Final answer:
To find the minimum required mass of CaCO3 to establish equilibrium, we used the equilibrium constant and the volume of the container to calculate the moles of CO₂ produced, which is equal to the moles of CaCO3 required. With the molar mass of CaCO3, we then calculated the mass. However, none of the provided options matched the calculated mass.
Step-by-step explanation:
The question asks for the minimum mass of CaCO3 required to establish equilibrium in a 6.50-L container, given that the equilibrium constant (Kc) for the decomposition reaction of CaCO3 is 0.50. The balanced equation for the decomposition of CaCO3 is:
CaCO3(s) → CaO(s) + CO₂(g)
Given that Kc = [CO₂] for the decomposition, and since the solids do not appear in the expression for Kc, to find the moles of CO₂ at equilibrium, we use the expression:
Kc = [CO₂]
Thus, the equilibrium concentration [CO₂] is 0.50 M. To find the number of moles of CO₂ in a 6.50 L container:
Moles CO₂ = 0.50 mol/L × 6.50 L = 3.25 mol
Since CaCO3 decomposes to CaO and CO₂ in a 1:1 ratio, the moles of CaCO3 required for the reaction is also 3.25 mol. The molar mass of CaCO3 is:
Molar mass CaCO3 = 40.08 (Ca) + 12.01 (C) + 3 × 16.00 (O) = 100.09 g/mol
The minimum mass of CaCO3 is then calculated by multiplying the moles of CaCO3 needed by the molar mass:
Minimum mass CaCO3 = 3.25 mol × 100.09 g/mol = 325.2925 g
However, all the options provided are less than the calculated mass. This may indicate a problem with the provided options or the need to approximate down to the smallest available option.
The closest option to the calculated mass, which is still an over-approximation, is option (d) 13.00 g. Therefore, with the given options, we would choose the highest available mass while acknowledging it is not the exact calculated value.