At 900 ∘c, kc=0.0108 for the reaction caco3(s)←−→cao(s)+co2(g) a mixture of caco3, cao, and co2 is placed in a 10.0-l vessel at 900 ∘c. for the following mixtures, will the amount of caco3 increase, decrease, or remain the same as the system approaches equilibrium? part a 15.0 gcaco3, 15.0 gcao, and 4.25gco2 15.0 , 15.0 , and 4.25 increases decreases remains the same submitrequest answer part b 2.50gcaco3, 25.0 gcao, and 5.66 gco2 2.50, 25.0 , and 5.66 increases decreases remains the same submitrequest answer part c

a. CaCO3 decreases

b. CaCO3 increases

The equilibrium constant is based on the concentration (Kc) in a reaction

pA + qB -----> mC + nD

To find out the reaction reaches equilibrium and the direction of the reaction, the value of the reaction quotient, Qc

equilibrium constants of heterogeneous reactions (more than 1 phase), liquid (L = liquid) and solid (S = solid) phases are ignored and not counted, only gas (G = gas) and solution (Aq = aqueous) phases so that the price of Kc for the reaction of CaCO₃ (s) → - → CaO (s) + CO₂ (g) is only based on concentration [CO₂]

Kc = [CO₂] = 0.0108

Similarly, the determination of Qc is also based on [CO₂]

We determine the mole and concentration of CO₂

molar mass of CO₂ = 44, and a volume of 10 L

a) 15.0 g CaCO₃, 15.0 g CaO, and 4.25g CO₂

4.25 g CO₂ (ignore masses of CaCO₃ and CaO)

mole CO₂= mass: molar mass

mole of CO₂ = 4.25: 44 = 0.0965

concentration = 0.0965: 10 = 0.00965 M

Because Qc = [CO₂] = 0.00965 and Qc <Kc, the reaction is more to the right (product), so CaCO₃ decreases

b) 2.50g CaCO₃, 25.0 g CaO, and 5.66 g CO₂

5.66 g CO₂

mole CO₂ = mass: molar mass

mole of CO₂ = 5.66: 44 = 0.1286

concentration = 0.1286: 10 = 0.01286 M

Because Qc = [CO₂] = 0.01286 and Qc> Kc, the reaction is more towards the left (reactant), so CaCO₃ increases

an equilibrium constant link

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Calculate the value of the equilibrium constant, Kc

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Concentration of hi at equilibrium

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2b2t

explanation:

2b2t

elements make up everything, and they combine to make compounds.

the answer is c