Welcome back guys.
In this new video, we're going to handle a topic that's essential for any type of calculations
we're going to do this semester.
It can be quite easy as long as we follow basic rules.
It's all about balancing chemical equations.
We're going to say when you're balancing an equation, always make sure that the number
and types of atoms on both sides of the arrow are equal.
For this example it says write balanced equations for each of the following by inserting the
correct coefficient in the blanks.
Coefficient is just the number that goes in front of each one of these compounds in a
balanced equation.
The approach we should always take is let's just write a list.
Both sides of the equation have the same elements.
For them to be balanced, they have to have the same exact numbers of each of these types
of elements.
On the left side, we have only one aluminum.
We also have – there's a two here, so we have two chlorines.
On the right, we still have one aluminum and we have now three chlorines.
You can see that the lists don't match, that's because the equation is not balanced.
What we have to do is just go down the list and see what doesn't match.
Our chlorines aren't matching, so that's what we want to balance.
We have to think what number can we put in front of both of those compounds to make them
both have the same number of chlorines.
Here we're going to say lowest common multiple is six between three and two.
If I put a three in front of that Cl2, it becomes distributed.
It's three times two, so this becomes six.
And here I'm going to put a two.
We're going to say two times the three gives us six here also.
Our chlorines now match.
But that two is also in front of the aluminum, so that's going to get multiplied by two as
well, so this becomes two.
Our lists are almost equal.
We just have to do one more step.
All we have to do now is just throw a two in front of this aluminum, making this two.
Now both our lists match.
Now they're balanced.
For the next one, let's take a look to see what groups we have.
What we should notice here is – you see this PO4 right here, it's right here as well.
It's a polyatomic ion.
It came from phosphate.
If you have polyatomic ions that look the same on both sides, just keep them all together
to help us balance.
We're going to say on the left side we have Ba.
We have PO4.
We're keeping it all together.
Counting it as the same.
We also have K. And look, you see this OH right here, that OH is right over here as
well.
We're going to have OH in this as well.
We'll also have the same exact elements and compounds on the right side.
Now all we have to do is just to see which ones match up and which ones are different.
On the left side, we have a three here with the Ba, barium.
We have three bariums.
We have a two on the outside, which is going to mean we have two phosphates.
We have one potassium, right here and one OH.
On the right side, Ba, we only have one Ba.
We only have one phosphate.
We have a little three here, so we have three potassiums.
Then we have a two out here, so we have two hydroxide ions.
Let's go down the list and match things up.
For barium, we have three on the left, but one on the right, so let's put a three here.
That three will get distributed.
This becomes three.
Then three times the two out here gives us six OHs.
For phosphate we have two on the left, but only one on the right, so we're going to put
a two here, which makes this two, but then it gets distributed with this three, so three
times two gives us six potassiums.
All we have to do now is our potassiums don't match, so I'm going to put a six here.
That's going to get distributed, making the potassium six and also my OH six.
Both our lists match up, so now it's balanced.
If you want, you could put a one here to show that there's a one there.
You don't really have to, but if you wanted to, you could just put it there.
This last one, this one's an example of combustion.
We have a carbon and hydrogen reacting with O2 to give us CO2 and water.
I'm going to minus myself from the shot guys, so that we have more space to work with.
Like before, we're just going to make a list.
Make sure things match up.
We have carbon, hydrogen, and oxygen here.
Carbon, hydrogen, and oxygen here.
What we're going to do now is we're going to say we have four carbons, ten hydrogens,
and then we have two oxygens.
On the other side, we have one carbon, two hydrogens.
We have two oxygens here and one oxygen over here, so this is three.
Now we just have to make sure things balance.
We have four carbons on the left, only one on the right, so we're going to throw a four
here.
This changes this to four, but then four times this two gives me eight oxygens, plus the
one oxygen we have here gives me nine.
Next, we want to look at the hydrogens.
One the left, we have ten, but on the right, we only have two, so we're going to throw
a five here.
Five gets multiplied with the two to give me ten hydrogens.
Then we have four times this two here gives me eight oxygens here.
And then five times one gives me give oxygens here.
So eight oxygens here, five oxygens here gives me 13 oxygens total.
Now what we have to do here is we need to balance this our further.
The only number I can put right here to give me 13 is if I put 6.5 because 6.5 times two,
the two right here, gives me 13 as well.
Now with combustion analysis, this happens sometimes.
This would not be the best answer because we can't have decimals or fractions as coefficients.
It has to be whole numbers.
What we're going to do here is we're just going to multiply this entire thing times
two.
There's a one here.
So all of the coefficients inside the brackets will get multiplied by two and that's going
to give me a new equation 2 C4H10 plus the 6.5 gets multiplied by the two on the outside,
so that' s 13O2 gives me 8 CO2 plus 10 H2O.
Just like that guys, we balanced everything.
Balancing can be really easy as long as you make your list and follow where the numbers
go.
Remember, balancing may seem very simple, but it's essential for any type of calculations
we're going to do later on because later on we're going to be tackling chemical calculations
where we need a balanced equation in order to do them.
Just keep practicing.
Look at different types of balancing problems and it can be really easy as long as you implement
this type of system.