00:01
So sulfur trioxide, s .o3, i'm assuming that's a solid.
00:07
Let's just go with it.
00:08
Dissolves in water, liquid, to form hydro.
00:17
I mean, not hydro, sulfuric acid, which is aqueous.
00:23
That is a strong acid.
00:26
Woo -hoo.
00:26
Okay.
00:27
So now our job is to balance this.
00:29
Do we need to do that? let's see.
00:32
On the left and on the right, there's one sulfur.
00:35
On the left and on the right, there's two hydrogens.
00:40
On the left and on the right, there's four oxygen.
00:43
Sweet.
00:44
No need to balance.
00:46
How much sulfuric acid can be produced from 13 .5 millimeters of water and 22 .1 grams, so we got to convert all these grams.
00:56
No, not many grams.
00:59
Got to convert everything to moles.
01:03
Okay, so we have 13 .5 milliliters water, and we're given that the density of water is 1 gram per milliliter.
01:16
Cool.
01:17
Cancel these units, we get 13 .5 grams of water.
01:22
We also know that the molar mass of water is 16 plus 2.
01:28
Point, let's see, 16 plus 1 .08 plus 1 is 16 .016 .016 .16, that's 18 .016 grams per one mole of water.
01:48
So then when we solve that, we get 13 .5 divided by 18 .016 is 0 .7493 moles of water.
02:07
So now let's do our sulfur trioxide.
02:10
We have 22 .1 grams of s .03.
02:16
The molar mass of s .o3 is 32 .06 plus 48, 80 .06.
02:27
06 grams for one mall of s .03.
02:31
So then we convert it to malls, and we get 0 .27 .7.
02:41
0 .7, 6 .0...