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.

Step by step solution

01

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

02

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

03

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

04

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.

05

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

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!