Q. 354.0( 78 Votes )

Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction,



What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl?

Answer :

Given:

Volume of HCl = 25 ml


Molarity of HCl = 0.75 M


Atomic Mass of Hydrogen = 1g


Atomic Mass of Chlorine = 35.5g


Molecular Mass of HCl = 1 + 35.5


= 36.5g


By Avogadro’s Law,


1 mole of HCl 36.5g of HCl


So 0.75 moles of HCl X g of HCl


X = 0.75×36.5


X = 27.35 g


Thus, 1000 mL of solution contains HCl = 27.375g


1000ml of HCl 27.357g of HCl


So 25 ml of HCl X g of HCl


X = [25×27.375]/1000


= 684.375/1000


=0.68438 g


From the chemical equation given in the question,


CaCO3+2 HCL CaCL2+H2O


We observe that


2 moles of HCl reacts with 1 mole of CaCO3


73 g of HCl reacts with 100g of CaCO3


0.68438g of HCl reacts with X g of CaCO3


X = [0.68483×100]/73


= 68.483/73


= 0.93812 g


So amount of CaCO3 required is 0.94g.


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