Well, start with the given grams of alkazelteser that you have. To start, compute the molar mass of it, which you can do by finding the sum of it's molecular formula. For example, molar mass has the units of \(\frac{mass}{mole}\), therefore, if you have H\(_2\), the molar mass of that is 1.008 \(\times\) 2 = 2.016 \(\frac{grams}{mol}\). Does that make sense?
And very similar to how you did in algebra. you are going to use that relationship to cancel out the units.
given grams of Alka-Seltz \(\times\) \(\frac{mol}{mass~of~alka-setlz}\)
notice that the units of grams of alka-seltz cancel out? You will be left with moles.
Then, take:
mol of Alka-seltz \(\times\) \(\frac{\sf \color{red}{3}mol~NaHCO_3}{\sf \color{blue}{1} mol~alka-seltz}\)
again, note that the units of mole cancel out. Now you have Mol's of NaHCO\(_3\)!
Finally, multiply by the molar mass of NaHCO\(_3\) to get the molar mass!
Now, based on your equation, it tells you that you have a 1:1 ratio of NaHCO\(_3\). So, mole of NaHCO\(_3\) = mol of CO\(_2\).
Now, this is the amount of CO\(_2\) you would get if All the NaHCO\(_3\) fully reacted. It is called the theoretical yield of CO\(_2\). It is the amount of CO\(_2\) you are expecting to get, assuming that all the NaHCO\(_3\) reacts.
Convert it to mass, SIMILARLY to how you did above.