Question

How do you calculate the theoretical yield of CaO in moles when 0.322 grams of CaCO₃ is reacted with 33.6 mL of   1.05 M HCl?

Answer 1

mass of CaCO3 = 0.322 g

moles of CaCO3 = 0.322 / 100 = 3.22 x 10^-4 mol

moles of HCl = 33.6 x 1.05 / 1000 = 0.03528 mol

CaCO3 +   2 HCl    -------------->   CaCl2 + H2O +   CO2

     1                2                                  1

3.22 x 10^-4      0.03528                           ??

here limiting reagent is CaCO3.

moles of CaCl2 formed = 3.22 x 10^-4

mass of CaCl2 formed = 3.22 x 10^-4 x 110.984

                                     = 0.0357 g

theoretical yield = 0.0357 g

Note : here you gave CaO . but the reaction between CaCO3 and HCl yield CaCl2 and H2O and CO2. so i calculated for CaCl2.