Question

CaCO₃(s) ⇄ CaO(s) + CO₂(g)

0.100 mol of CaCO₃ and 0.100 mol CaO are placed in an 10.0 L evacuated container and heated to 385 K. When equilibrium is reached the pressure of CO₂ is 0.220 atm. 0.300 atm of CO₂ is added, while keeping the temperature constant and the system is allowed to reach again equilibrium. What will be the final mass of CaCO₃?

7.47 g

12.54 g

18.01 g

2.00 g

10.01 g

Answer 1

Answer:

Step 1: write the balanced chemical equation

CaCO₃(s) <-------------> CaO(s) + CO₂(g)

Step 2: Calculate the amount before adding any additional CO2

K = P_co2    [ all other elements are in solid condition so the value will be 1 ]
K = 0.220 atm

By Using  ideal gas formula, we have

n = PV/RT = ( 0.220 atm ×10 L ) / (0.08206 L-atm mol^-1 K^-1 × 385 K) = 0.0696 mol

This is the amount of CaCO3 which has been converted to CaO before pumping-in additional 0.300 atm CO2(g).

Step 3: Calculate the amount after adding the additional CO2

After pumping-in additional 0.300 atm CO2(g), the equilibrium CO2 pressure is still 0.220 atm . Therefore all this additional CO2 would completely convert to CaCO3.

n = PV/RT = ( 0.300 atm ×10 L ) / (0.08206 L-atm mol^-1 K^-1 × 385 K) = 0.0949 mol

Hence the total CaCO3 after equilibrium is re-established is:

0.100 mol - 0.0696 mol + 0.0949 mol = 0. 1253 mol  

Step 4: calculate the final mass of CaCO3

Now convert mole into mass of CaCO3

We know, mass = moles × molar mass of CaCO₃ = 0.1253 mol × 100.0869 g/mol = 12.54 g