Q35 of 37 Page 22

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?

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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