Question

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 \(0.382 \mathrm{~L}\) has a partial pressure of \(\mathrm{O}_{2}\) of \(3.5 \times 10^{-6}\) torr at \(27^{\circ} \mathrm{C}\), what mass of magnesium will react according to the following equation? $$ 2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s) $$

Short answer

The mass of magnesium needed to react with the remaining oxygen in the enclosure is approximately \(3.50 \times 10^{-9} \mathrm{~g}\).

Step-by-step solution

Convert given pressure to atm

The given pressure is in torr, so we need to convert it into atm for use in the ideal gas law, since R typically uses units of atm. 1 atm = 760 torr: $$ 3.5 \times 10^{-6} \text{ torr} \times \frac{1 \text{ atm}}{760 \text{ torr}} \approx 4.61 \times 10^{-9} \text{ atm} $$

Use the ideal gas law to find moles of O₂

Use the ideal gas law (PV=nRT) to determine the number of moles of O₂ in the enclosure. We know the pressure (P = 4.61 x 10⁻⁹ atm), the volume (V = 0.382 L), and the temperature (T = 27°C = 300K), as well as the ideal gas constant (R ≈ 0.0821 L atm K⁻¹ mol⁻¹). $$ n = \frac{PV}{RT} = \frac{(4.61 \times 10^{-9} \text{ atm})(0.382 \text{ L})}{(0.0821 \text{ L atm K}^{-1} \text{mol}^{-1})(300 \text{ K})} \approx 7.22 \times 10^{-11} \text{ moles} $$

Use stoichiometry to find moles of Mg

From the balanced chemical equation, we know that 2 moles of Mg react with 1 mole of O₂. So we can find the moles of Mg needed: $$ \text{moles of Mg} = (7.22 \times 10^{-11} \text{ moles}) \times \frac{2 \text{ moles of Mg}}{1 \text{ mole of O}₂} = 1.44 \times 10^{-10} \text{ moles} $$

Convert moles of Mg to mass

We can now convert the moles of Mg required to mass, by using the molar mass of Mg (24.31 g/mol): $$ \text{mass of Mg} = (1.44 \times 10^{-10} \text{ moles}) \times (24.31 \text{ g/mol}) \approx 3.50 \times 10^{-9} \text{ g} $$ So the mass of magnesium needed to react with the remaining oxygen in the enclosure is approximately \(3.50 \times 10^{-9} \mathrm{~g}\).

Key concepts

Understanding the Ideal Gas Law

The ideal gas law is a crucial tool in understanding the behavior of gases under various conditions and is represented by the equation PV = nRT.

First, we have 'P', standing for pressure, typically measured in atmospheres (atm) or torr. In the exercise, we initially had the pressure in torr, but since the gas constant ('R') is given in terms of atm, a conversion was necessary.

To apply the law effectively, it's essential to maintain the consistency of units; hence, the pressure was converted to atm as follows: \[\begin{equation}3.5 \times 10^{-6} \text{ torr} \times \frac{1 \text{ atm}}{760 \text{ torr}} \approx 4.61 \times 10^{-9} \text{ atm}\end{equation}\]

Then, 'V' represents volume in liters (L), 'n' is the amount of gas in moles, 'T' is the temperature in Kelvin (which requires converting from Celsius), and 'R' is the ideal gas constant, specific to the units in use. By isolating 'n' from the ideal gas law, the exercise determines the moles of oxygen present:

\[\begin{equation}n = \frac{PV}{RT}\end{equation}\]

Understanding each component of this equation is pivotal as it's applicable across various stoichiometry problems involving gases.

Molar Mass Calculation

To convert moles into grams, which are more tangible quantities in a laboratory setting, we use the concept of molar mass. Molar mass is the weight in grams of one mole of a particular substance and is usually derived from the periodic table of elements.

In the resolved problem, we needed to find the molar mass of magnesium (Mg), which is 24.31 g/mol. This conversion factor transforms moles into mass, allowing us to quantify the material needed for reactions or produced as products:

\[\begin{equation}\text{mass} = \text{moles} \times \text{molar mass}\end{equation}\]

With this equation, the mass of magnesium required for the reaction was calculated. Deep understanding of molar mass calculation is key for students to handle stoichiometry problems accurately, ensuring they can apply it to various elements and compounds.

Chemical Reactions Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It relates to the coefficients in a balanced chemical equation, which reflect the proportion by moles of each substance involved.

In the given example, we see the stoichiometric relationship in the equation:

\[\begin{equation}2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s)\end{equation}\]

This tells us for every mole of oxygen (O₂), two moles of magnesium (Mg) are required. After determining the moles of oxygen through the ideal gas law, stoichiometry is then used to find the corresponding moles of magnesium needed for a complete reaction. The conversion is straightforward:

\[\begin{equation}\text{moles of Mg} = (\text{moles of O₂}) \times \frac{2 \text{ moles of Mg}}{1 \text{ mole of O₂}}\end{equation}\]

This critical aspect of stoichiometry allows students to extend beyond single-compound calculations and solve more complex scenarios involving multiple reactants and products.