Question

What amount of HCL is required to completely react with 200 gram of calcium carbonate

Answer 1

for complete neutralisation
no of equivalent of half= no of equivalent of caco3
moles of hcl× n factor =moles of caco3× n factor
= moles of hcl×1= 2×2
moles of hcl requires 4
so mass required is moles ×molecular weight
= 4×36.5
=146g