2HCl (aq)+CaCO3 (s)->H20(l)+CaCl2(aq)+CO2(g). If using 40cm^3 of 2.5mol.dm^-3 Hcl and 5.67g of CaCO3, determine the limiting reagent and how much CO2(g) could be theoretically produced by this reaction.

First, we have to convert the starting measurements for the HCl and CaCO3 content into moles so we can compare the ratios to the formula. 40cm^3=0.04dm^30.04dm^32.5mol.dm^-3=0.1 moles of HClmass/molar mass= moles5.67g/100=0.0567 moles of CaCO30.1 moles of HCl will fully react to 0.05 moles of CaCO30.0567>0.05 so HCl is the limiting reagent 0.1 moles of HCl fully reacting will produce 0.05 moles of CO2 as the coefficient of HCl is 2 and the coefficient of CO2 is 1.0.05 moles CO244 g/mol=2.2 grams of CO2(g)