if a 20.0-g sample of caco3 is put into a 10.0-l container and heated to 800.8c, what percentage by mass of the caco3 will react to reach equilibrium?



Answer :

Kp is the equilibrium constant of the reaction in accordance with the partial pressure of the gas. The mass percentage of calcium carbonate is 2.5%.

What is the mass percentage?

The mass percentage is the ratio of the moles reacted to that of the moles present initially multiplied by  100.

The reaction of calcium carbonate can be shown as,

CaCo3 ----------------->  CaO   + CO2

The formula for the equilibrium constant is given as,

Kp= Kc[tex](RT)^{n}[/tex]

Here,

n = 1

Gas constant (R) = 8.314 J /mol K

Temperature (T) = 1073 K

Kp = 1.16

      = 1.3 x [tex]10^{-4}[/tex]

Substituting values in the above equation:

             x =0.005

From the ICE table of the reaction the value of moles of calcium carbonate (x) is calculated as:

Mass% = 0.005/0.2x 100%

              2.5%

             

Therefore, 2.5% is the mass percentage.

Learn more about mass percentage from given link

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