Question

Calcium carbonate reacts with dilute hydrochloric acid as given below:

\[\ce{CaCO3 + 2HCl -> CaCl2 + H2O + CO2}\]

1. What is the mass of 5 moles of calcium carbonate? (Relative molecular mass of calcium carbonate is 100)
2. How many moles of HCl will react with 5 moles of calcium carbonate?
3. What is the volume of carbon dioxide liberated at S.T.P. at the same time?

Answer 1

a. 1 mole of CaCO₃ = 100 g

∴ 5 moles of CaCO₃ weighs = 100 × 5 = 500 g

Hence, the mass of 5 moles of CaCO₃ will be 500 g.

b. 1 mole of CaCO₃ requires 2 moles of HCl.

∴ 5 moles of CaCO₃ will require 2 × 5 = 10 moles of HCl

c. 1 mole of CaCO₃ produces 1 mole of CO₂ and 1 mole occupies 22.4 l of volume.

∴ 5 moles of CaCO₃ will produce 5 moles of CO₂ and 5 moles will occupy 22.4 × 5 = 112 L