When 0.100 mol of CaCO3 (s) and 0.100 mol of CaO (s) are placed in an evacuated container with a volume of 10.0-L and heated to 385 K, PCO2 = 0.220 atm after equilibrium is established: CaCO3(s) ⇌ CaO (s) + CO 2 (g) An additional 0.300 atm of CO2 (g) is pumped in. What is the total mass (in grams) of CaCO3 after equilibrium is re-established?
This problem seems too simple. If I've screwed up I'm sure another tutor will let me know.
CaCO3(s) ==> CaO(s) + CO2(g)
Kp = pCO2 = 0.22 atm.
So if we add 0.3 atm EXTRA, the reaction must shift to use up the extra 0.3.
Use PV = nRT and solve for n = number of moles CO2 and that will be the as moles CaCO3. Then add the extra moles toltghe 0.100 mol to start to find total CaCO3 after equilibrium is re-established.
To solve this problem, we need to use the ideal gas law equation and the stoichiometry of the reaction.
First, let's calculate the number of moles of CO2 initially present in the container:
n_CO2_initial = PCO2_initial * V / RT
n_CO2_initial = 0.220 atm * 10.0 L / (0.0821 L·atm/mol·K * 385 K)
n_CO2_initial = 0.00608 mol
Since the reaction stoichiometry tells us that 1 mol of CaCO3 produces 1 mol of CO2, we know that 0.00608 mol of CaCO3 reacted initially.
The total number of moles of CaCO3 present after equilibrium is established can be calculated using the following equation:
n_CaCO3_final = n_CaCO3_initial - n_CO2_initial
n_CaCO3_final = 0.100 mol - 0.00608 mol
n_CaCO3_final = 0.09392 mol
Finally, we can calculate the mass of CaCO3 using the molar mass of CaCO3:
mass_CaCO3 = n_CaCO3_final * molar mass_CaCO3
The molar mass of CaCO3 is calculated as follows:
molar mass_CaCO3 = (molar mass_Ca + molar mass_C + 3 * molar mass_O)
Substituting the values:
molar mass_CaCO3 = (40.08 g/mol + 12.01 g/mol + 3 * 16.00 g/mol)
molar mass_CaCO3 = 100.09 g/mol
Now we can calculate the mass of CaCO3:
mass_CaCO3 = 0.09392 mol * 100.09 g/mol
mass_CaCO3 ≈ 9.39 g
Therefore, the total mass of CaCO3 after equilibrium is re-established is approximately 9.39 grams.
To find the mass of CaCO3 after equilibrium is re-established, we need to use the ideal gas law equation and the balanced chemical equation. Here's how you can solve it step by step:
1. First, let's calculate the initial pressure of CO2 (g) at equilibrium. The given PCO2 = 0.220 atm is the partial pressure of CO2 when equilibrium is established. Therefore, the initial pressure of CO2 is 0.220 atm.
2. Next, we need to calculate the moles of CO2 at equilibrium. We know the pressure and volume of the container, so we can use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. Rearranging the equation, n = PV / RT.
- P = 0.220 atm (initial pressure of CO2 at equilibrium)
- V = 10.0 L (volume of the container)
- R = 0.0821 L * atm / (mol * K) (ideal gas constant)
- T = 385 K (temperature)
Substitute the values into the equation: n = (0.220 atm * 10.0 L) / (0.0821 L * atm / (mol * K) * 385 K).
Calculate the moles of CO2 at equilibrium.
3. Since the balanced chemical equation tells us that 1 mol of CaCO3 produces 1 mol of CO2, the moles of CaCO3 will be the same as the moles of CO2 in the system. Consequently, the moles of CaCO3 are the same as the moles of CO2 at equilibrium.
4. Calculate the mass of CaCO3 using the molar mass of CaCO3. The molar mass of CaCO3 is the sum of the molar masses of each element present in the compound:
- Molar mass of Ca = 40.08 g/mol
- Molar mass of C = 12.01 g/mol
- Molar mass of O = 16.00 g/mol
Add up the molar masses: molar mass of CaCO3 = 40.08 g/mol + 12.01 g/mol + (16.00 g/mol * 3) = ...
5. Multiply the moles of CaCO3 by the molar mass of CaCO3 to get the mass of CaCO3 in grams.
I hope this explanation helps you understand how to solve the problem!