For the reaction below, kp 5 1.16 at 800.8c. caco3(s) 34 cao(s) 1 co2(g) 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?

The mass percentage of calcium carbonated reacted is 2.5%.

Explanation:

The reaction is:

Thus the Kp of the equilibrium will be:

Kp = partial pressure of carbon dioxide [as the other are solid]

Moles of calcium carbonate initially present =

Let us apply ICE table to the equilibrium given:

Initial                       0.2                       0          0

Change                 -x                            +x        +x

Equilibrium           0.2-x                         x          x

Kp = partial pressure of carbon dioxide

Kp = Kc(RT)ⁿ

where n = difference in the number of moles of gaseous products and reactants

for given reaction n = 1

R = gas constant = 8.314 J /mol K

T = temperature = 800 ⁰C = 1073 K

Putting values

Kc =

Kc =

On calculating

x =  0.005

where x = the moles of calcium carbonate dissociated or reacted.

Percentage of the moles or mass reacted = %

- both b and d.

-since b and d are the closest coastlines then that makes it the most affected.

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