You weigh out 1.70 g of sodium carbonate monohydrate, dissolve it in water and react it with 15.00 mL of 2.00 M HCl. Write an equation for the reaction and calculate the expected mass of carbon dioxide that will be produced.

Na2CO3.H2O + 2HCl ==> 2H2O + CO2 + 2NaCl

mols Na2CO3.H2O = grams/molar mass = about 0.014 but you need to do it better than that.
mols HCl = M x L = about 0.030

mols CO2 that could be produced from Na2CO3 = 0.014
mols CO2 that could be produced from HCl = about 0.03
In limiting reagent problems the SMALLER number always wins; therefore, the carbonate is the limiting reagent and HCl is the excess reagent.
Therefore, 0.014 mols CO2 will be produced.
grams CO2 = mols CO2 x molar mass CO2 = ?