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 0.452 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? $$2\mathrm{Mg}(s)+{\mathrm{O}}_2(g)⟶2\mathrm{MgO}(s)$$

Short answer

The mass of magnesium required to react with the oxygen in the enclosure is approximately $3.38\times {10}^{-9}$ g.

Step-by-step solution

Convert given values to appropriate units

First, we need to convert the given values to appropriate units for calculation. The temperature of the system should be in Kelvin, and the pressure should be in atm (if we use R in atm.L/(mol.K)). - Convert temperature to Kelvin: T = 27°C + 273.15 = 300.15 K - Convert pressure to atm: P = $3.5\times {10}^{-6}$ torr × (1 atm / 760 torr) ≈ $4.61\times {10}^{-9}$ atm

Apply the Ideal Gas Law

Using the Ideal Gas Law (PV = nRT), we can calculate the amount of oxygen in moles. - V = 0.452 L (given) - R = 0.0821 atm.L/(mol.K) Solve for n: n = PV/(RT) = ($4.61\times {10}^{-9}$ atm × 0.452 L) / (0.0821 atm.L/(mol.K) × 300.15 K)

Calculate the amount of oxygen in moles

Now let's calculate the amount of oxygen in moles using the numbers we obtained: n = ($4.61\times {10}^{-9}$ atm × 0.452 L) / (0.0821 atm.L/(mol.K) × 300.15 K) ≈ $6.95\times {10}^{-11}$ mol

Find the amount of magnesium required using stoichiometry

From the balanced chemical equation, we can see that 2 moles of magnesium react with 1 mole of oxygen: 2 Mg + O₂ → 2 MgO. Using stoichiometry, we can calculate the amount of magnesium required in moles: Amount of Mg in moles = 2 × Amount of O₂ in moles = 2 × $6.95\times {10}^{-11}$ mol ≈ $1.39\times {10}^{-10}$ mol

Calculate the mass of magnesium required

Now we can determine the mass of magnesium required by multiplying the number of moles by the molar mass of magnesium (24.31 g/mol): Mass of Mg = Amount of Mg in moles × Molar mass of Mg = $1.39\times {10}^{-10}$ mol × 24.31 g/mol ≈ $3.38\times {10}^{-9}$ g Therefore, the mass of magnesium required to react with the oxygen in the enclosure is approximately $3.38\times {10}^{-9}$ g.

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental concept in chemistry that allows us to relate the pressure, volume, temperature, and number of moles of a gas. The law is represented by the equation $PV=nRT$, where:

• $P$ is the pressure of the gas,
• $V$ is the volume of the gas,
• $n$ is the number of moles,
• R is the ideal gas constant, and
• $T$ is the temperature in Kelvin.

In this exercise, we use the Ideal Gas Law to find the amount of oxygen gas present in the enclosure. This is achieved by solving for $n$, the number of moles of oxygen. First, we need to ensure all units are compatible. It's important to convert the temperature from Celsius to Kelvin by adding 273.15. Also, converting pressure from torr to atm is essential since the gas constant $R$ is often given in atm.L/(mol.K). By plugging in the converted values into the equation, we can solve for $n$, the moles of oxygen present. This paves the way for further calculations to determine the amount of magnesium needed for the reaction.

Chemical Reactions

In chemical reactions, reactants transform into products through a process involving rearrangement of atoms. The accuracy of stoichiometric relationships in reaction equations is crucial for quantitative predictions about the amounts of substances consumed and produced. In this exercise's context, the balanced chemical reaction is:$$2\text{Mg}(s)+{\text{O}}_2(g)⟶2\text{MgO}(s)$$This tells us that two moles of magnesium react with one mole of oxygen to form two moles of magnesium oxide. The stoichiometry of the reaction is essential because it dictates the proportion in which reactants combine and products form. By understanding the stoichiometric coefficients from the balanced equation, we can calculate how much magnesium is required if we know how much oxygen is available. This step involves direct application of the mole ratio derived from the balanced equation. Each stoichiometric calculation is rooted in these ratios, ensuring the chemical equation is balanced to reflect real-world chemical processes.

Moles Calculation

Moles are a basic unit in chemistry used for expressing amounts of a substance. The concept of a mole allows chemists to count and use macroscopic quantities of material based on atomic and molecular scale measurements. In the present problem, once we have determined the number of moles of oxygen using the Ideal Gas Law, stoichiometry allows us to find the amount of magnesium needed. For this:

• Determine the amount of oxygen in moles - a key outcome of utilizing the Ideal Gas Law in computations.
• Use the mole ratio from the balanced equation (2 mol Mg : 1 mol ${\text{O}}_2$) to find the moles of magnesium.
• Multiply the moles of magnesium by its molar mass (24.31 g/mol) to get the mass in grams.

This calculation demonstrates how converting between moles and mass requires multiplication by a substance's molar mass and is a practice fundamental to solving many stoichiometry problems. A good grasp of mole calculations and stoichiometric conversions is thus essential for conducting precise chemical calculations.