Magnesium can be used as a "getter" in evacuated enclosures to react with the last traces of oxygen. (The magnesium is usually heated by passing an electric current through a wire or ribbon of the metal.) If an enclosure of $5.67 \mathrm{~L}\( has a partial pressure of \)\mathrm{O}_{2}\( of \)7.066 \mathrm{mPa}\( at \)30^{\circ} \mathrm{C}$, what mass of magnesium will react according to the following equation? $$2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s).$$

We will use the ideal gas law to find the moles of oxygen gas: \(PV = nRT\), where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. We are given the pressure (\(P = 7.066 \,\text{mPa}\)), the volume (\(V = 5.67 \,\text{L}\)), and the temperature (\(T = 30^{\circ} \mathrm{C}\)). Note that we need to convert these values into appropriate units. First, let's convert the pressure from mPa to atm: \(1\, \text{Pa} = 9.869 \times 10^{-6}\, \text{atm}\) and \(1\, \text{mPa} = 10^{-3}\, \text{Pa}\), so: $$ P = 7.066 \,\text{mPa} \times \frac{10^{-3}\, \text{Pa}}{1\, \text{mPa}} \times \frac{9.869 \times 10^{-6}\, \text{atm}}{1\, \text{Pa}} \approx 6.98 \times 10^{-8} \,\text{atm}. $$ Now, we need to convert the temperature from Celsius to Kelvin: \(T_{K} = T_{C} + 273.15\), so: $$ T_{K} = 30^{\circ}\mathrm{C} + 273.15 \approx 303.15\,\text{K}. $$ Now, we can plug in the values into the ideal gas law equation and find the moles of oxygen gas, using \(R = 0.0821 \frac{\text{L}\, \text{atm}}{\text{mol}\, \text{K}}\): $$ n_{O_{2}} = \frac{PV}{RT} = \frac{(6.98 \times 10^{-8}\,\text{atm})(5.67\,\text{L})}{(0.0821\, \frac{\text{L}\, \text{atm}}{\text{mol}\, \text{K}})(303.15\,\text{K})} \approx 1.71 \times 10^{-9}\, \text{mol}. $$

Using the stoichiometry of the given chemical equation, we can find the moles of magnesium needed to react with the moles of oxygen gas: $$ 2 \mathrm{Mg} + \mathrm{O}_{2} \longrightarrow 2 \mathrm{MgO}. $$ Since 2 moles of magnesium react with 1 mole of oxygen gas, we apply the stoichiometry and find the moles of magnesium needed: $$ n_{Mg} = 2 \times n_{O_{2}} = 2 \times (1.71 \times 10^{-9}\, \text{mol}) \approx 3.42 \times 10^{-9}\, \text{mol}. $$

Now, we have the moles of magnesium needed for the reaction. To find the mass, we will multiply the moles with the molar mass of magnesium. The molar mass of magnesium is approximately 24.31 g/mol, so: $$ m_{Mg} = n_{Mg} \times Molar\, Mass_{Mg} = (3.42 \times 10^{-9}\, \text{mol}) \times (24.31\, \text{g/mol}) \approx 8.32 \times 10^{-8}\, \text{g}. $$ So, the mass of magnesium needed to react with the oxygen inside the enclosure is approximately \(8.32 \times 10^{-8}\, \text{g}\).