For the reaction below, Kp = 1.16 at 800.°C.

CaCO3(s) CaO(s) + CO2(g)
If a 21.7-g sample of CaCO3 is put into a 10.6 L container and heated to 800°C, what percentage by mass of the CaCO3 will react to reach equilibrium?

Try this.

Kp = 1.16 = pCO2

How many mols CO2 can be formed in a 10.6L container @ 800 C.
n = PV/RT = 1.16*10.6/0.08206*1073 = about 0.14 but you need to do it more accurately. You had 21.7/100 = 0.217 mols.
% decomposed = (0.14/0.217)*100 = ?

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

First, let's write the balanced chemical equation:
CaCO3(s) CaO(s) + CO2(g)

Next, we need to convert the given mass of CaCO3 into moles. To do this, we need to know the molar mass of CaCO3, which is calculated by summing the atomic masses of its constituent elements.
Molar mass of CaCO3 = 40.08 g/mol (Ca) + 12.01 g/mol (C) + (3 x 16.00 g/mol) (3 O) = 100.09 g/mol

Now we can calculate the moles of CaCO3:
moles of CaCO3 = mass of CaCO3 / molar mass of CaCO3
moles of CaCO3 = 21.7 g / 100.09 g/mol

Given that the container has a volume of 10.6 L, we can use the ideal gas law to calculate the equilibrium partial pressure of CO2, assuming the reaction goes to completion:
PV = nRT

R is the ideal gas constant (0.0821 L·atm/(mol·K))
T is the temperature in Kelvin (800°C = 1073 K)

Since the number of moles of CO2 produced is equal to the number of moles of CaCO3 reacted, we can use the moles of CaCO3 calculated earlier to find the moles of CO2.

moles of CO2 = moles of CaCO3

Now, we can calculate the partial pressure of CO2 using the ideal gas law:
P = (n * R * T) / V

Substituting the known values into the equation, we can find the partial pressure of CO2.

Finally, we can use the equation for the equilibrium constant (Kp) to find the percentage by mass of CaCO3 that will react to reach equilibrium.

Kp = (P(CO2) / (P(CaCO3))^

Rearranging the equation:
(P(CaCO3))^ = (P(CO2)) / Kp

Now, we can calculate:
percentage by mass of CaCO3 = (moles of CaCO3 remaining / moles of CaCO3 initially) x 100%

To solve this problem, we need to use the ideal gas law and the formula for percentage by mass.

Step 1: Convert the mass of CaCO3 from grams to moles.
The molar mass of CaCO3 is calculated as follows:
Ca = 40.08 g/mol
C = 12.01 g/mol
O = 16.00 g/mol
Adding these up gives us:
40.08 + 12.01 + (16.00 x 3) = 100.08 g/mol
We can now calculate the number of moles of CaCO3:
21.7 g / 100.08 g/mol = 0.2167 mol

Step 2: Convert the volume from liters to moles.
Since we have a 10.6 L container, we can assume that 1 mol of CO2 occupies 22.4 L at standard conditions. We can use this information to calculate the number of moles of CO2:
10.6 L / 22.4 L/mol = 0.4732 mol

Step 3: Set up the equilibrium expression using the given Kp value.
The equilibrium expression for the reaction is:
Kp = [CO2]
Using the ideal gas law, we can relate the partial pressure of CO2 to the number of moles:
PV = nRT,
where:
P = pressure in atm,
V = volume in liters,
n = number of moles,
R = ideal gas constant (0.0821 L.atm/mol.K), and
T = temperature in Kelvin.

Since we know the volume (V), the temperature (T = 800 + 273 = 1073 K), and we have calculated the number of moles of CO2 (n = 0.4732 mol), we can solve for the partial pressure of CO2 (P).
P = nRT / V = (0.4732 mol)(0.0821 L.atm/mol.K)(1073 K) / 10.6 L = 4.825 atm

Step 4: Calculate the reaction quotient (Qp) using the partial pressure of CO2.
The reaction quotient is given by Qp = [CO2].
Qp = 4.825 atm

Step 5: Calculate the percentage by mass of CaCO3 that will react to reach equilibrium.
Percentage by mass = (mass of reactant reacted / initial mass of reactant) x 100%
We need to calculate the mass of CaCO3 that will react. Assuming x grams of CaCO3 reacts, we can set up the following equation based on the stoichiometry of the reaction:
0.2167 mol of CaCO3 reacts to produce 0.2167 mol of CO2.

From the equation, we can see that 1 mole of CaCO3 produced 1 mole of CO2, so the mole ratio is 1:1.

Therefore, 0.2167 mol of CaCO3 reacts to produce 0.2167 mol of CO2.

Using the molar mass of CaCO3 (100.08 g/mol), we can calculate the mass of CaCO3 that reacts:
x g = (0.2167 mol / 1 mol) x (100.08 g/mol) = 21.69 g (approx)

Finally, we can calculate the percentage by mass of CaCO3 that will react:
% mass = (21.69 g / 21.7 g) x 100% = 99.54%

Therefore, approximately 99.54% by mass of the CaCO3 will react to reach equilibrium.