00:01
Okay, we are given the following chemical equation, which i will jot down quickly, for consideration.
00:14
This is my precipitate.
00:30
Okay, we're going to calculate the mass of the limiting reactants for each trial.
00:37
Okay, so my mass of my unknown is given as 1 .104, 1 .104 grams.
00:57
My actual weight mass of my n .a .2 .s .o .4 is 1 .45 grams.
01:16
The mass of my precipitate, which is barium sulfate, is 1 .137 grams.
01:26
Calculate the mass of the limiting reactant that reacted.
01:32
Okay.
01:34
The mass of the limiting reactant that reacted.
01:37
So i'm assuming this is barium chloride, i think.
01:47
So let's go with 1 .137 grams of barium sulfate should produce, and first we're going to go to barium chloride needed, and then we'll do na2 -s -o -4 needed.
02:13
We'll calculate each of these and see what we come up with.
02:21
Actual mass weight of n .a2n .o4, mass or precipitate, and mass of unknown.
02:26
So this is all the information i have.
02:28
I'll see what i can figure out.
02:31
Beryn sulfate.
02:34
The molar mass of barym sulfate is 233, i think, 233 .39.
02:54
And i have a one -to -one mole ratio for barium sulfate to barym chloride.
03:05
My molar mass of barium chloride would be 208 .23.
03:26
And this will equal one clear clear, 1 .137.
03:32
Is that right? yes.
03:35
Times 208 .23 divided by 23 .39.
03:41
This would give me 1 .014 grams of bacl.
03:50
To would have been needed.
03:54
Now let's do something similar to our sodium sulfate.
04:04
Whoops see, that's not what i needed there...