In order to find the limiting and excess reactant in a question, the first thing we need to do is set up the properly balanced equation:
CaCO3 + 2 HCl -> CaCl2 + H2O + CO2
Now we know that the molar ratio between CaCO3 and HCl is 1:2, this will be important in order to determine the limiting and excess reactant, and we also have:
10.0 grams of CaCO3, molar mass 100.086g/mol
15.0 grams of HCl, molar mass 36.46g/mol
Now we need to find out how many moles of reactant we have and see if by their molar ratio, these values are proportional
CaCO3:
100.086g = 1 mol
10.0g = x moles
x = 0.099 moles of CaCO3 in 10 grams
By the molar ratio, we should have 2 times more moles of HCl than CaCO3, and 0.099*2 = 0.198 moles of HCl, let's check if that is what we actually have
HCl:
36.46g = 1 mol
15.0g = x moles
x = 0.411 moles, this value is way more than we actually need, making it HCl the excess reactant and CaCO3 will be the limiting reactant