Question

Percentage purity of calcium carbonate test method

Answer 1

CaCO3+2HCl⟶CaCl2+H2O+CO2CaCOX3+2HCl⟶CaClX2+HX2O+COX2

Number of milli equivalence of HClHCl used initially is 1515

Number of milli moles of HClHCl used initially is 152152

22 moles of HClHCl reacts with 11 mole of CaCO3CaCOX3.

So amount of calcium carbonate is 154×10−3 mol=38 g154×10−3 mol=38 g

But answer given is 75%
Hope its help you