The mass of caco3 required to react completely with 20 ml of 1 M hcl as per the reaction caco3+2hcl->cacl2+co2+h2o is

Let us first calculate the mass of HCl that is present in 20 mL of 1 M HCl solution. Molarity of a solution is given by the following relation

Molarity = number of moles of solute / Volume of solution (in litres)

1 = n / 0.020

This gives n = 0.02

Now, number of moles is given by the following realtion

number of moles = mass of substance / molar mass of substance

So, the mass of HCl will be

0.02 = mass of HCl  / 36.5

mass of HCl = 0.730g

Now, a balanced chemical equation showing the reaction of HCl and CaCO3 is as follows

 CaCO3 + 2HCl  →  CaCl2 + H2O + CO2

According to the above equation, 100g of CaCO3 is required to react completely with 71g of HCl. That is 

 73g of HCl  → 100g of CaCO3

 1g of HCl  → (100 / 73) g of CaCO3

 So, 0.73g would require = (100 / 73) X 0.73 g of CaCO3

 = 1g of CaCO3

Thus 1g of CaCO3 would be required to completely react with 20ml of 1 M HCl.