Final answer:
The student is asked to calculate the minimum mass of CaCO3 required to achieve equilibrium at a given temperature and equilibrium constant. To solve this problem, one must use the ideal gas law and stoichiometry, starting with the Kc value to find the moles of CO2 produced at equilibrium, then converting those moles into grams of CaCO3.
Step-by-step explanation:
To determine the minimum mass of CaCO3 required to establish equilibrium in a 7.14-L container with a given Kc, we must first understand that the decomposition of CaCO3 is represented by the reaction: CaCO3(s) → CaO(s) + CO2(g). The equilibrium constant (Kc) for this reaction is given as 0.438, which equals the concentration of CO2 at equilibrium since the concentration of solids does not change and is not included in the expression.
To begin, we must calculate the number of moles of CO2 at equilibrium. This can be obtained by using the ideal gas law (PV = nRT), assuming standard conditions where P = 1 atm, V = 7.14 L, R = 0.0821 L·atm/mol·K, and T = 298 K (25°C). With those conditions, the moles of CO2 (n) produced would equal to Kc since Kc = [CO2]
After that, we calculate the mass of CaCO3 needed to produce this amount of CO2. The molar mass of CaCO3 is 100.1 g/mol, so we can find the mass by multiplying the moles of CO2 by the molar mass of CaCO3. Here, we will not provide the exact calculations and values, but the method of calculating it is described. Once you do the math, this will give you the mass of CaCO3 in grams.