Question

3- 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?

Answer 1

Complete decomposition reaction of the mixture can be writen as follows,

Al2(CO3)3 + CaCO3 ------------> Al2O3 + CaO + 4CO2

Molecular weight of Al2(CO3)3 =233.9898 g/mol

Molecular weight of CaCO3 = 100.0869 g/mol

total weight of the mixture given as 8.68 g, therefore number of moles of mixture = 8.46/(233.9898 +100.0869) = 0.0259 moles

now from above equation we can say that 1 mole of mixture gives 4 moles of CO2

so 0.0259 moles of mixture will give 0.1012 moles of CO2

We can calculate volume of gas libearted at from the equation PV = nRT,

R = 62.3637 L·Torr/mol·K or L·mmHg/mol·K

V = [0.1012*62.3637*(800+273.15)]/752 =9.0064 L

now to this vessle after completion of decomposition 246.2 ml i.e 0.2462 L of HCl is added.

So total volume of reaction mixture will be 9.0064+0.2462 = 9.25 L

9.25 L is the value for maximum volume of the reaction vessle.