I am seeking verification that I have done this problem correctly!
Consider the balanced equation: Na2SO4 + 2C = Na2S + 2CO2.
What is the theoretical yield of Na2S when 5.00g Na2SO4 react?
(5.00g * 1mol * 2molCO * 44gCO)/(142g * 1mol * 1molCO) = 440/142 = 3.098gCO

Are you sure you wrote the question as intended? Because you're
proposed solution is grams CO not g Na2S...

I'll answer it in a comment if no answerer jumps on this in a few days.

The question is written correctly... however, I don't know if my
answer is. The CO could be wrong. I am having a lot of difficulty even
knowing what parameters the answer should be in!

Think of it this way -- for every mole of Na2SO4 that reacts, you make
1 mole of Na2S if the reaction goes to completion.

The molecular weight of Na2SO4 is 142.04 gm/mol, so 5 grams of this
salt amounts to 5gm / 142.04 gm/mol = 0.0352 moles.  According to the
reaction stoichiometry, you will get the same number of moles of
product Na2S.  The molecular weight of Na2S is 78.04 gm/mol, so 0.0352
moles of this salt has a mass of 78.04 gm/mol * 0.0352 mol = 2.747 gm.

5 grams of Na2SO4 will react to form 2.747 grams of Na2S.

The answer is 2.748g