ecuatia reactiei:
1mol 1mol
CaCO3 + 2HCl---> CaCl2 + H2O+CO2↑
(40Ca+12gC+3x16gO)............................................22,4l
m= 2x100g...................................................................44,8 l
m,CaCO3pur= 200g
p/100= m,pur/m,impur-------> p=200x100/240= 83,33%