- All right now we have another combustion reaction.
Instead of ethylene, we now have ethane, C2H6,
has two carbons and six hydrogen
atoms in each molecule of ethane,
and it is reacting.
It's ethane gas, it is reacting with molecular oxygen
in gaseous form and they combust to
form carbon dioxide gas and liquid water,
and like we've seen in previous examples,
this chemical equation is not balanced.
How can we tell?
Well here on the left hand side we have two carbons.
Here on the right hand side we have one carbon.
Here on the left hand side we have six hydrogens.
Here on the right hand side we only have two hydrogens.
Here on the left hand side we have two oxygens.
On the right hand side we have two plus, three oxygens,
so none of the elements here are balanced.
But like we did in the example of ethylene,
whenever you see this where you have several
somewhat complex molecules involved,
it's good to save the element that
is in a molecule by itself for last,
because you can just tweak this
to change the number of oxygens
without it having any other side effects
on the number of carbons or hydrogens.
So what I'm going to do is I'm going to first balance,
like we've done before,
the carbons and the hydrogens,
which are going to have implications on the oxygens
because if I change the number here,
it's going to change the number of oxygens.
If I change the number here it's going to
change the number of oxygens.
But lucky for me I have this dioxygen molecule
on the left hand side that I can just tweak at the end
to balance the entire chemical equation.
So let's start with, you know last time
we started with carbon, let's start with hydrogen this time.
Just for kicks.
So over here I have six hydrogens
on the left hand side, of the entire left hand side
I only have six hydrogen atoms.
On the right hand side I only have two right now,
so if I want to have six I would multiply these two by three
so now I have three water molecules,
each of them have two hydrogen atoms,
so I'm going to have six, six hydrogen atoms
on the right hand side.
Fair enough.
Now let's go to the carbon.
Remember I'm saving oxygen for last.
Carbon on the left hand side I have two carbons.
How many carbons do I have on the right hand side?
Well right now I only have one,
but I can change that very easily.
Instead of having one molecule of carbon dioxide,
I can have two molecules of carbon dioxide.
And so now my carbons are balanced,
two carbons, two carbons.
And now let's go to the oxygens.
So right now, right now on, I'll do this in a mauve color,
right now on the left hand side I have two oxygens,
but on the right hand side what do I have?
Let's see, I have two times two,
so this right over here is four oxygens,
and then I have three water molecules,
each of them have one oxygen atom,
so three times one, so this is going to be three
right over here.
So on the entire right hand side I have seven oxygen atoms
and on the left I only have two.
So what can I do here?
What can multiply by two to get to seven?
Two times what is equal to seven?
Well two times three-and-a-half is equal to seven.
So two times three-and-a-half is equal to seven.
Remember I have two here, I'm saying two times
something is equal to seven.
I want to get to the four plus the three.
Well I multiply it by three point five,
and now I have seven, seven oxygen atoms
on both sides of my chemical equation.
But like we've seen in previous videos,
it is not standard to just leave a three-and-a-half here,
it's kind of this weird notion of
three-and-a-half molecules,
we like to have whole numbers here.
So how do we make sure we have all whole number
coefficients in front of our molecules?
Well we could just multiply everything by two,
then this thing is going to become a seven,
this thing is going to become a two,
this is going to become a four,
this is going to become a six,
so let's just do that.
I'll write the whole reaction over again.
So I have my ethane, and I won't write
actually what state it's in just to save some time,
plus some molecular oxygen.
They combust, they yield,
so these are the reactants,
the products are carbon dioxide gas
and liquid water, and liquid water.
So let's see, you multiply, if you say that there
was a one out here before,
we're going to multiply that by two to get two.
We had a three point five here, multiply that by two,
you get a seven here.
We had a two right over here, multiply that by two
you get a four.
Once again I'm just multiplying all the coefficients
by two just like you did in algebraic equations
in your algebra classes.
And then finally three times two is six.
And we're all balanced.
We were balanced here but we didn't
have whole number coefficients.
Multiplying everything by two gave us
the whole number coefficients and we are much happier.