Question

An 8.68 gram sample of a mixture of aluminum carbonate and calcium carbonate is placed in a vessel equipped with a massless, frictionless piston. The initial volume can be assumed to be zero. The sample is heated to 800^oC. This results in complete decomposition to the metal oxides and carbon dioxide. After the reaction, the metal oxides are titrated to equivalence with 246.2 ml of 0.84 M HCl. If the atmospheric pressure is 752 mm Hg, what is the maximum value for the volume of the reaction vessel?

Answer 1

Consider that both metal carbonates decompose to give metal oxides (Al₂O₃ and CaO) at 800 °C.

Metal oxides reacts with HCl to give AlCl₃ and CaCl₂ .

(Note: CaCO₃ decomposes completely above 825 °C).

Here, the data of mass of sample is not required for the calculation of vessel. Volume of vessel increased is due to amount of CO₂ liberated at 800 °C. Here is the relation,

Moles of CO₂ liberated = Moles of CO₃^{– 2} = moles of O ^{– 2} = 2 moles of Cl ^{– 1} replaced in oxide salts

(note: 2 moles of Cl ^-1 replaces one mole of oxide in metal and which is equivalent to amount of CO₂ leberated)

Thus, Moles of CO₂ liberated = ½ (moles of HCl reacted) = ½ (246.2 x 0.84) = 0.1034 moles

Presuur of CO₂ = 752 mm of Hg,

R = 62.3 litre-mm of Hg/K^-1.mole^-1.

T = 800 °C = 1073 K

Volume of CO₂, V = nRT/P (Ideal gas equation)

                                = (0.1034 x 62.3 x 1073) / 752

                               = 9.19 litres

                                = 9190 mL = maximum volume of vessel at 800 °C and 752 mm of Hg