00:01
So we're going to determine the limiting reactant in terms of the theoretical yield of lead sulfate that each species can produce, knock them out with one go.
00:10
So we have 55 milliliters, which is 0 .055 liters of 0 .102 molar potassium sulfate.
00:21
So that's 0 .102 moles in 1 liter of k2 s .4.
00:28
And then looking at the balanced equation, it's a one -to -one ratio of lead sulfate to potassium sulfate.
00:47
And then we're going to look at the molar mass of lead sulfate to get the grams.
00:52
Oh, that should not be an equal sign.
00:55
So the molar mass is 303 .6 grams in one mole.
01:05
And we're going to just, i'll do the calculations at the end, but we're just going to do the same thing for the other.
01:13
Reactant.
01:14
So we have 35 milliliters or 0 .035 liters and it's 0 .114 moles of lead acetate in 1 liter.
01:37
And then we have a one to one ratio for that as well.
01:58
And then we're going to multiply by the molar mass again.
02:07
So whichever one of these two comes out to a smaller number is going to be our limiting reaction and then we'll get our theoretical yield, which will be that number...