For the reaction below, kp = 1.16 at 800.°c. caco3(s) equilibrium reaction arrow cao(s) + co2(g) if a 25.0-g sample of caco3 is put into a 14.4 l container and heated to 800°c, what percentage by mass of the caco3 will react to reach equilibrium?

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:

answer: they produce more slow than unicellular organisms.