Mass Percentage Question?

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

From the above reaction it is evident that 1 mole of CaCO3 reacts with 2 moles of HCl

OR 1 mole of HCl will react with 0.5 moles of CaCO3.

So find out the moles of HCl which has reacted with CaCO3

From the given conditions

Vol.of excess HCl = 6.80 x 0.125/0.15 = 5.7 mL

So the vol. of HCl used in reaction = 50.0 - 5.7 = 44.3 mL = 0.0443 L

Thus the number of moles of HCl which reacted with CaCO3 = 0.0443 x 0.15 = 0.0066

So the number of moles of CaCO3 which reacted with 0.0066 moles of HCl = 0.5 x 0.0066 = 0.0033

Thus the mass of CaCO3 = 100 x 0.0033 = 0.33 g (100 is the molar mass of CaCO3)

Thus in 0.450 g sample pure CaCO3 = 0.33

Therefore % purity = 100 x 0.33/0.450 = 73.3