00:06
We are asked what mass of silver chloride can be prepared when we are given 190 .0 milliliters of silver nitrate and 160 milliliters of calcium chloride.
00:32
Let me see if i have a balanced.
00:34
I do.
00:35
We're also asked to calculate the concentration of each ion after precipitation.
01:06
Okay, let's begin.
01:07
First thing we're going to do is figure out which one of these is an access and i will be, or limiting i should say.
01:15
And i'm going to do this by finding how many grams of silver chloride each could prepare.
01:21
Let's begin.
01:23
1 .90 .0 milliliters of silver nitrate.
01:35
First, let's change our milliliters to liters.
01:41
And i forgot to write down of 0 .30 molar.
01:48
Of 0 .14 molar.
01:52
There we go.
02:07
And here i will have a 2 to 2 mole ratio for my precipitate and let me get my molar mass of ag, cl2, 143 .32.
02:43
This will equal 0 .19 times 0 .3 and skip my twos times 143 .32.
02:56
And this will equal 8 .169 grams of agcl.
03:10
And i'm going to just jot this down for maybe future use.
03:17
Okay, next, similar calculation, but this time we're going to be using 160 .0 milliliters of calcium chloride.
03:34
Again, we'll convert milliliters to liters.
03:40
Then i'll multiply this by 0 .14 moles per liter.
03:51
And this time i will have a 1 to 2 mole ratio.
04:05
And we will end with the same part that we used for our first one, the molar mass of silver chloride.
04:19
And see what this equals.
04:21
The lower these two numbers will be our theoretical yield.
04:32
And this is 0 .16 times 0 .14 times 2 times 143 is 6 .421 grams of ag cl.
04:57
Okay, so my first question was, what mass of silver chloride? this is the mass of silver chloride.
05:08
So this is my theoretical yield.
05:10
Now the concentrations of each remaining ion, calcium chloride is limiting, is the limiting reactant, is all consumed.
05:44
No, well, i shouldn't say that.
05:50
All the chloride is consumed.
05:59
Chloride concentration is zero.
06:11
Okay, for the calcium ion, which is in this a spectator, the concentration will be 0 .160 mill.
06:32
I'm just going to go to liters this time, times 0 .14 moles per liter.
06:43
Then we're going to divide that by my new volume, which will be 0 .190 liters.
06:51
Plus 0 .160 liters...