In order to solve this question we need to set up the properly balanced equation first:
CaCO3 + 2 HCl -> CaCl2 + H2O + CO2
Now that our reaction is properly balanced, we have to find the limiting reactant in order to find how much mass of CaCl2 will be produced
We have:
10.0 g of CaCO3, molar mass = 100.086 g/mol
15.0 g of HCl, molar mass = 36.46 g/mol
Now we need to find the number of moles of the reactants, and see if the amount available of the other reactant is the one needed to react.
CaCO3:
100.086g = 1 mol
10g = x moles
x = 0.099 moles of CaCO3
According to the molar ratio, the CaCO3 and HCl will have a relationship in which we will always need 2 times more HCl than CaCO3, therefore if we have 0.099 moles of CaCO3, we will need 0.198 moles of HCl in order to proceed with the reaction, but we don't know if that is the right amount, let's check:
HCl:
36.46g = 1 mol
15.0g = x moles
x = 0.411 moles of HCl, this means that we have an excess of HCl, since we only need 0.198 moles in order to react with CaCO3. Therefore Calcium carbonate is the limiting reactant.
Now that we know which one is the limiting reactant, we can find the mass produced of CaCl2.
Again the molar ratio is important here, but know, we will always have the same number of moles for CaCO3 and CaCl2, 1:1 will be the molar ratio, therefore if we have 0.099 moles of CaCO3, we will also have 0.099 moles of CaCl2, using its molar mass, 110.978g/mol, we can find the final mass
CaCl2:
110.978g = 1 mol
x grams = 0.099 moles
x = 11.0 grams (it is actually 10.986 grams, but I have rounded up to 11 grams)