At \(900^{\circ} \mathrm{C}, K_{c}=0.0108\) for the reaction $$\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$$ A mixture of \(\mathrm{CaCO}_{3}, \mathrm{CaO},\) and \(\mathrm{CO}_{2}\) is placed in a \(10.0-\mathrm{L}\) vessel at \(900^{\circ} \mathrm{C}\). For the following mixtures, will the amount of \(\mathrm{CaCO}_{3}\) increase, decrease, or remain the same as the system approaches equilibrium? (a) \(15.0 \mathrm{~g} \mathrm{CaCO}_{3}, 15.0 \mathrm{~g} \mathrm{CaO},\) and \(4.25 \mathrm{~g} \mathrm{CO}_{2}\) (b) \(2.50 \mathrm{~g} \mathrm{CaCO}_{3}, 25.0 \mathrm{~g} \mathrm{CaO},\) and \(5.66 \mathrm{~g} \mathrm{CO}_{2}\) (a) \(30.5 \mathrm{~g} \mathrm{CaCO}_{3}, 25.5 \mathrm{~g} \mathrm{CaO},\) and \(6.48 \mathrm{~g} \mathrm{CO}_{2}\)

To calculate the initial concentration of \(\mathrm{CO}_{2}\), first convert the mass of \(\mathrm{CO}_{2}\) into moles using the molar mass of \(\mathrm{CO}_{2} = 12.01 + 16.00\times2 = 44.01\ \mathrm{g/mol}\). Then divide the number of moles by the volume of the reaction vessel, which is 10.0 L.

After calculating the initial concentration of \(\mathrm{CO}_{2}\), plug into the equation for \(Q_c\) for the given reaction: $$ Q_c = \frac{[\mathrm{CO}_{2}]}{[\mathrm{CaO}] ([\mathrm{CaCO}_{3}])}$$ We can ignore the concentrations of solids (\((s)\), such as \(\mathrm{CaO}\) and \(\mathrm{CaCO}_3\)) in calculating \(Q_c\) as they don't affect the equilibrium. Thus, we can simplify our comparison to whether the concentration of the product, \([\mathrm{CO}_{2}]\), is greater than, equal to, or less than \(K_c\).

Based on the comparison between \(Q_c\) and \(K_c\), we can determine what will happen to the amount of \(\mathrm{CaCO}_{3}\): - If \(Q_c > K_c\), the reaction will proceed towards the left to achieve equilibrium and the amount of \(\mathrm{CaCO}_{3}\) will increase. - If \(Q_c = K_c\), the reaction is at equilibrium, and the amount of \(\mathrm{CaCO}_{3}\) will remain constant. - If \(Q_c < K_c\), the reaction will proceed towards the right to achieve equilibrium, and the amount of \(\mathrm{CaCO}_{3}\) will decrease. We repeat the steps for the three situations given in the exercise: (a) 1. Convert mass to moles for CO2: \(4.25\ \mathrm{g\ CO_{2}} \times \frac{1\ \mathrm{mol\ CO_{2}}}{44.01\ \mathrm{g\ CO_{2}}} = 0.0966\ \mathrm{mol\ CO_{2}}\) 2. Calculate the concentration of CO2: \(\frac{0.0966\ \mathrm{mol\ CO_{2}}}{10\ \mathrm{L}}=0.00966\ \mathrm{M}\) 3. Compare \(Q_c\) and \(K_c\): \(0.00966 < 0.0108\), so the amount of \(\mathrm{CaCO}_{3}\) will decrease. (b) 1. Convert mass to moles for CO2: \(5.66\ \mathrm{g\ CO_{2}} \times \frac{1\ \mathrm{mol\ CO_{2}}}{44.01\ \mathrm{g\ CO_{2}}} = 0.1286\ \mathrm{mol\ CO_{2}}\) 2. Calculate the concentration of CO2: \(\frac{0.1286\ \mathrm{mol\ CO_{2}}}{10\ \mathrm{L}}=0.01286\ \mathrm{M}\) 3. Compare \(Q_c\) and \(K_c\): \(0.01286 > 0.0108\), so the amount of \(\mathrm{CaCO}_{3}\) will increase. (c) 1. Convert mass to moles for CO2: \(6.48\ \mathrm{g\ CO_{2}} \times \frac{1\ \mathrm{mol\ CO_{2}}}{44.01\ \mathrm{g\ CO_{2}}} = 0.1472\ \mathrm{mol\ CO_{2}}\) 2. Calculate the concentration of CO2: \(\frac{0.1472\ \mathrm{mol\ CO_{2}}}{10\ \mathrm{L}}=0.01472\ \mathrm{M}\) 3. Compare \(Q_c\) and \(K_c\): \(0.01472 > 0.0108\), so the amount of \(\mathrm{CaCO}_{3}\) will increase.