An 8.68 gram sample of a mixture of aluminum carbonate and calcium carbonate is

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 800oC. 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?

Explanation / Answer

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

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.