[webguy54]

3SA + 1 TEA --> 1 (ES)₃A + 3H₂O

MW
SA=284 g/mol
TEA=149 g/mol
(ES)₃A=947 g/mol
H₂O=18 g/mol

Amount
SA=300 g
TEA=80g

First I divided:
300g / 284 g/mol = 1.06 mol
80g / 149 g/mol = 0.537 mol

For SA, I subtracted:
1.06 mol - 1.06 mol = 0

For TEA, I subtracted:
0.537 mol - X = ?

Then I cross-multiplied:
3 / 1.06 mol = 1 / X
X = 0.353 mol

0.537 mol - 0.353 mol = 0.184 mol
0.184 mol * 149 g/mol = 27.4 g

From here I can tell that SA is the limiting reagent.

Next, I had to calculate how much (ES)₃A was produced.  From an example in class, it looks like I would multiply:
0.353 mol * 947 g/mol = 334.3 g (ES)₃A

QUESTION: Why would I multiply 947 g/mol by 0.353 mol of TEA?

[ARGOS++]

[ARGOS++]

                             Copy of the lost replay No. 1

Dear webguy54;

Do you know why you subtracted and then “cross-multiplied”?  (I don’t believe you know.)
You determined with it:   With how many moles you are able to fulfil the given RxN.

Maybe you study the Recipe:   How to do Stoichiometry under:

"Stoichiometry Problem (#23‘025)”

Then you would result in:
B.)        3SA      +      1 TEA       ------>     1(ES)₃A      +    3H₂O         Excess
C1.)      284                 149                           947                18              149
Cm.)     300g                 80g                         333.45g                           27.7g
C2.)      1.056            0.5379                     1 * 0.3521                       1 *  0.1858
D.)       0.3521          0.5379                        0.3521                        (0.5379 – 0.3521)

I hope to have been of some help to you.
Good Luck!
                    ARGOS⁺⁺