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

Further explanation

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

Qc = Kc, the reaction equilibrium ---> reactant ⇌ product Qc <Kc, reaction from left to right (product) until it reaches equilibrium (Qc = Kc) ---> reactant → product Qc> Kc, the reaction from right to left (reactant) until it reaches equilibrium (Qc = Kc). If Qc> Kc, reactant ← product

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

Learn more

an equilibrium constant link

link

Calculate the value of the equilibrium constant, Kc

link

Concentration of hi at equilibrium

link

Water isn't wet by itself, but it makes other materials wet when it sticks to the surface of them.