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

Calculate the moles of Mo required

We know that 1000 kg of molybdenum is required, we'll convert that to moles using its molar mass: $$ 1000\: \text{kg Mo} × \frac{1\: \text{mol Mo}}{95.94\: \text{g Mo}} × \frac{1000\: \text{g}}{1\: \text{kg}} = 1.042×10^4\: \text{mol Mo} $$

02

Determine the moles of MoS₂ and O₂ required

Using the given balanced equations, calculate moles of MoS₂ and O₂ required to produce 1.042×10^4 moles of Mo. 1 mole of MoS₂ produces 1 mole of Mo. $$ \text{Moles of MoS₂} = 1.042×10^4\: \text{mol MoS₂} $$ 7/2 moles of O₂ are required to produce 1 mole of MoS₂ $$ \text{Moles of O₂} = 1.042×10^4\: \text{mol MoS₂} × \frac{7}{2\: \text{mol O₂}}{1\: \text{mol MoS₂}} = 3.647×10^4\: \text{mol O₂} $$

03

Calculate the moles of H₂ required

Determine the moles of H₂ required to produce 1.042×10^4 moles of Mo. 3 moles of H₂ are required to produce 1 mole of Mo $$ \text{Moles of H₂} = 1.042×10^4\: \text{mol Mo} × \frac{3\: \text{mol H₂}}{1\: \text{mol Mo}} = 3.125×10^4\: \text{mol H₂} $$

04

Calculate the volume of O₂ required

We know the moles of O₂ required and will use the ideal gas law equation to find the volume of O₂: $$ PV = nRT \\ V = \frac{nRT}{P} $$ where: \(P = 1.00\: \text{atm}\) \(n = 3.647×10^4\: \text{mol O₂}\) \(R = 0.0821\: \frac{\text{L\: atm}}{\text{molK}}\) \(T = 17{ }^{\circ} \mathrm{C} = 290\: \mathrm{K}\) $$ V = \frac{3.647×10^4\: \text{mol O₂} × 0.0821\: \frac{\text{L\: atm}}{\text{mol\: K}}× 290\: \text{K}}{1.00\: \text{atm}} = 8.678×10^5\: \text{L O₂} $$

05

Calculate the volume of air required

Air contains 21% oxygen by volume, so we'll find the volume of air required using the volume of O₂ calculated in the previous step. $$ \text{Volume of air}=\frac{\text{Volume of O₂}}{0.21}=\frac{8.678\times10^{5}\: \text{L O₂}}{0.21}= 4.133\times10^{6}\: \text{L air} $$

06

Calculate the volume of H₂ required

Calculate the volume of H₂ required using the ideal gas law equation and moles of H₂ (calculated in Step 3): $$ V = \frac{3.125×10^4\: \text{mol H₂} × 0.0821\: \frac{\text{L\: atm}}{\text{mol\: K}}× 290\: \text{K}}{1.00\: \text{atm}} = 7.471×10^5\: \text{L H₂} $$

07

State the volumes of air and H₂ required

The volumes of air and hydrogen gas required to produce 1000 kg of pure molybdenum from MoS₂ are: Air: \(4.133\times10^{6}\: \text{L}\) Hydrogen gas: \(7.471\times10^{5}\: \text{L}\)

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