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.