Question

For the reaction below, Kp 5 1.16 at 800.8C. CaCO3(s) 34 CaO(s) 1 CO2(g) If a 20.0-g sample of CaCO3 is put into a 10.0-L container and heated to 800.8C, what percentage by mass of the CaCO3 will react to reach equilibrium?

Answer 1

To calculate the percentage by mass of CaCO3 that will react to reach equilibrium, we need to use the equilibrium constant (Kp) and the given conditions. By converting the mass of CaCO3 to moles and then determining the moles of CO2 produced, we can calculate the percentage by mass. Approximately 43.8% by mass of CaCO3 will react to reach equilibrium.

Step-by-step explanation:

To calculate the percentage by mass of CaCO3 that will react to reach equilibrium, we need to use the equilibrium constant (Kp) and the given conditions.

First, we convert the mass of CaCO3 to moles using its molar mass (100.1 g/mol). The number of moles of CaCO3 is then used to determine the moles of CO2 produced, since the reaction has a 1:1 mole ratio. Finally, we divide the mass of CO2 produced by the initial mass of CaCO3 and multiply by 100 to calculate the percentage.

Using the given data, we have:

1. Mass of CaCO3 = 20.0 g
2. Molar mass of CaCO3 = 100.1 g/mol
3. Moles of CaCO3 = mass / molar mass = 20.0 g / 100.1 g/mol = 0.199 moles
4. Moles of CO2 = 0.199 moles
5. Mass of CO2 = moles x molar mass = 0.199 moles x 44.01 g/mol = 8.77 g
6. Percentage by mass = (mass of CO2 / initial mass of CaCO3) x 100 = (8.77 g / 20.0 g) x 100 = 43.8%

Therefore, approximately 43.8% by mass of CaCO3 will react to reach equilibrium.