The oxides of Group \(2 \mathrm{~A}\) metals (symbolized by \(\mathrm{M}\) here) react with carbon dioxide according to the following reaction: $$ \mathrm{MO}(s)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{MCO}_{3}(s) $$ A \(2.85-\mathrm{g}\) sample containing only \(\mathrm{MgO}\) and \(\mathrm{CuO}\) is placed in a \(3.00-\) L container. The container is filled with \(\mathrm{CO}_{2}\) to a pressure of 740 . torr at \(20 .{ }^{\circ} \mathrm{C}\). After the reaction has gone to completion, the pressure inside the flask is 390 . torr at \(20 .{ }^{\circ} \mathrm{C}\). What is the mass percent of \(\mathrm{MgO}\) in the mixture? Assume that only the \(\mathrm{Mg} \mathrm{O}\) reacts with \(\mathrm{CO}_{2}\).

We know the initial pressure (740 torr) and final pressure (390 torr) of CO₂ in the container. To find the moles reacted, we can use the ideal gas law. We calculate the difference in moles, as only the reacted CO₂ affects the pressure change. First, we convert the pressures into atmospheres, as the ideal gas law requires pressure in atm. \(P_{initial}\) = \(\frac{740}{760}\) atm \(P_{final}\) = \(\frac{390}{760}\) atm Now, we can use the ideal gas law to find the initial and final moles of CO₂: \(PV = nRT\) Where: P is the pressure V is the volume n is the moles R is the gas constant (0.0821 L atm/mol K) T is the temperature in Kelvin We know the volume (3 L) and temperature (20°C = 293 K) of the container. So, we can now calculate the initial and final moles of CO₂: \(n_{initial} = \frac{P_{initial}V}{RT}\) \(n_{final} = \frac{P_{final}V}{RT}\) And the difference in moles of CO₂: \(Δn_{CO₂} = n_{initial} - n_{final}\)

Now that we have the moles of CO₂ that reacted, we can use stoichiometry to find the moles of MgO in the sample. We know that one mole of MgO reacts with one mole of CO₂: \(MgO + CO₂ \to MgCO₃\) Hence, the moles of MgO in the sample are equal to \(Δn_{CO₂}\) moles. \(n_{MgO} = Δn_{CO₂}\)

We know the moles of MgO and the molar mass of MgO (40.30 g/mol). So, we can find the mass of MgO in the sample: \(mass_{MgO} = n_{MgO} \times Molar_mass_{MgO}\)

Finally, we can calculate the mass percent of MgO in the mixture using the mass of MgO and the given total mass of the mixture (2.85 g): \(mass\%_{MgO} = \frac{mass_{MgO}}{mass_{total}} × 100\%\)