Calcium carbonate reacts with dilute hydrochloric acid as given below:

CaCO₃ + 2HCl ⟶ CaCl₂ + H₂O + CO₂

(a) What is the mass of 5 moles of calcium carbonate? (Relative molecular mass of calcium carbonate is 100)

(b) How many moles of HCl will react with 5 moles of calcium carbonate?

(c) What is the volume of carbon dioxide liberated at S.T.P. at the same time?

(a) Given,

1 mole of CaCO₃ = molecular mass of CaCO₃ = 100 g

∴ 5 moles of CaCO₃ weighs = 100 x 5 = 500 g

(b) 1 mole of CaCO₃ requires 2 moles of HCl.

∴ 5 moles of CaCO₃ will require 2 x 5 = 10 moles of HCl

(c) 1 mole of CaCO₃ produces 1 mole of CO₂ and 1 mole occupies 22.4 l of volume.

∴ 5 moles of CaCO₃ will produce 5 moles of CO₂ and 5 moles will occupy 22.4 x 5 = 112 L