Dilute HCl is reacted with 4.5 moles of calcium carbonate.
Calculate :
(i) The mass of 4.5 moles of CaCO3.
(ii) The volume of CO2 liberated at stp.
(iii) The mass of CaCl2 formed ?
(iv) The number of moles of the acid HCl used in the reaction [relative molecular mass of CaCO3 is 100 and of CaCl2 is 111.]

CaCO₃ + 2HCl ⟶ CaCl₂ + H₂O + CO₂

(i) Given,

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

∴ 4.5 moles of CaCO₃ weighs = 100\over 1 × 4.5 = 450 g

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

∴ 4.5 moles of CaCO₃ will produce 4.5 moles of CO₂ and 4.5 moles will occupy 22.4 × 4.5 = 100.8 l

(iii) 1 mole CaCO₃ produces 111 g. CaCl₂

∴ 4.5 moles of CaCO₃ will produce = 111 × 4.5 = 499.5 g.

(iv)

∴ number of moles of HCl used = 2 × 4.5 = 9 moles.

Hence, mass of CaCO₃ = 450 g,

vol. of CO₂ liberated = 100.8 lits.,

mass of Cl₂ formed = 499.5 g,

and moles of HCl used = 9 moles.