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 $\dfrac{100}{1}$ x 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 x 4.5 = 100.8 l

(iii) 1 mole CaCO₃ produces 111 g. CaCl₂

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

(iv)

$\begin{matrix}\text{CaCO}_3 & : & \text{HCl} & \ \text{ 1 mole} & : & 2 \text{ moles} \ \text{ 4.5 mole} & : & x \text{ moles} \ \end{matrix}$

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

Hence, mass of CaCO₃ = 450 g, vol. of CO₂ liberated = 100.8 lits., mass of Cl₂ formed = 499.5 g, moles of HCl used = 9 moles.