When 4 g of CaCO3 and sand mixture is treated with excess of HCl , 0.88 g of CO2 is produced.  Calculate % weight of CaCO3.

CaCO3+2HCl→CO2+H2O+CaCl2 Sand +HCl→ No reaction  Number of moles of CO2=0.8844=0.02 mole∴ number of moles of CaCO3=0.02 moles ∴% mass of CaCO3=0.02×100 Weight of mixture ×100                                              =0.02×1004×100=50%