The reaction CaCO_3(s) CaO(s) + CO_2(g) has K degree = 0.244 at 800 degree C. A

Question

The reaction CaCO_3(s) CaO(s) + CO_2(g) has K degree = 0.244 at 800 degree C. A 4.00-L vessel at 800 degree C initially contains only CO_2 (g) at pressure P. If 0.500 g of CaO(s) is added to the container, find the equilibrium amounts of CaCO_3 (s). CaO(s), and CO_2 (g) if the initial CO_2 pressure P is (a) 100 torr; (b) 195 torr; (c) 659 torr. R= 82.06 cm^3.atm/mol.K. T(K) = T(degree C) + 273.1. Assume ideal gas behavior. Please enter the equilibrium amounts in mol in 3 significant figures, e.g. if the answer is 0.09004 mol. enter '0.0900'. a) nCaCO_3 = mol, nCaO = mol, nCO_2 = mol. b) nCaCO_3 = mol, nCaO = mol = nCO_2 = mol. c) nCaCO_3 = mol, nCaO= mol, nCO_2= mol.

Explanation / Answer

Let ua assume the Ideal gas behaviour,

So

Pv = n RT

Answer:

a) P = 100 torr

1 Torr = 0.00131579 atm, therefore 100 torr = 0.131579 atm

V = 4 L = 4000 cm3

R = 82.06 cm3. atm / mol. K (Given / Gas Constant)

T = 800 ° C = 800 + 273.1 = 1073.1 K

n = Pv / RT = 0.131579 atm * 4000 cm3 / ( 82.06 (cm3. atm / mol. K) * 1073.1 K) = 0.0059 moles of CO2

From the reaction, 1 mole of CaCO3 forms 1 mole of CO2 and CaO,

So, 0.0059 moles of CO2 formed by 0.0059 moles of CaCO3 and CaO.

In mass basis:

= 0.0059 * 100 (MW of CaCO3) = 0.59 g of CaCO3

= 0.0059 * 56 (MW of CaO) = 0.3304 g of CaO + 0.5 g (added to the container) = 0.8304 g of CaO.

In the same procedure, we have to do for rest of the pressures.

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