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Molybdenum is an important industrial metal, often derived from the mineral molybdenite, which is chemically represented as MoS \(_2\). The production of metallic molybdenum involves a two-step chemical process. First, molybdenite is oxidized to form molybdenum trioxide (MoO \(_3\)) and sulfur dioxide (SO \(_2\)) using oxygen from the air. This reaction can be represented by the balanced equation:

• MoS \(_2(s)\) + \(\frac{7}{2}\) O \(_2(g)\) \Rightarrow MoO \(_3(s)\) + 2 SO \(_2(g)\)

The next step is the reduction of molybdenum trioxide to metallic molybdenum. This is done using hydrogen gas, according to the following balanced equation:

• MoO \(_3(s)\) + 3 H \(_2(g)\) \Rightarrow Mo \(_(s)\) + 3 H \(_2O(l)\)

This reduction reaction requires hydrogen gas to convert MoO \(_3\) to Mo. Both reactions are assumed to be 100% efficient, meaning all products are successfully converted as described. This makes stoichiometry crucial in calculating the exact amounts of reactants required.

In chemical reactions involving gases, it’s important to determine the volume of gases needed under specific conditions, such as temperature and pressure. According to the ideal gas law, the volume of a gas (V) is directly proportional to the number of moles (n) and temperature (T), and inversely proportional to the pressure (P). The equation is represented as:\[PV = nRT\]Where \(R\) is the ideal gas constant (0.0821 L\cdot atm/mol\cdot K). Using this principle, we calculate the volumes of oxygen (as part of air) and hydrogen needed for the production of molybdenum. Given that air is 21% oxygen by volume, the total volume of air required is larger than the volume of pure oxygen. Using the calculated moles of O \(_2\) (3.648 x 10\(^4\) mol) and hydrogen (3.1269 x 10\(^4\) mol), we can determine their respective volumes. The temperatures must be converted to Kelvin (17°C = 290.15 K) for calculations to be accurate. For oxygen, we need more air due to the fraction of oxygen present in air, represented as:

• Volume of Air = \(\frac{\text{Number of moles of O}_2}{\text{Volume fraction of O}_2}\)
• Volume of H \(_2 = \frac{\text{Number of moles of H}_2 \times R \times T}{P}\)

Through calculations, the necessary volumes of air and hydrogen are found to be 1.395 x 10\(^6\) L and 7.99 x 10\(^5\) L respectively, showcasing how stoichiometric principles guide gas consumption.

Chemical reactions are transformations that convert one set of chemical substances into another. In the context of molybdenum production, these reactions are transformations of molybdenite to molybdenum trioxide, and then to molybdenum metal. Each reaction in chemistry needs to be balanced, meaning the same number of each type of atom must be present on both sides of the equation. This reflects the law of conservation of mass. In our exercise, two different reactions are combined to achieve a single goal: obtaining pure molybdenum from its mineral form. Reactants react directly with oxygen and hydrogen under specified conditions to produce the desired solid and gaseous products. Typically, conditions like temperature and pressure are managed to optimize the reaction efficiency.

• Oxidation Reaction: Molybdenite converts to molybdenum trioxide and sulfur dioxide.
• Reduction Reaction: Molybdenum trioxide converts to metallic molybdenum and water.

Understanding the steps involved in these transformations, including calculating the stoichiometry, is crucial for students and industry professionals alike, ensuring that they comprehend the molecular interaction and the necessity of precise reagent quantities.