Problem 55 - Chemistry The Central Science In Si Units

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

Step-by-Step Solution

Answer

The mass of magnesium required is approximately \(7.78 \times 10^{-8} \text{ g} \).

Step 1: Understand the Given Information

We are given the volume of the enclosure as 5.67 L, the partial pressure of oxygen (6_2) as 7.066 mPa, and the temperature as 30°C. We need to find the mass of magnesium (Mg) required for the complete reaction with the oxygen in the container according to the reaction: \[ 2 \mathrm{Mg}(s) + \mathrm{O}_2(g) \rightarrow 2 \mathrm{MgO}(s) \].

Step 2: Convert Pressure to Appropriate Units

First, convert the pressure from millipascal (mPa) to atmospheres (atm) because 1 atm = 101325 Pa. \[ 7.066 \text{ mPa} = 7.066 \times 10^{-3} \text{ Pa} \]\[ \text{Pressure in atm} = \frac{7.066 \times 10^{-3}}{101325} = 6.98 \times 10^{-8} \text{ atm} \]

Step 3: Using the Ideal Gas Law

Use the ideal gas law \( PV = nRT \) to find the moles of \( \mathrm{O}_2 \). Where \( R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1} \) and \( T = 30^{\circ} \text{C} = 303 \text{ K} \).\[ n = \frac{PV}{RT} = \frac{(6.98 \times 10^{-8} \text{ atm}) \times (5.67 \text{ L})}{0.0821 \times 303} \approx 1.60 \times 10^{-9} \text{ mol} \]

Step 4: Stoichiometry of the Reaction

From the balanced chemical equation, 2 moles of Mg react with 1 mole of \( \mathrm{O}_2 \). Using stoichiometry, calculate the moles of Mg required: \[ \text{Moles of Mg} = 2 \times \text{Moles of } \mathrm{O}_2 = 2 \times 1.60 \times 10^{-9} \approx 3.20 \times 10^{-9} \text{ mol} \]

Step 5: Convert Moles of Mg to Mass

Mg has a molar mass of 24.305 g/mol. Convert moles to grams:\[ \text{Mass of Mg} = 3.20 \times 10^{-9} \text{ mol} \times 24.305 \text{ g/mol} \approx 7.78 \times 10^{-8} \text{ g} \]

Step 6: Conclusion

The mass of magnesium required to react with the oxygen in the enclosure is approximately \(7.78 \times 10^{-8} \text{ g} \).

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental concept in chemistry. It helps us understand how gases behave under different conditions. The law is expressed as \( PV = nRT \). Here:

• \( P \) is the pressure of the gas,
• \( V \) is the volume it occupies,
• \( n \) is the number of moles of the gas,
• \( R \) is the ideal gas constant (0.0821 L atm K\(^{-1}\) mol\(^{-1}\)),
• \( T \) is the temperature in Kelvin.

To apply the Ideal Gas Law, we usually convert temperatures to Kelvin and pressures to atmospheres, if needed. For example, if you are given temperature in Celsius, you simply add 273.15 to convert it to Kelvin. Similarly, pressures in millipascal can be converted to atmospheres for convenience. Using the ideal gas law allows us to calculate unknown variables such as the number of moles of a gas in a container, as was needed in the given exercise to determine the moles of oxygen present based on the given pressure and volume.

Chemical Reactions

Chemical reactions describe how substances interact to form new products. This is depicted by chemical equations. In the exercise, the equation \(2 \text{Mg}(s) + \text{O}_2(g) \rightarrow 2 \text{MgO}(s)\) is used to illustrate the reaction between magnesium and oxygen.In every chemical reaction:

• Reactants are substances that start the reaction, here it's magnesium (Mg) and oxygen (O\(_2\)).
• Products are substances formed by the reaction, in this case, magnesium oxide (MgO).

Balancing a chemical equation is crucial. It shows the ratio of moles of reactants to products based on the conservation of mass. In the provided equation, two moles of magnesium react with one mole of oxygen to produce two moles of magnesium oxide. Using this balanced equation, we perform stoichiometry to determine how much of each substance is needed or produced in a reaction.

Molar Mass Calculations

Molar mass calculations are essential when converting between moles and grams. The molar mass is the mass of one mole of a substance, which is expressed in grams per mole (g/mol). It is found by summing the atomic masses of all atoms in a compound from the periodic table. For magnesium (Mg), its molar mass is small due to its position on the periodic table. Specifically, the molar mass of Mg is 24.305 g/mol, as given in the exercise. To find the mass of a substance from its moles:

• First, calculate the number of moles, as shown in the previous sections.
• Then, multiply the number of moles by the molar mass to get the mass in grams.

The calculation used in the exercise shows how a small amount of oxygen in a large volume still results in a minute mass of reaction material, emphasizing the precision needed in chemical computations.