Question

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

1. Write the equation for the reaction.
2. What is the mass of 4.5 moles of calcium carbonate? (Relative molecular mass of calcium carbonate is 100).
3. What is the volume of carbon dioxide liberated at STP?
4. What mass of calcium chloride is formed? (Relative molecular mass of calcium chloride is 111).
5. How many moles of HCl are used in this reaction?

Answer 1

(i) \[\ce{\underset{100}{CaCO3} + \underset{111}{2HCl} -> CaCl2 + H2O + CO2 ^}\]

(ii) 1 mole of calcium carbonate weighs 100 g

∴ 4.5 moles of calcium carbonate weighs 100 × 4.5 = 450 g

(iii) 1 mole of calcium carbonate liberates CO₂ = 22.4 l

∴ 4.5 moles of calcium carbonate liberates CO₂ = 22.4 × 4.5

= 100.8 litres

(iv) \[\ce{\underset{100}{CaCO3} + \underset{111}{2HCl} -> CaCl2 + H2O + CO2 ^}\]

From 100 g CaCO₃; CaCl₂ formed is 111 g

From 1 g of CaCO₃; CaCl₂ formed is `111/100`g

From 450 g of CaCO₃; CaCl₂ formed is `111/100 xx 450`

= 499.5 g of CaCl₂

(v) 1 mole of CaCO₃ uses 2 moles of HCI

∴ 4.5 moles of CaCO₃ use HCI = 2 × 4.5

= 9 moles