4.5 moles of calcium carbonate are reacted with dilute hydrochloric acid.

(i) Write the equation for the reaction.

(ii) What is the mass of 4.5 moles of calcium carbonate? (Relative molecular mass of calcium carbonate is 100)

(iii) What is the volume of carbon dioxide liberated at STP?

(iv) What mass of calcium chloride is formed? (Relative molecular mass of calcium chloride is 111).

(v) How many moles of HCl are used in this reaction?

(i) CaCO₃ + 2HCl ⟶ CaCl₂ + H₂O + CO₂

(ii) 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

(iii) 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

(iv) 1 mole CaCO₃ produces 111 g. CaCl₂

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

(v)

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

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

Hence, moles of HCl used = 9 moles.