Question

Metallic molybdenum can be produced from the mineral molybdenite, \(\mathrm{MoS}_{2}\). The mineral is first oxidized in air to molybdenum trioxide and sulfur dioxide. Molybdenum trioxide is then reduced to metallic molybdenum using hydrogen gas. The balanced equations are $$ \begin{aligned} \mathrm{MoS}_{2}(s)+\frac{2}{2} \mathrm{O}_{2}(g) & \longrightarrow \mathrm{MoO}_{3}(s)+2 \mathrm{SO}_{2}(g) \\ \mathrm{MoO}_{3}(s)+3 \mathrm{H}_{2}(g) & \longrightarrow \mathrm{Mo}(s)+3 \mathrm{H}_{2} \mathrm{O}(l) \end{aligned} $$ Calculate the volumes of air and hydrogen gas at \(17^{\circ} \mathrm{C}\) and \(1.00\) atm that are necessary to produce \(1.00 \times 10^{3} \mathrm{~kg}\) pure molybdenum from \(\mathrm{MoS}_{2}\). Assume air contains \(21 \%\) oxygen by volume and assume \(100 \%\) yield for each reaction.

Short answer

To produce 1.00 x 10^3 kg pure molybdenum at 17°C and 1.00 atm, approximately 4,133,867 L of air and 748,595 L of hydrogen gas are required.

Step-by-step solution

Convert Mass of Molybdenum to Moles

To convert the mass of molybdenum to moles, we will use the molar mass of molybdenum, which is 95.96 g/mol. Moles of Mo = (1.00 x 10^3 kg) * (1000 g/kg) / (95.96 g/mol) Moles of Mo ≈ 10417 mol

Determine the Moles of Oxygen and Hydrogen

Now, using the balanced equations, we can calculate the moles of oxygen and hydrogen required for the reaction. From the balanced equations, 1 mole of MoS2 needs 3.502 mol of O2 and 3 mol of H2 to be fully converted to Mo. We know that moles of Mo = 10417 mol Moles of O2 = 3.502 × 10417 mol ≈ 36491 mol Moles of H2 = 3 × 10417 mol ≈ 31250 mol

Calculate the Volume of Oxygen Needed

Since O2 makes up 0.21 of air volume at 1.00 atm, we can calculate the volume of air needed for the reaction. Using the ideal gas law equation: PV = nRT, where P = pressure, V = volume, n = number of moles, R = gas constant (0.0821 L atm/mol K), and T = temperature in Kelvin (17°C + 273 = 290 K). 38491 mol of O2 requires a volume of air V_air, V_O2 = n(RT/P) V_O2 = 36491 mol × (0.0821 L atm/mol K) × (290 K) / (1.00 atm) V_O2 ≈ 868110 L As the O2 content of air is 21%, V_air = V_O2 / 0.21 V_air ≈ 4133867 L

Calculate the Volume of Hydrogen Needed

Using the ideal gas law equation again, we will calculate the volume of hydrogen needed for the reaction at 1.00 atm and 290 K. V_H2 = n(RT/P) V_H2 = 31250 mol × (0.0821 L atm/mol K) × (290 K) / (1.00 atm) V_H2 ≈ 748595 L So, 748595 L of hydrogen gas is needed.

Final Results

The volumes of air and hydrogen gas necessary to produce 1.00 x 10^3 kg pure molybdenum at 17°C and 1.00 atm are: Volume of air ≈ 4133867 L Volume of hydrogen gas ≈ 748595 L

Key concepts

Chemical Reactions

Chemical reactions involve breaking bonds in reactants and forming new bonds to create products. In the production of molybdenum from molybdenite, two main chemical reactions occur.

The first chemical reaction oxidizes molybdenite (\( \mathrm{MoS}_{2} \)) in the presence of oxygen to form molybdenum trioxide (\( \mathrm{MoO}_{3} \)) and sulfur dioxide (\( \mathrm{SO}_{2} \)). The balanced equation for this reaction is:

• \( \mathrm{MoS}_{2}(s) + \frac{2}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{MoO}_{3}(s) + 2 \mathrm{SO}_{2}(g) \)

The next reaction involves reducing \( \mathrm{MoO}_{3} \) with hydrogen gas to produce metallic molybdenum and water:

• \( \mathrm{MoO}_{3}(s) + 3 \mathrm{H}_{2}(g) \rightarrow \mathrm{Mo}(s) + 3 \mathrm{H}_{2} \mathrm{O}(l) \)

Understanding these reactions is crucial as they define the steps needed to convert the original mineral into usable metal, highlighting both oxidation and reduction processes.

Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry that relates the pressure, volume, temperature, and amount of gas. It is given by the formula \( PV = nRT \), where:

• \( P \) is the pressure of the gas
• \( V \) is the volume
• \( n \) is the number of moles
• \( R \) is the ideal gas constant \((0.0821 \text{ L atm/mol K})\)
• \( T \) is the temperature in Kelvin

This law allows us to calculate the volume of gas needed or produced in a chemical reaction if we know the other variables. For instance, to find the volume of air required for the reaction, we solve for \( V \) by rearranging the equation:

\[ V = \frac{nRT}{P} \]Where \( n \) is the number of moles of oxygen. Understanding the ideal gas law is essential in determining the conditions under which gases react or are produced.

Molar Mass Calculation

Molar mass is the mass of one mole of a given substance, expressed in grams per mole (g/mol). It is a critical concept in stoichiometry, which involves quantitative relationships in chemical reactions.

To convert a known mass to moles, you need the substance's molar mass. For example, the molar mass of molybdenum (\( \mathrm{Mo} \)) is 95.96 g/mol.
To calculate the number of moles from a given mass of molybdenum, the formula is:

\[ \text{Moles of } \mathrm{Mo} = \frac{ ext{mass (g)}}{ ext{molar mass (g/mol)}} \]In the provided problem, converting 1000 kg (or 1,000,000 g) of \( \mathrm{Mo} \) results in the number of moles as follows:

\[ \text{Moles of } \mathrm{Mo} = \frac{1,000,000 \text{ g}}{95.96 \text{ g/mol}} \approx 10,417 \text{ moles} \]This step forms the foundation of determining how much reactant or product is involved in or results from a chemical reaction.

Oxygen and Hydrogen Gas

Oxygen and hydrogen gas play crucial roles in the chemical production processes. Both gases are diatomic, meaning they naturally exist as two-atom molecules (\( \mathrm{O}_2 \) and \( \mathrm{H}_2 \)).

In the context of molybdenum production:

• Oxygen gas is a reactant in the oxidation of \( \mathrm{MoS}_{2} \), effectively meaning it provides the necessary oxygen to convert \( \mathrm{MoS}_{2} \) to \( \mathrm{MoO}_{3} \) and \( \mathrm{SO}_{2} \).
• Hydrogen gas is used to reduce \( \mathrm{MoO}_{3} \) to metallic molybdenum. It is essential for stripping the oxygen from \( \mathrm{MoO}_{3} \), producing water as a by-product.

Both reactions demonstrate the use of oxygen as an oxidizing agent and hydrogen as a reducing agent. Understanding the role of these gases is vital in comprehending the chemical transformations necessary to obtain pure metallic elements from naturally occurring minerals.