Question

Dilute hydrochloric acid (HCl) is reacted with 4.5 moles of calcium carbonate. Give the equation for the said reaction. Calculate :

1. The mass of 4.5 moles of CaCO₃ . 

2. The volume of CO₂ liberated at stp. 

3. The mass of CaCl₂ formed ? 

4. The number of moles of the acid HCl used in the reaction (relative molecular mass of CaCO₃ is 100 and of CaCl₂ is 111].

Answer 1

(ii) 1 mole of CaCO₃ produces 1 mole of CO₂

4.5 moles of CaCO₃ will produce 4.5 moles of moles of CO₂

∴ 1 mole of CO₂ occupies 22.4 l at S.T.P.

∴ 4.5 moles of CO₂ will = 4.5 x 22.4 l = 100.80 l

(iii) 1 mole = 22.4

1 mole of CaCO₂ gives 111 g of CaCl₂

4.5 mole of CaCO₃ gives 111 x 4.5 = 499.5 g

(iv) 1 mole uses 2 moles of HCl

4.5 mole uses 4.5 x 2 = 9 moles