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 $$\operatorname{MoS}_{2}(s)+\frac{7}{2} \mathrm{O}_{2}(g) \longrightarrow \operatorname{MoO}_{3}(s)+2 \mathrm{SO}_{2}(g)$$ $$\mathrm{MoO}_{3}(s)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{Mo}(s)+3 \mathrm{H}_{2} \mathrm{O}(l)$$ 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

The required volumes of air and hydrogen gas at 17°C and 1.00 atm for producing 1000 kg of pure molybdenum are 4.134 × 10^7 L and 7.419 × 10^6 L, respectively.

Step-by-step solution

Calculate molar masses

Molar Mass of Mo = 95.94 g/mol Molar Mass of MoS_2 = 95.94 g/mol + (2 × 32.07 g/mol) = 160.08 g/mol

Convert mass of molybdenum to moles

\( 1000 kg = 10^6 g \) Moles of Mo = \(\frac{10^6\, g}{95.94\, g/mol} = 1.042 \times 10^4\, mol\) Now, using stoichiometry, let's calculate the number of moles of MoS2, O2, and H2 required.

Moles of MoS2, O2, and H2 required

From the balanced chemical equations, we have: MoS2 : Mo : O2 : MoO3 : H2 : H2O = 1 : 1 : \(\frac{7}{2}\) : 1 : 3 : 3 Moles of MoS2 = 1.042 × 10^4 mol (as Mo and MoS2 have a 1:1 stoichiometry in the reaction) Moles of O2 = \(\frac{7}{2} \times 1.042 \times 10^4\, mol = 3.647 \times 10^4\, mol\) Moles of H2 = \(\ 3 \times 1.042 \times 10^4\, mol = 3.126 \times 10^4\, mol\) Now we will calculate the volumes of air and hydrogen gas at the given temperature and pressure conditions, knowing the number of moles of O2 and H2 needed. We will use the Ideal Gas Law, \(PV=nRT\), 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 in Kelvin. Given: P = 1 atm, T = 17°C = 290 K, R = 0.0821 L atm/mol K

Calculate the volume of oxygen gas required

\(V_{O_2} = \frac{n_{O_2}RT}{P} = \frac{3.647 \times 10^4\, mol \cdot 0.0821\, L\, atm\, mol^{-1} K^{-1} \cdot 290 K}{1\, atm} = 8.683 \times 10^6 L\)

Calculate the volume of air required

Since the air contains 21% oxygen by volume: Required air volume = \( \frac{Volume\, of\, O_2\, required}{Percentage \,of\, oxygen \,in\, air} = \frac{8.683 \times 10^6 L}{0.21} = 4.134 \times 10^7 L\)

Calculate the volume of hydrogen gas required

\(V_{H_2} = \frac{n_{H_2}RT}{P} = \frac{3.126 \times 10^4\, mol \cdot 0.0821\, L\, atm\, mol^{-1} K^{-1} \cdot 290 K}{1\, atm} = 7.419 \times 10^6 L\) In conclusion, the required volumes of air and hydrogen gas at 17°C and 1.00 atm for producing 1000 kg of pure molybdenum are 4.134 × 10^7 L and 7.419 × 10^6 L, respectively.

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental principle in chemistry used to relate the four parameters of a gas: pressure, volume, temperature, and the number of moles. This law is expressed mathematically as \(PV = nRT\), where:

• \(P\) represents the pressure of the gas in atmospheres (atm).
• \(V\) stands for the volume of the gas in liters (L).
• \(n\) denotes the number of moles of the gas.
• \(R\) is the ideal gas constant, 0.0821 L atm/mol K.
• \(T\) is the temperature in Kelvin (K).

To convert temperature from Celsius to Kelvin, simply add 273.15 to the Celsius temperature. The Ideal Gas Law is particularly useful when you need to find one of these four variables, provided the others are known. In our problem, we used it to determine the volumes of oxygen and hydrogen gases required.

Chemical Reactions

Chemical reactions are processes where reactants are transformed into products. They are represented through balanced chemical equations, showing the ratios of molecules involved. In the given problem, two reactions are key:

• The oxidation of molybdenite: \[\mathrm{MoS}_2(s) + \frac{7}{2} \mathrm{O}_2(g) \rightarrow \mathrm{MoO}_3(s) + 2 \mathrm{SO}_2(g) \]
• The reduction of molybdenum trioxide: \[\mathrm{MoO}_3(s) + 3 \mathrm{H}_2(g) \rightarrow \mathrm{Mo}(s) + 3 \mathrm{H}_2\mathrm{O}(l)\]

Each equation must adhere to the principle of conservation of mass, ensuring the same number of each type of atom on both sides of the equation. Understanding these balanced equations helps us determine the amounts of reactants needed and products formed.

Moles Calculation

Moles are a standard unit used in chemistry to measure the amount of a substance. A mole represents \(6.022 \times 10^{23}\) entities (atoms, molecules, etc.) known as Avogadro's number. To convert between mass and moles, you use the substance's molar mass, expressed in grams per mole (g/mol). For example, the molar mass of molybdenum (Mo) is 95.94 g/mol.
To find the moles of molybdenum from a given mass, use the formula:\[\text{Moles of Mo} = \frac{\text{mass of Mo (g)}}{\text{molar mass of Mo (g/mol)}}\]In the exercise, we converted 1000 kg (or \(10^6\) g) of Mo into moles to determine how much of other reactants were needed for the reactions. Knowing the number of moles allows us to effectively apply stoichiometry in chemical equations.

Gas Volume Calculation

Calculating the volume of a gas involves using the Ideal Gas Law, once you know the number of moles, temperature, and pressure. In this problem, we calculated the required volumes of oxygen and hydrogen gases for the production of molybdenum.
For each gas, we used the formula derived from the Ideal Gas Law:\[V = \frac{nRT}{P}\]Where \(n\) is the number of moles, \(R\) is the gas constant, \(T\) is temperature in Kelvin, and \(P\) is pressure. The problem also considered that air contains 21% oxygen by volume, affecting how much air needed to be measured out to ensure sufficient oxygen.
These calculations are essential when scaling chemical reactions to industrial or experimental settings. By knowing the precise volumes needed, you can ensure efficiency and effectiveness in your process.