QQuestion
Calcium carbonate reacts with dilute hydrochloric acid as given below:
\mathrm{CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2}(a) What is the mass of 5 moles of calcium carbonate? (Relative molecular mass of calcium carbonate is 100)
(b) How many moles of HCl will react with 5 moles of calcium carbonate?
(c) What is the volume of carbon dioxide liberated at S.T.P. at the same time?
✓Answer
(a) Given,
1 mole of CaCO3 = molecular mass of CaCO3 = 100 g
∴ 5 moles of CaCO3 weighs = 100 x 5 = 500 g
(b) 1 mole of CaCO3 requires 2 moles of HCl.
∴ 5 moles of CaCO3 will require 2 x 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 x 5 = 112 L
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