A chemist weighed out 5.14 g of a mixture containing unknown amounts of \(\mathrm{BaO}(s)\) and \(\mathrm{CaO}(s)\) and placed the sample in a \(1.50-\mathrm{L}\) flask containing \(\mathrm{CO}_{2}(g)\) at $30.0^{\circ} \mathrm{C}\( and 750 . torr. After the reaction to form BaCO_ \)_{3}(s)$ and \(\mathrm{CaCO}_{3}(s)\) was completed, the pressure of \(\mathrm{CO}_{2}(g)\) remaining was 230 . torr. Calculate the mass percentages of \(\mathrm{CaO}(s)\) and \(\mathrm{BaO}(s)\) in the mixture.

Firstly, we need to find the moles of \(\mathrm{CO}_{2}(g)\) that have reacted. We can use the ideal gas law to do this, which states: \(PV = nRT\) where: P: pressure (atm) V: volume (L) n: moles of gas molecules R: gas constant = 0.0821 L atm/mol K T: temperature (K) First, we need to convert the pressure from torr to atm, and temperature from Celsius to Kelvin. \(P_{1} = (750\,\text{torr})\left(\frac{1\,\text{atm}}{760\,\text{torr}}\right) = 0.9868 \,\text{atm}\) \(T_{1} = 30.0 + 273.15 = 303.15\,\mathrm{K}\) Now, we can determine the moles of \(\mathrm{CO}_{2}(g)\) initially: \(n_{1} = \frac{P_{1}V}{RT_{1}} = \frac{(0.9868\,\text{atm})(1.50\,\text{L})}{(0.0821\,\text{L atm/mol K})(303.15\,\text{K})} = 0.0603\,\text{mol}\) We do the same for the final pressure: \(P_{2} = (230\,\text{torr})\left(\frac{1\,\text{atm}}{760\,\text{torr}}\right) = 0.3026\,\text{atm}\) And the moles of \(\mathrm{CO}_{2}(g)\) after the reaction: \(n_{2} = \frac{P_{2}V}{RT_{1}} = \frac{(0.3026\,\text{atm})(1.50\,\text{L})}{(0.0821\,\text{L atm/mol K})(303.15\,\text{K})} = 0.0187\,\text{mol}\) We can now find the moles of \(\mathrm{CO}_{2}(g)\) that have reacted: \(\Delta n_{\mathrm{CO}_2} = n_{1} - n_{2} = 0.0603\,\mathrm{mol} - 0.0187\,\mathrm{mol} = 0.0416\,\mathrm{mol}\)

Now that we have the moles of \(\mathrm{CO}_{2}(g)\) that reacted, we can set up stoichiometry equations to solve for the moles of \(\mathrm{BaO}(s)\) and \(\mathrm{CaO}(s)\). Let x = moles of \(\mathrm{BaO}(s)\) and y = moles of \(\mathrm{CaO}(s)\). The balanced chemical equations for the reactions are as follows: \(\mathrm{BaO(s)}+\mathrm{CO}_{2}(g) \rightarrow \mathrm{BaCO}_{3}(s)\) \(\mathrm{CaO(s)}+\mathrm{CO}_{2}(g) \rightarrow \mathrm{CaCO}_{3}(s)\) Both \(\mathrm{BaO}(s)\) and \(\mathrm{CaO}(s)\) react with \(\mathrm{CO}_{2}(g)\) in a 1:1 stoichiometric ratio. Therefore, we can create the equations: \(x + y = \Delta n_{\mathrm{CO}_2} = 0.0416\,\mathrm{mol}\) \[137.33x + 56.08y = 5.14\,\mathrm{g} \] Where 137.33 g/mol is the molar mass of \(\mathrm{BaO}(s)\) and 56.08 g/mol is the molar mass of \(\mathrm{CaO}(s)\).

Now, we can solve for x and y using Substitution or Elimination Method. For this example, we will use the Substitution Method. From equation 1, we can isolate x: \(x = 0.0416 - y\) Now, we substitute this expression of x into equation 2: \(\begin{aligned} 137.33(0.0416 - y) + 56.08y = 5.14\,\mathrm{g} \end{aligned}\) Solve for y: \(y \approx 0.0369\,\text{mol}\) Now, substitute the value of y back into the expression for x: \(x \approx 0.0416 - 0.0369 \approx 0.0047\,\text{mol}\)

Now that we have the moles of \(\mathrm{BaO}(s)\) and \(\mathrm{CaO}(s)\), we can determine the mass percentages in the mixture: \(\text{Mass Percent of BaO} = \frac{\text{mass of BaO}}{\text{total mass}} \times 100\) \(\text{Mass Percent of BaO} = \frac{(0.0047\,\text{mol})(137.33\,\text{g/mol})}{5.14\,\text{g}} \times 100 \approx 12.62\%\) \(\text{Mass Percent of CaO} = \frac{\text{mass of CaO}}{\text{total mass}} \times 100\) \(\text{Mass Percent of CaO} = \frac{(0.0369\,\text{mol})(56.08\,\text{g/mol})}{5.14\,\text{g}} \times 100 \approx 87.38\%\) So, the mass percentages of \(\mathrm{CaO}(s)\) and \(\mathrm{BaO}(s)\) in the mixture are approximately 87.38% and 12.62%, respectively.