Respuesta :
The CO₂ increases the mass of CaCO₃(s) and CaO in the container based on the equilibrium constant to give approximately 10.65 g of CaCO₃(s).
Response:
The total mass of CaCO₃(s) after equilibrium is reestablished approximately 10.65 grams
What is the equation for equilibrium constant?
The given reaction is presented as follows;
Number of moles of CaCO₃(s) = 0.100 mol
Number of moles of CaO(s) = 0.100 mol
Volume of the container = 10.0 L
Temperature to which the container is heated = 385 K
Pressure of the CO₂ after equilibrium = 0.220 atm
The chemical reaction is presented as follows;
CaCO₃(s) ⇌ CaO(s) + CO₂(g)
Additional CO₂(g) added = 0.240 atm.
Solution:
P·V = n·R·T
Which gives;
Therefore;
Concentration of the CO₂ = [CO₂] = 0.0696 ÷ 10 = 0.00696
The number of moles of CaCO₃(s) after the reaction = 0.1 - 0.0696 = 0.0304
The number of moles of CaO(s) after the reaction = 0.1 + 0.0696 = 0.1696
Number of moles in 0.24 atm of CO₂ is found as follows;
The number of moles of CaCO₃ after the reaction is therefore;
= 0.0304 + 0.076 = 0.1064
The molar mass of CaCO₃ = 100.0869 g/mol
Which gives;
Mass of CaCO₃ = 100.0869 g/mol × 0.1064 moles ≈ 10.65 g
The total mass of CaCO₃(s) after equilibrium ≈ 10.65 grams
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