Question

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?

Answer 1

(i) The balanced chemical equation for the reaction between calcium carbonate and dilute hydrochloric acid is:
CaCO₃ (s) + 2HCl (aq) $\to$ CaCl₂ (aq) + H₂O (l) + CO₂ (g)

(ii) Molecular mass of calcium carbonate = 100 a.m.u.
Mass of one mole of calcium carbonate = 100 a.m.u.
Mass of 4.5 moles of calcium carbonate = (4.5 $\times$ 100) a.m.u. = 450 a.m.u.

(iii) According to the balanced chemical reaction, 4.5 moles of calcium carbonate will liberate 4.5 moles of carbon dioxide gas.
Volume occupied by 1 mole of gaseous carbon dioxide at STP = 22.4 dm³
Volume occupied by 4.5 moles of gaseous carbon dioxide at STP = (22.4 $\times$ 4.5) dm³ = 100.8 dm³

(iv) According to the balanced chemical reaction, 4.5 moles of calcium carbonate will form 4.5 moles of calcium chloride.
Molecular mass of calcium chloride = 111 a.m.u.
Mass of one mole of calcium chloride = 111 a.m.u.
Mass of 4.5 moles of calcium chloride = (4.5 $\times$ 111) a.m.u. = 499.5 a.m.u.

(v) According to the balanced chemical reaction, for one mole of calcium carbonate reacted, two moles of HCl are used.
Number of moles of HCl used by 4.5 moles of calcium carbonate = (2 $\times$ 4.5) mol = 9 mol