00:01
For this problem, we have two calculations to do.
00:11
We'll find the theoretical yield given product info, or excuse me, given reactant info.
00:31
In order to do this, we're going to need the molar masses of everything.
00:35
So aluminum, we will use 26 .99.
00:41
I'll show you 26 .98.
00:52
For chlorine, we will use 70 .9.
01:01
One i'll use if your instructor wants you to use more sig figs do it and aluminum chloride i'll use 133 .34 so there's all my preliminary information let's start with a and for a we're given one gram of each and we'll start with 26 in this box or in this part of my grid i'll be putting the molar or excuse me the mole ratio and finally we'll use the mass.
02:38
This will equal 1 times 130.
02:47
And that will give me 4 .94 and 4 .9.
02:59
For my second problem, we're going to do the same mechanics for our reaction just with a very few different numbers.
03:16
Notice that my mole ratio is 2 to 3 this time, 1 times 2 times 133 .34, divided 70 .91 .3 .8.
04:00
So for this problem, our theoretical yield is the lower of these two values.
04:23
Now we're going to do the same exact thing except two different numbers.
04:34
This time i have 5 .5 grams of aluminum and i have 19 .8 grams of chlorine.
05:09
The only thing that's going to be different will be these first two numbers...