Answer:
76.0%
Explanation:
Let's consider the following reaction.
CaCO₃(s) ⇄ CaO(s) + CO₂(g)
At equilibrium, the equilibrium constant Kp is:
Kp = 1.16 = pCO₂ ⇒ pCO₂ = 1.16 atm
We can calculate the moles of CO₂ at equilibrium using the ideal gas equation.
From the balanced equation, we know that 1 mole of CO₂ is produced by 1 mole of CaCO₃. Taking into account that the molar mass of CaCO₃ is 100.09 g/mol, the mass of CaCO₃ that reacted is:
The percentage by mass of the CaCO₃ that reacted to reach equilibrium is: