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

Answer – We are given, mass of mixture of aluminum carbonate and calcium carbonate = 8.68 g , P = 752 mm Hg = 0.989 atm

Temp, T = 800 +273 = 1073 K , [HCl] = 0.84 M , volume = 246.2 mL

We know the decomposition of the both carbonate form the CO2 and metal oxide

Al2(CO3)3 -----> Al2O3 + 3 CO2­

CaCO3 -----> CaO + CO3

So metal oxide titrate with 246.2 ml of 0.84 M HCl, so first we need to calculate the moles of HCl

Moles of HCl = 0.84 M * 0.2462 L

                       = 0.207 moles

We know there are total moles of HCl = total moles of oxide reacted

So, moles of oxide = 0.207 moles

There are 2 moles of metal oxide form and 4 moles of CO2

So, 2 moles of metal oxide = 4 moles of CO2

0.207 moles of metal oxide = ?

= 0.827 moles of CO2

Now using the Ideal gas law –

PV = nRT

So, V = nRT/P

         = 0.827 moles *0.0821 L.atm.mol-1.K-1*1073 K / 0.989 atm

           = 73.6 L

the maximum value for the volume of the reaction vessel is 73.6 L

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