150ml of 0.1M HCl is required to react completely with 1gm of a sample of limestone. Calculate the% purity of calcium carbonate.

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

Clearly, for 2 moles of HCL you require 1 mole of CaCO3..

150ml of 0.1M HCl means..0.015 moles of pure HCl..

so it will react with (0.015)/2 moles of pure CaCO3..i.e 0.0075 moles..i.e 0.75gm..

so out of 1gm 0.75gm is pure ..so % purity = 75%

CaCO₃=100 g MW

150 ml 0.1 M HCl=.015 moles HCl

CaCO₃ 1 mole reacts with 2 moles HCl

so amt of lime stone=0.015/2 moles=0.015/2*100g=0.75 grams

so 0.75 g limestone/1 g sample

percanetage purity=75

moles of HCl=150*0.10*10^-3 =0.015 moles

Now according to the stoichoimetric equation:

2HCl+CaCO₃ ---> CaCl ₂ + H₂O +CO₂

 from the eqaution:

1CaCO3 mole   2mole HCl

.: Mole of CaCO3 reacted =0.015/2 moles =0.015/2 *100 gms =0.75 gms

.:% purity =0.75/1 *100 =75%

thnx.