00:01
To find the percent yield, we first have to identify the limiting reactants.
00:05
To do that, we're going to divide each of the masses we were given by the respective molar mass of that compound.
00:12
So this 1 .87 grams.
00:14
We're going to divide by 60 .05 grams for every one mole, which is the molar mass of the acetic acid.
00:25
This tells us that we have 0 .0311 moles of acetic acid, doing the same for.
00:34
The alcohol, dividing by its molar mass, 88 .15 grams for every mole, tells us that we have 0 .062 moles of the second reactant.
00:51
Because these reactants reactants react in a one -to -one mole -to -mole ratio, we can easily see that there is less alcohol available, and that tells us that this is the limiting reactant.
01:09
It's going to run out first.
01:11
Thus, the limiting reactant tells us how much product is produced.
01:15
Again, the reactant and the product are in one -to -one multi -mole ratios.
01:21
So if i had 0 .0262 moles of the reactant, i'm going to make 0 .0262 moles of the product, c7h14 -02.
01:35
That's how much is formed...