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.

Step by step solution

01

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

02

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.

03

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

04

Calculate the volume of oxygen gas required

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

05

Calculate the volume of air required

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

06

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.

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