At \(125^{\circ} \mathrm{C}, K_{\mathrm{p}}=0.25\) for the reaction $$2 \mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$.A 1.00 -L flask containing \(10.0 \mathrm{g}\) NaHCO \(_{3}\) is evacuated and heated to \(125^{\circ} \mathrm{C}\) a. Calculate the partial pressures of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) after equilibrium is established. b. Calculate the masses of \(\mathrm{NaHCO}_{3}\) and \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) present at equilibrium. c. Calculate the minimum container volume necessary for all of the \(\mathrm{NaHCO}_{3}\) to decompose.

Given the mass of NaHCO3, we need to find its moles by dividing the mass by its molar mass. Molar mass of NaHCO3 = 23 (Na) + 1 (H) + 12 (C) + 16 * 3 (O) = 84 g/mol Initial mass of NaHCO3 = 10 g So, the initial moles of NaHCO3 = 10 g ÷ 84 g/mol = \( \frac{5}{42}\) mol

We know the Kp for the given reaction at 125°C. We can create the equilibrium expression according to the balanced equation as follows: \[ K_{p} = \dfrac{P_{CO_{2}} \times P_{H_{2}O}}{(P_{NaHCO_{3}})^{2}}\] Let x represent the partial pressures of CO2 and H2O (since their stoichiometry is 1:1). Since the NaHCO3 is a solid, its activity does not depend on its partial pressure, and we will assume P_NaHCO3 = 1. At equilibrium: Kp = 0.25 The equilibrium expression becomes: \( 0.25 = \dfrac{x^2 \times 1}{1^2}\)

We can use the equation derived in step 2 to find the partial pressures: \[ x^2 = 0.25\] Solving for x we get, \[ x = \sqrt{0.25} = 0.5 \] So, the partial pressures of CO2 and H2O at equilibrium are 0.5 atm.

We can use the ideal gas law to find the moles of CO2 and H2O at equilibrium: \[ PV = nRT\] Given the volume of the flask (1.00 L) and the temperature (125°C = 398.15K), we can calculate the moles: \[ n_{CO_{2}} = n_{H_{2}O} = \dfrac{P_{CO_{2}} V_{flask}}{RT} =\dfrac{0.5\, atm \times 1.00\, L}{0.0821\, L\, atm\, mol^{-1} K^{-1} \times 398.15\, K}\] n_CO2 = n_H2O = 0.01518 mol

Since for every mole of CO2 and H2O produced, 2 moles of NaHCO3 decompose, we can calculate the remaining moles of NaHCO3 and moles of Na2CO3: Moles of NaHCO3 decomposed = 2 * (moles of CO2) = 0.03036 mol Remaining moles of NaHCO3 = initial moles of NaHCO3 - moles of NaHCO3 decomposed = \( \frac{5}{42}\) - 0.03036 = 0.08929 mol Moles of Na2CO3 = moles of CO2 = 0.01518 mol

Now that we have the moles of NaHCO3 and Na2CO3 at equilibrium, we can find their masses: Mass of NaHCO3 = moles of NaHCO3 * molar mass of NaHCO3 = 0.08929 mol * 84 g/mol = 7.50 g Molar mass of Na2CO3 = 2 * 23 (Na) + 12 (C) + 16 * 3 (O) = 106 g/mol Mass of Na2CO3 = moles of Na2CO3 * molar mass of Na2CO3 = 0.01518 mol * 106 g/mol = 1.61 g

Finally, we will calculate the minimum volume required for complete decomposition of NaHCO3. For this, we should find the volume needed to convert all of the initial NaHCO3 into Na2CO3: Initial moles of NaHCO3 = \( \frac{5}{42} \) mol 2 moles of NaHCO3 produce 1 mole of CO2, so the moles of CO2 produced = \( \frac{5}{84} \) mol Now we can calculate the minimum volume required for complete decomposition using the ideal gas law at 125°C: \[ V = \dfrac{nRT}{P} = \dfrac{(\frac{5}{84}\,mol) \times (0.0821\, L\,atm\,mol^{-1}\,K^{-1}) \times (398.15\,K)}{0.5\, atm}\] Minimum volume required = 4.00 L In summary, the results for each part are: a. Partial pressures of CO2 and H2O at equilibrium are 0.5 atm. b. Masses of NaHCO3 and Na2CO3 at equilibrium are 7.50 g and 1.61 g, respectively. c. The minimum container volume necessary for complete decomposition of NaHCO3 is 4.00 L.