A 4.00 -g sample of a mixture of \(\mathrm{CaO}\) and \(\mathrm{BaO}\) is placed in a 1.00-L vessel containing \(\mathrm{CO}_{2}\) gas at a pressure of 730 torr and a temperature of \(25^{\circ} \mathrm{C}\). The \(\mathrm{CO}_{2}\) reacts with the \(\mathrm{CaO}\) and \(\mathrm{BaO},\) forming \(\mathrm{CaCO}_{3}\) and \(\mathrm{BaCO}_{3} .\) When the reaction is complete, the pressure of the remaining \(\mathrm{CO}_{2}\) is 150 torr. (a) Calculate the number of moles of \(\mathrm{CO}_{2}\) that have reacted. (b) Calculate the mass percentage of \(\mathrm{CaO}\) in the mixture.

To do this, we use the Ideal Gas Law: \(PV = nRT\), where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. It is important to use consistent units, so we'll use P in atm, V in L, n in moles, R = 0.0821 Latm/molK, and T in K. First, convert the pressures from torr to atm: - Initial pressure: \(730 \, \frac{torr}{1} = 730 \, \frac{torr}{1} \cdot \frac{1 \, atm}{760 \, torr} = 0.9605 \, atm\) - Final pressure: \(150 \, \frac{torr}{1} = 150 \, \frac{torr}{1} \cdot \frac{1 \, atm}{760 \, torr} = 0.1974 \, atm\) Next, convert the temperature from Celsius to Kelvin: - Temperature: \(25^{\circ}C + 273.15 = 298.15 K\) Now, using the Ideal Gas Law, we can find the initial moles (n_initial) and final moles (n_final) of CO2: Initial moles of CO₂ (\(n_{initial}\)): \(n_{initial} = \frac{P_{initial} V}{RT} = \frac{0.9605 \, atm \cdot 1.00 \, L}{0.0821 \frac{L \, atm}{mol \, K} \cdot 298.15 \, K} = 0.0394 \, mol\) Final moles of CO₂ (\(n_{final}\)): \(n_{final} = \frac{P_{final} V}{RT} = \frac{0.1974 \, atm \cdot 1.00 \, L}{0.0821 \frac{L \, atm}{mol \, K} \cdot 298.15 \, K} = 0.00807 \, mol\)

We can now find the moles of CO2 that have reacted by taking the difference between the initial and final moles of CO2: \(\Delta n = n_{initial} - n_{final} = 0.0394 \, mol - 0.00807 \, mol = 0.0313 \, mol\)

One mole of CO2 reacts with one mole of CaO or one mole of BaO, resulting in one mole of CaCO3 or BaCO3, respectively. We can use stoichiometry to find the mass of reacted CaO and BaO. Let x be the mass of reacted CaO and (4.00 - x) be the mass of reacted BaO. \(\frac{x}{56.08} + \frac{4.00 - x}{153.33} = 0.0313 \, mol\) where 56.08 g/mol is the molar mass of CaO, and 153.33 g/mol is the molar mass of BaO.

Now, we solve the equation in Step 3 for x (mass of CaO): \(x = 2.18 \, g\)

Finally, we can find the mass percentage of CaO in the mixture: Mass percentage of CaO = \(\frac{2.18 \, g}{4.00 \, g} \times 100 \% = 54.5\%\)