00:02
Hi there, let's go ahead and balance some equations.
00:05
We are looking at problem number 12, which gives us several equations to balance.
00:10
As i write these, i am going to leave the state of matter out because while it's important, it is not really part of the balancing.
00:19
So let's get started with the first one here.
00:21
Remember when we balance an equation, we want to add coefficients so that we end up with an equal number of each type of atom on both sides of the arrow.
00:32
So our first equation, this is letter a, looks like this.
00:37
I am going to jump in and start with the nitrogen, because if i start with the lithium, i'm going to have to come back and change what i did.
00:46
So let's go ahead and start with the nitrogen.
00:48
There are two on the reactant side, only one on the product side.
00:51
So i'm going to add a two as a coefficient in front of the li3n.
00:57
Now this gives me six lithium, so i can come back to the reactant side and add a six there.
01:03
Now this is balanced.
01:04
There are six lithium on each side of the equation and two nitrogens.
01:10
Let's move on to letter b.
01:13
In letter b, we have ticl4, and it's reacting with water to produce ti .o2 and hcl.
01:40
Okay, so first of all, the ti, there's one on each side, so it is balanced right now.
01:45
There are four chlorines on the reactant side and only one on the product side.
01:51
So i need to add a four in front of hcl to give me four chlorines.
01:56
That gives me four hydrogen.
01:58
So i'm going to need to come back to the reactant side and put a two in front of h2o to give me four hydrogen.
02:05
Now i have two oxygen on the reactant side and two oxygen on the product side.
02:11
This one is now balanced.
02:17
Let's look at letter c.
02:21
Let us see, we have nh4, n03, and it is decomposing to form n2, o2, and h2.
02:40
All right, starting out with nitrogen.
02:47
It appears i need to add a two here so that the nitrogen will be balanced with two on each side.
02:56
Actually, there were already two on the reactant side.
02:59
Let me erase that because we have a nitrogen here and a nitrogen here.
03:07
So there are already two nitrogen's there.
03:09
So nitrogen is balanced as of right now.
03:13
Let's look at the hydrogen, though.
03:15
There are four hydrogen on the reactant side and only two on the product side.
03:21
So i need to add a two on the product side in front of h2o.
03:25
That gives me four hydrogen.
03:28
Now for the oxygen, i have two oxygen in h2o and another two oxygen in the o2.
03:37
On the reactant side, there are only three.
03:40
Ideally, this is not correct, i'm just putting this here temporarily.
03:46
Ideally, if i had half of an oxygen molecule, that would give me three oxygen atoms on the product side and three on the reactant side and this would be balanced.
03:57
However, we know that we cannot have coefficients that are not whole numbers.
04:03
So borrowing from math, what we can do to get rid of that half is we can multiply every coefficient through by two.
04:11
It should remain balanced and all of the coefficients will then be whole numbers.
04:16
Let's go ahead and do that.
04:18
We get 2 in h4, n03.
04:26
We get 2n2, 2 times 1⁄2 gives me just 1 .02, and 2 times 2 gives me 4h2o.
04:36
Now we can check our work and we can see that we have 4 nitrogen on each side.
04:40
We have a total of 6 oxygen and 8 hydrogen.
04:45
This is now balanced.
04:48
We could have kept going by trial and error.
04:51
But once we recognize it's balanced, if we could only add a coefficient that's a half or some form of a half, we can add that temporarily and then multiply all the coefficients through by two to get rid of that half.
05:04
It's a little trick we can do to speed the process up when we get to that point.
05:10
All right, let's look at letter d.
05:13
We have ca3p2.
05:20
That is reacting with water.
05:25
When it does that, we get caa, oh2 and ph3.
05:39
Right away, i see that hydrogen is in both of my products, so i typically want to save hydrogen for last.
05:45
When i'm trying to balance, if something occurs in both reactants or both products, you want to try to reserve that for last.
05:52
So starting out with the calcium, there are three calcium on the reactant side.
05:57
I would need to add a coefficient of three in front of calcium hydroxide to give me three calceums.
06:02
There are two phosphorus, so i will add a two, in front of ph3.
06:08
All right, now let's address our oxygens or our hydrogens, either or.
06:16
We see that, let me look at the oxygen.
06:19
There's three times two or a total of six oxygen.
06:24
So that would mean i would need a six in front of h2o.
06:26
Let's check and see if our hydrogen are now balanced.
06:30
There are six hydrogen here because three times two is six, and we have another six hydrogen here because two times.
06:38
Times 3 is 6.
06:39
So sure enough, that gives us 12 hydrogen, and this is balanced.
06:49
That would be our balanced equation for d.
06:55
Let's move on to letter e.
07:00
In letter e, we have aluminum hydroxide, aloh3, and it's reacting with h2s -o -4.
07:21
This is a neutralization reaction because al -o -h -3 is a base and h -2 -s -4 is an acid.
07:27
So we are going to get a salt.
07:29
And that is al2, s .o4, three, and water as our other product.
07:42
All right, let's go ahead and balance.
07:44
Starting with aluminum, one on the reactant side, two on the product side.
07:49
So i'm adding a coefficient of two there...