When 0.100 mol CaCO3(s) and 0.100 mol CaO(s) are placed
in an evacuated sealed 10.0 L container and heated to 385 K, P(CO2) = 0.220 atm after equilibrium is established.
CaCO3(s) <~> CaO(s) + CO2(g)
0.05 atm CO2(g) is then removed from the container. What is the total mass (in g) of CaCO3 after equilibrium is reestablished?
_____ g
To solve this problem, we can use the ideal gas law and the stoichiometry of the given reaction.
Step 1: Calculate the initial moles of CO2(g)
Given that the initial pressure of CO2(g) is 0.220 atm, and the volume of the container is 10.0 L, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure of CO2(g) = 0.220 atm
V = volume of the container = 10.0 L
n = moles of CO2(g) (unknown)
R = ideal gas constant = 0.0821 L·atm/(mol·K)
T = temperature in Kelvin = 385 K
Rearranging the equation, we have:
n = PV / RT
Substituting the given values:
n = (0.220 atm) * (10.0 L) / (0.0821 L·atm/(mol·K) * 385 K)
n ≈ 0.0672 mol
Step 2: Determine the coefficient ratio between CO2 and CaCO3
Looking at the balanced equation:
CaCO3(s) ⇌ CaO(s) + CO2(g)
We can see that the stoichiometric coefficient of CO2 is 1, meaning that for every mole of CaCO3 reacted, 1 mole of CO2 is produced.
Step 3: Calculate the moles of CaCO3 reacted
Since the moles of CO2 removed is given as 0.05 atm, we can use the ideal gas law to calculate its moles:
n = PV / RT
n = (0.05 atm) * (10.0 L) / (0.0821 L·atm/(mol·K) * 385 K)
n ≈ 0.0155 mol
Since the stoichiometric coefficient of CaCO3 is also 1, this means that 0.0155 mol of CaCO3 reacted.
Step 4: Calculate the remaining moles of CaCO3
To find the remaining moles of CaCO3 after equilibrium is reestablished, we subtract the moles of CaCO3 reacted from the initial moles of CaCO3:
Remaining moles of CaCO3 = Initial moles of CaCO3 - Moles of CaCO3 reacted
Remaining moles of CaCO3 = 0.100 mol - 0.0155 mol
Remaining moles of CaCO3 ≈ 0.0845 mol
Step 5: Calculate the total mass of CaCO3
The molar mass of CaCO3 is:
Ca = 40.08 g/mol
C = 12.01 g/mol
O = 16.00 g/mol
Molar mass of CaCO3 = (1 * Ca) + (1 * C) + (3 * O)
Molar mass of CaCO3 = (1 * 40.08 g/mol) + (1 * 12.01 g/mol) + (3 * 16.00 g/mol)
Molar mass of CaCO3 ≈ 100.09 g/mol
To find the total mass of CaCO3, we multiply the remaining moles of CaCO3 by the molar mass:
Total mass of CaCO3 = Remaining moles of CaCO3 * Molar mass of CaCO3
Total mass of CaCO3 = 0.0845 mol * 100.09 g/mol
Total mass of CaCO3 ≈ 8.457 g
Therefore, the total mass of CaCO3 after equilibrium is reestablished is approximately 8.457 g.
To find the total mass of CaCO3 after equilibrium is reestablished, we need to use the stoichiometry of the reaction and the ideal gas law.
First, let's calculate the initial number of moles of CO2 in the container. We know that the initial pressure of CO2 (P(CO2)) is 0.220 atm, and the volume (V) is 10.0 L. Using the ideal gas law equation (PV = nRT), we can rearrange it to solve for moles (n):
n = PV / RT
R represents the ideal gas constant, which is 0.0821 L·atm/mol·K, and T is the temperature in Kelvin (385 K). Plugging in the values:
n = (0.220 atm)(10.0 L) / (0.0821 L·atm/mol·K)(385 K) = 0.066 mol CO2
Now, since the reaction is in a sealed container, removing CO2 will cause a shift to the right, resulting in more CaCO3 being converted to CaO and CO2 to reestablish equilibrium.
From the balanced equation, we know that 1 mol of CaCO3 produces 1 mol of CaO and 1 mol of CO2.
Since 0.066 mol of CO2 was removed, 0.066 mol of CaCO3 must react to produce that amount of CO2. Therefore, the total number of moles of CaCO3 after equilibrium is:
0.100 mol - 0.066 mol = 0.034 mol
To find the total mass of CaCO3, we need to multiply the number of moles by its molar mass. The molar mass of CaCO3 is:
40.08 g/mol (Ca) + 12.01 g/mol (C) + (3 * 16.00 g/mol) (O) = 100.09 g/mol
Finally, calculating the mass:
Mass = Number of moles x Molar mass
Mass = (0.034 mol) x (100.09 g/mol) = 3.4026 g
Therefore, the total mass of CaCO3 after equilibrium is reestablished is approximately 3.4026 grams.
I would do this.
...........CaCO3 ==> CaO(s) + CO2
I.........0.1 mol...0.1 mol...0
C...........-x.......+x.......+x
E..........0.1-x....01+x......+x
You know x is 0.220 atm. Use PV = nRT and solve for n = number of mols. x = approx 0.07 mols and (CO2) = approx 0.07/10L = approx 0.007 M.
Do another ICE chart starting the equilibrium conditions. Removing 0.05 atm CO2 means the equilibrium will shift to the right and mols CO2 removed is approx 0.016 (again, you need to redo all this numbers) and the approx 0.016 again comes from PV = nRT.
Plug those values into the new ICE chart and calculate (CO2) from the Kc value shown above. Convert to mols and subtract from the new value of CaCO3. That should give you the mols CaCO3 remaining. Convert mols CaCO3 to grams.
Post your work if you get stuck.