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