Question

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

(i) Write the equation for the reaction.

(ii) wWhat 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 as STP?

(iv) What mass of calcium chloride is formed? (Relative molecular mass of calcium carbonate is 111.)

(v) How many moles of HCI are used in this reaction?

Answer 1

(i) The equation for the reaction:

${\mathrm{CaCO}}_{3\left(\mathrm{s}\right)}+2{\mathrm{HCl}}_{\left(\mathrm{aq}\right)}\to {\mathrm{CaCl}}_{2\left(\mathrm{aq}\right)}+{\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)}+{\mathrm{CO}}_{2\left(\mathrm{g}\right)}$
(ii) Molecular mass of ${\mathrm{CaCO}}_3$= 100 a.m.u.
Mass of one mole of ${\mathrm{CaCO}}_3$= 100 a.m.u.
Mass of 4.5 moles of ${\mathrm{CaCO}}_3$= (4.5 × 100) a.m.u. = 450 a.m.u.

Hence, the mass of 4.5 moles of calcium carbonate is 450 a.m.u.
(iii) According to the balanced chemical reaction:
4.5 moles of ${\mathrm{CaCO}}_3$ will liberate 4.5 moles of + $C{O}_2$ gas.
The volume occupied by 1 mole of + $C{O}_2$ at STP = 22.4 dm³
The volume occupied by 4.5 moles of + $C{O}_2$ at STP = (22.4 × 4.5) dm³
= 100.8 dm³

Hence, 100.8 dm³ is the volume of carbon dioxide liberated as STP
(iv) According to the balanced chemical reaction:
4.5 moles of ${\mathrm{CaCO}}_3$ will form 4.5 moles of ${\mathrm{CaCl}}_2$.
Molecular mass of ${\mathrm{CaCl}}_2$= 111 a.m.u.
Mass of one mole of ${\mathrm{CaCl}}_2$= 111 a.m.u.
Mass of 4.5 moles of ${\mathrm{CaCl}}_2$= (4.5 × 111) a.m.u.
= 499.5 a.m.u.

Hence, the mass of calcium chloride is 499.5 a.m.u

(v) According to the balanced chemical reaction:
1 mole of ${\mathrm{CaCO}}_3$ reacted and 2 moles of $\mathrm{HCl}$ are used.
Number of moles of HCl used by 4.5 moles of ${\mathrm{CaCO}}_3$ = (2 × 4.5) mol
= 9 mol

Hence, 9 moles of $\mathrm{HCl}$ are used in this reaction.