00:02
Hi there.
00:03
In this question, we have three equations to balance.
00:07
The first one is s .o .2 plus o2 produces s .o .3.
00:19
All right.
00:20
So the sulfur is balanced right now.
00:22
But the oxygens, we see there are four on the reactant side and an odd number of three on the product side.
00:29
So i want to make that product side even.
00:33
So i'm going to add a two there.
00:34
So that gives me six oxygen.
00:36
But let's look first.
00:37
Of the sulfur.
00:38
It gives me two sulfur.
00:40
So i need to put a two over here.
00:42
So that gives me two times two, which is four, plus two is six on the left for oxygen, and two times three, which is six on the right for the s .o3.
00:52
All right, so this one is balanced.
00:58
Next, we have h2s, and it is reacting with oxygen to form s .o2 and h2.
01:16
Okay, again, we have that even odd problem, where the only thing we can have are even numbers of oxygens on the left side, but on the right side we currently have an odd number.
01:26
So i want to try to make an even number of oxygen on the right side and see if that helps.
01:31
So i'm going to put a two there.
01:34
So now we have two times two or four hydrogen, so i would need a two in front of h2s, which gives me two sulfur, which means i needed two in front of s02.
01:47
Okay, well, it seems like it's created more of a mess, but let's see.
01:51
Two times two is four oxygen plus two more is six oxygen on the right side.
01:58
Ah, on the left side, if we put a three in front of o2, we'll have six oxygen.
02:05
And this is the smallest set of whole number coefficients because two, twos, and the three are not all divisible by any number.
02:16
Three is a prime number.
02:18
All right.
02:19
Let's go on to this last one.
02:21
Now this last one is a bit of a bit of a challenging one because we have h2s and we have s -o2.
02:32
So we have sulfur in two places and then we have s -8 plus h -2o.
02:42
So what we see here is that on the product side, the water is in a 2 -1 ratio of h -2o.
02:50
Regardless what coefficient i put in front of here, the hydrogen is in a 2 -1 ratio to the oxygen.
02:56
So i need to look for numbers that will give me the hydrogen on the left to the oxygen on the left in a 2 to 1 ratio.
03:06
In other words, i need to have twice as many of h2s as i have s02.
03:14
So let's try 4 and 2.
03:16
Well, 4 and 2 only gives me 6 sulfur, and i have 8 sulfur on the product side.
03:21
Let's go ahead with 6 and 3.
03:24
Well, 6 and 3 would give me 9 sulfur.
03:29
Trying to match up with eight or perhaps some multiple of eight.
03:34
Let's keep going here.
03:35
Let's try eight and four...