00:01
So in this problem, we're going to mix h2s .o4 with some barium chloride.
00:11
Let's go ahead and write our balanced equation so that we can see what our precipitate will be.
00:16
So we'll make some barium sulfate and some hcl.
00:22
We'll go ahead and balance that.
00:24
And since hcl is not a precipitate, our barium sulfate.
00:29
This is going to be our precipitate right here.
00:33
So now we use that balanced equation to answer some questions.
00:37
So this is a limiting reactant problem, and i know that because i've been given amounts of both h2s -o -4 and barium chloride.
00:44
So let's start with the barium chloride.
00:47
I've been given the molarity and the volume.
00:52
So let's go ahead and find out how many moles of barium chloride i've started with.
01:02
And then for the h -2 -s -4, i have 25 milliliters, and i've been given the density.
01:11
1 .107 grams per milliliter.
01:17
So that's 27 .7 grams.
01:21
And it's 15 % of that is actually our h2s .4 from the solution.
01:28
So that will give us 4 .22 grams of h2s .4.
01:36
And i'll keep going and go ahead and change that to moles using the molar mass.
01:41
So the molar mass of h2s .o4 is 98.
01:46
0 .09.
01:49
So what i want to do here is keep going, though, and find out how many grams of baso4 i would produce in both of these cases.
01:59
So one of these is going to be a limiting reactant.
02:02
So this is our h2 -s -o -4.
02:06
This is our barium sulfate...