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 \(5.67 \mathrm{~L}\) 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? $$2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s).$$

Short answer

The mass of magnesium needed to react with the oxygen inside the enclosure is approximately \(8.32 \times 10^{-8}\, \text{g}\).

Step-by-step solution

Calculate moles of Oxygen gas

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 \(R = 0.0821 \frac{\text{L}\, \text{atm}}{\text{mol}\, \text{K}}\): $$ n_{O_{2}} = \frac{PV}{RT} = \frac{(6.98 \times 10^{-8}\,\text{atm})(5.67\,\text{L})}{(0.0821\, \frac{\text{L}\, \text{atm}}{\text{mol}\, \text{K}})(303.15\,\text{K})} \approx 1.71 \times 10^{-9}\, \text{mol}. $$

Determine moles of Magnesium

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: $$ n_{Mg} = 2 \times n_{O_{2}} = 2 \times (1.71 \times 10^{-9}\, \text{mol}) \approx 3.42 \times 10^{-9}\, \text{mol}. $$

Calculate mass of Magnesium

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 \(8.32 \times 10^{-8}\, \text{g}\).

Key concepts

Ideal Gas Law

The Ideal Gas Law is a critical equation in chemistry, often used to relate the pressure, volume, and temperature of a gas to the number of moles present. The formula is expressed as \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is temperature (in Kelvin). This law assumes a perfect or "ideal" gas, which is a simplified model where gas particles have no volume and do not interact except for elastic collisions.
To utilize this law in calculations, it's essential to ensure all units are compatible.
For example, pressure should be in atmospheres (atm), volume in liters (L), and temperature in Kelvin (K).

• Convert temperature from Celsius to Kelvin by adding 273.15.
• Pressure unit conversions may involve converting from millipascals (mPa) to pascals (Pa), then to atmospheres (atm).
• The gas constant \( R \) commonly used is \( 0.0821 \frac{\text{L} \cdot \text{atm}}{\text{mol} \cdot \text{K}} \).

Using the Ideal Gas Law helps us find the moles of a gas, which is crucial in determining how much of a chemical is present in mole form under specific conditions.

Chemical Reactions

Chemical reactions involve the transformation of one set of chemical substances into another. This process is represented by chemical equations, showcasing the reactants and products.
In our example, magnesium reacts with oxygen to form magnesium oxide, and the balanced reaction is: \[ 2\ \text{Mg} + \text{O}_2 \longrightarrow 2\ \text{MgO} \]Balancing equations is vital as it ensures the law of conservation of mass is obeyed, meaning atoms are neither created nor destroyed.
In this reaction, two moles of magnesium (\( 2 \text{Mg} \)) are needed for each mole of oxygen \( \text{O}_2 \) to form two moles of magnesium oxide \( \text{MgO} \).
This stoichiometric ratio guides us in calculations to find how much of a reactant is required to completely use another.
The role of chemical reactions in stoichiometry links to mole calculations, as we derive moles of one substance from another based on these balanced equations.

Mole Calculations

In stoichiometry, mole calculations are foundational for quantifying substances involved in chemical reactions. The mole is a standard unit used to express the amount of a chemical substance, equivalent to Avogadro's number \( 6.022 \times 10^{23} \) particles. To find how many moles you have from a given gas volume, utilize the Ideal Gas Law.
In this exercise:

• Determined moles of \( \text{O}_2 \) using \( n = \frac{PV}{RT} \).
• From moles of \( \text{O}_2 \), used stoichiometry from the balanced equation to calculate moles of \( \text{Mg} \).

The relationship between moles of different substances is often deduced from mole ratios in the balanced chemical equation, enabling conversion from moles of one reagent or product to another.
Accurate mole calculations are critical to finding amounts needed or produced in reactions, such as determining mass of a reactant.

Molar Mass

Molar Mass is the mass of one mole of a substance, expressed in grams per mole (g/mol). It enables conversion between the mass of a substance and the number of moles.
This concept is indispensable for determining how much of a chemical appears in a reaction when given mass or when calculating mass from moles.
For instance, magnesium, used in this exercise, has a molar mass of approximately 24.31 g/mol.
Here's how you use it:

• Calculate the mass of magnesium by multiplying moles of magnesium \((3.42 \times 10^{-9} \text{ mol})\) by its molar mass \((24.31 \text{ g/mol})\).
• Resulting mass: \(m_{\text{Mg}} = n_{\text{Mg}} \times \text{molar mass}_{\text{Mg}}\).

This calculation yields the mass required for reacting a given amount of oxygen, as shown in the exercise.
Grasping molar mass helps efficiently bridge the gap between laboratory masses and theoretical moles in stoichiometry.