00:01
We're given here a pair of balanced reactions that happen in sequence, and we're asked, given that they both have a 92 % actual yield, how much of the ch2cl are we going to end up with, starting with 112 grams of methane and excess clothing.
00:24
So, first of all, we need to calculate our number of moles, we're going to need some molecular weights.
00:28
This is 16 .042.
00:34
And we're also going to need the molecular weight of ch3cl, which is 50 .484.
00:49
And we're also going to need our formula, which is molecular weight, it's mass, divided by number of moles.
00:58
So number of moles is just going to be our mass, 112 grams, divided by 116 .042 grams per mole.
01:16
So we're consuming 6 .98 moles of methane.
01:22
And since our stoichiometry is 1 to 1, the equations are balanced here, we expect 6 .98 moles here.
01:31
Of our intermediate product.
01:34
And so we can calculate our expected mass, so that's our theoretical, is just going to be molecular weight, 50 .484 grams per mole, times our 6 .98 moles.
01:58
So our expected mass is 300 ,000.
02:08
And 52 .46 grams.
02:14
But our actual mass is going to be 92 % of that...