> 00:10:10.775
you just multiply both sides by
some number to get rid of

00:10:10.775 --> 00:10:13.030
the fraction, of and
then you're done.

00:10:13.030 --> 00:10:13.590
All right.

00:10:13.590 --> 00:10:15.850
Let's do another one.

00:10:15.850 --> 00:10:17.950
So this one looks all hairy.

00:10:17.950 --> 00:10:24.190
I have this iron oxide plus
sulfuric acid yields all this

00:10:24.190 --> 00:10:25.260
hairy stuff.

00:10:25.260 --> 00:10:30.000
But the key here to realize is
that this group right here,

00:10:30.000 --> 00:10:35.170
this sulfate group,
stays together.

00:10:35.170 --> 00:10:35.480
Right?

00:10:35.480 --> 00:10:39.580
You have this SO4 there and
you have this SO4 there.

00:10:39.580 --> 00:10:42.740
So to really simplify things for
our head, you can kind of

00:10:42.740 --> 00:10:44.010
treat that like an atom.

00:10:44.010 --> 00:10:45.410
So let's make a substitution.

00:10:45.410 --> 00:10:48.230
We'll substitute that for x.

00:10:48.230 --> 00:10:52.230
So if we rewrite this as--
I'll do it in a vibrant

00:10:52.230 --> 00:11:03.390
color-- iron oxide plus H2--
there's only one sulfate

00:11:03.390 --> 00:11:07.690
group here-- x.

00:11:07.690 --> 00:11:15.070
And then that yields
two irons, this

00:11:15.070 --> 00:11:15.940
molecule with two irons.

00:11:15.940 --> 00:11:18.070
And it has three of these
sulfate groups.

00:11:21.140 --> 00:11:22.390
And then plus water.

00:11:26.450 --> 00:11:28.870
All I did is replace the
sulfate with an x.

00:11:28.870 --> 00:11:32.040
And now we can treat that x like
an atom, and we can just

00:11:32.040 --> 00:11:33.830
balance the equation.

00:11:33.830 --> 00:11:36.040
Let's see, on the left-hand
side, how

00:11:36.040 --> 00:11:37.300
many irons do we have?

00:11:37.300 --> 00:11:39.000
We have two irons here.

00:11:39.000 --> 00:11:40.820
We have two irons there.

00:11:40.820 --> 00:11:45.190
So the irons look balanced,
at first glance.

00:11:45.190 --> 00:11:47.630
Let's deal with the oxygens.

00:11:47.630 --> 00:11:52.280
So if we have three oxygens
here-- let me do it in a

00:11:52.280 --> 00:11:59.660
different color-- and we only
have one oxygen here.

00:11:59.660 --> 00:12:03.480
Remember, there were some
oxygens in this x group, but

00:12:03.480 --> 00:12:05.590
the sulfate group stayed
together, so we can just treat

00:12:05.590 --> 00:12:08.700
those separately.

00:12:08.700 --> 00:12:10.820
We want to have three oxygens
on the right-hand side, as

00:12:10.820 --> 00:12:13.250
well, so let's stick
a three here.

00:12:13.250 --> 00:12:14.500
Oops.

00:12:19.710 --> 00:12:21.040
So let me put that
three there.

00:12:21.040 --> 00:12:23.960
So now we have three
oxygens, as well.

00:12:23.960 --> 00:12:28.733
And then finally, let's
look at the hydrogens.

00:12:35.950 --> 00:12:37.160
Let's see, how many hydrogens
do we have here.

00:12:37.160 --> 00:12:41.270
We have six hydrogens now,
on the right-hand side.

00:12:41.270 --> 00:12:42.840
Six hydrogens here.

00:12:42.840 --> 00:12:43.725
Three times two.

00:12:43.725 --> 00:12:46.770
And we want to have six
hydrogens here, so we have to

00:12:46.770 --> 00:12:48.910
have three of these molecules.

00:12:48.910 --> 00:12:52.200
And then finally, let's look at
the sulfate group, that x.

00:12:52.200 --> 00:12:53.790
We have three x's here.

00:12:57.090 --> 00:13:01.350
And lucky for us, we have
three x's over there.

00:13:01.350 --> 00:13:03.420
So our equation has
been balanced.

00:13:03.420 --> 00:13:05.760
And if we want to write it back
in terms of the sulfate

00:13:05.760 --> 00:13:09.490
terms, we can just un-substitute
the x, and we're

00:13:09.490 --> 00:13:16.450
left with one molecule of iron
oxide plus three molecules of

00:13:16.450 --> 00:13:21.540
sulfuric acid, H2SO4 -- I just
un-substituted the x with SO4

00:13:21.540 --> 00:13:28.570
-- yields one of these
molecules, SO4 3.

00:13:32.270 --> 00:13:39.540
I just un-substituted the x plus
three molecules of water.

00:13:39.540 --> 00:13:40.670
Let's do one more.

00:13:40.670 --> 00:13:43.800
And then I think we'll
be all balanced out.

00:13:43.800 --> 00:13:47.750
Carbon dioxide plus hydrogen gas
yields methane plus water.

00:13:47.750 --> 00:13:51.700
So let's deal with the carbon
first. I have one carbon here,

00:13:51.700 --> 00:13:52.540
one carbon there.

00:13:52.540 --> 00:13:53.940
The carbons look happy.

00:13:53.940 --> 00:13:54.890
Let's look at the oxygen.

00:13:54.890 --> 00:13:56.920
I have two oxygens here.

00:13:56.920 --> 00:13:58.990
I have one oxygen there.

00:13:58.990 --> 00:14:02.350
So I want two oxygens here so
let me stick a two there.

00:14:02.350 --> 00:14:04.300
And then let's deal with the
hydrogens last, because this

00:14:04.300 --> 00:14:06.650
is the easiest one to play with
because it doesn't affect

00:14:06.650 --> 00:14:11.030
any of the other atoms. So if
I have two here and I have

00:14:11.030 --> 00:14:14.060
four, plus four here, right?

00:14:14.060 --> 00:14:15.865
I have four hydrogens there.

00:14:15.865 --> 00:14:18.160
And I have four hydrogens
there.

00:14:18.160 --> 00:14:20.790
So I need eight hydrogens
on the left-hand side.

00:14:20.790 --> 00:14:22.440
So I just put a four there.

00:14:22.440 --> 00:14:24.350
We're all done balancing.

00:14:24.350 --> 00:14:27.260
Anyway, hopefully you
found that useful.