00:01
We're given the following a balanced chemical equation as to balance it and do a stochialometry calculation.
00:10
I have to go find this because it's not coming across very nicely.
00:15
Hang on.
00:19
This is problem 149.
00:21
Let me find it.
00:27
There we go.
00:28
C .a .c .2, c .o .c.
00:30
And c .a .c .03.
00:32
Okay.
00:33
I have 1, 2, 3 c0 is on this side.
00:38
If i put a 3 here, then it'll give me 3, 4, 5.
00:41
And i'll have a one calcium and one calcium.
00:51
One, two, three, four, five c's.
00:55
One, there's five c's and three o's.
01:00
Caca, one, two, three for five, one, two, four five, three o's, and three o's.
01:07
I think that should be good.
01:11
When the reaction, balanced reaction is complete, 135 .4 grams of the calcium carbonate is produced, and 38 .5 grams of calcium carbide is left over 150.
02:05
Assuming the reaction on 100 % yield, what masses were the two reactants at the beginning? well, we can start that by figuring out this is how many grams of c .c .o3 was actually produced.
02:30
And we're told that this was a 100 % yield.
02:40
So i could use my 135 .4.
02:45
Grams of calcium carbonate to go to moles of calcium carbonate, or moles of calcium carbide and moles of carbon monoxide at the beginning.
03:05
C .a.
03:06
C .o .3.
03:07
Why do i feel like this is 100? it is.
03:19
100 .09.
03:36
Let me use my mole ratio.
03:39
And first i'm going to go to moles of calcium carbide.
03:47
I have a one -to -one mole ratio.
03:50
And then my calcium carbide cac2 has a molar mass of 64 .10.
04:08
These are all cacs, too.
04:11
And i'll get this in a moment, but my next question, i'm going to do the same calculation with a different substance.
05:16
I just took some benadryl for my allergies.
05:18
They get my conk out very soon.
05:23
My other product is three moles of co.
05:27
It'll be 28 .01 grams for mole.
05:33
Let me do my math on this.
05:34
135 .4 times 64 .10 divided by 100 .0...