00:02
We are told that we have 65 kilograms of lithium oxide as well as 80 .0 kilograms of water.
00:13
We need to recognize this immediately as a limiting reactant problem when asked for the product because we don't know which of these two reactions is going to run out first.
00:24
So to figure out which reactants is limiting, first we're going to find the number of moles of each reactants.
00:36
For these, we need their molar masses.
00:50
Let's see for lithium oxide and water is 18 .02.
00:59
Now we'll do the math.
01:01
So remember that for mass to mole conversions, we need grams.
01:07
Instead of 65 kilograms, i'm going to start with 65 ,000 grams of l -i -20.
01:16
Divide by its molar mass, 29 .89 grams of l -i -20.
01:22
In every one mole.
01:29
This is equal to 2 ,175 moles.
01:35
Same treatment for water.
01:38
80 ,000 grams of water.
01:46
Change to moles for every one mole of water.
01:49
I have 18 .02 grams.
01:53
The same.
01:54
This is equal to 4440 moles, water, i too.
02:06
In our next step, what we're going to do is we're going to compare these values to find the limiting reactants.
02:23
Here's how i'll do that.
02:25
First, i'll give myself a little bit of room.
02:29
I'm going to look at my coefficients.
02:32
So i see that for every one mole of lithium, i need one mole of water.
02:38
So i need equal amounts of each.
02:40
And you don't have to write this all out if you can kind of see it already, but i will for the sake of clarity...