Metallic molybdenum can be produced from the mineral moIybdenite, 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 $$\begin{array}{l}\operatorname{MoS}_{2}(s)+\frac{7}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{MoO}_{3}(s)+2 \mathrm{SO}_{2}(g) \\\\\mathrm{MoO}_{3}(s)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{Mo}(s)+3 \mathrm{H}_{2} \mathrm{O}(l)\end{array}$$ 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 MoS \(_{2}\). Assume air contains \(21 \%\) oxygen by volume, and assume \(100 \%\) yield for each reaction.

To calculate the number of moles of the metallic molybdenum (Mo), use the molar mass of Mo and the given mass: Given mass of Mo = 1.00 x 10^3 kg = 1.00 x 10^6 g Molar mass of Mo = 95.95 g/mol Number of moles of Mo = (Given mass of Mo) / (Molar mass of Mo) Number of moles of Mo = (1.00 x 10^6) / 95.95 = 1.0423 x 10^4 mol

According to the balanced equations: 1 mol of MoS\(_{2}\) produces 1 mol of MoO\(_{3}\), which produces 1 mol of Mo. Number of moles of MoS\(_{2}\) = Number of moles of MoO\(_{3}\) = Number of moles of Mo = 1.0423 x 10^4 mol

For oxygen, from the balanced equations: 1 mol of MoS\(_{2}\) requires 3.5 mol of O\(_{2}\) Number of moles of O\(_{2}\) = 3.5 x Number of moles of MoS\(_{2}\) = 3.5 x 1.0423 x 10^4 mol = 3.648 x 10^4 mol For hydrogen, from the balanced equations: 1 mol of MoO\(_{3}\) requires 3 mol of H\(_{2}\) Number of moles of H\(_{2}\) = 3 x Number of moles of MoO\(_{3}\) = 3 x 1.0423 x 10^4 mol = 3.1269 x 10^4 mol

Temperature and Pressure are given as 17°C, and 1.00 atm. Convert the temperature to Kelvin: T = 17°C + 273.15 = 290.15 K Assuming air contains 21% O\(_{2}\), the volume fraction of oxygen = 0.21. The total moles of air required = Number of moles of O\(_{2}\) / Volume fraction of O\(_{2}\) Total moles of air = 3.648 x 10^4 mol / 0.21 = 173714.2857 mol Using the Ideal Gas Law, PV = nRT, we can find the volume of air and hydrogen: Volume of air = (Number of moles of air x R x T)/ P Volume of air = (173714.2857 mol x 0.0821 L atm/mol·K x 290.15 K) / 1 atm = 1.395 x 10^6 L Volume of H\(_{2}\) = (Number of moles of H\(_{2}\) x R x T) / P Volume of H\(_{2}\) = (3.1269 x 10^4 mol x 0.0821 L atm/mol·K x 290.15 K) / 1 atm = 7.99 x 10^5 L The volumes of air and hydrogen gas needed are 1.395 x 10^6 L and 7.99 x 10^5 L, respectively.