I not sure if I did it right, but I got 15.9. here is the question- If 40.0 grams of H3PO4 reacts with 60.0 grams of MgCO3 according to this reaction,
2H3PO4 + 3MgCO3 —> Mg3 (PO4)2 + 3CO2 + 3H2O (already balanced)
Then calculate the volume of carbon dioxide gas produced at STP.

You're close, but not quite. I've gone through the math, and I know where you got your answer, so I can point to precisely where you went wrong.

Look at the number of moles of H3PO4 and MgCO3 you start with. You have just over 0.4 mol H3PO4 and just over 0.7mol MgCO3 (right?).

In the reaction given, the ratio of H3PO4 consumed to MgCO3 consumed is 2:3. That means to fully react 2 moles of H3PO4, it will require 3 moles of MgCO3. To fully react 0.4 moles of H3PO4, you need 0.6 moles of MgCO3. We have this much and more hanging around, so that means that the H3PO4 is the limiting reagent (or limiting reactant).

All of the H3PO4 will be used up, but not all of the MgCO3. So, you have to use the coefficients from the reaction equation to determine the ratio of CO2 produced to H3PO4 consumed, then multiply that number of moles by the magical 22.4 litres.

Also, be sure to include the units (litres) in your answer.

Good job otherwise, though!

Thank you so much dmatos!! That helped me a lot.