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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: $$ 1000\: \text{kg Mo} 脳 \frac{1\: \text{mol Mo}}{95.94\: \text{g Mo}} 脳 \frac{1000\: \text{g}}{1\: \text{kg}} = 1.042脳10^4\: \text{mol Mo} $$

02

Determine the moles of MoS鈧� and O鈧� required

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

03

Calculate the moles of H鈧� required

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

04

Calculate the volume of O鈧� required

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

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 O鈧� calculated in the previous step. $$ \text{Volume of air}=\frac{\text{Volume of O鈧倉}{0.21}=\frac{8.678\times10^{5}\: \text{L O鈧倉}{0.21}= 4.133\times10^{6}\: \text{L air} $$

06

Calculate the volume of H鈧� required

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

07

State the volumes of air and H鈧� required

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that helps us understand the quantitative relationships between reactants and products in a chemical reaction. It is based on the principle that matter is conserved in a reaction. This means the number of atoms of each element is the same on both sides of the equation.

In the exercise, stoichiometry allows us to determine the quantities of molybdenum disulfide (\( \text{MoS}_2 \)) and the amounts of oxygen (\( \text{O}_2 \)) needed to produce the desired amount of molybdenum.

• First, we calculate the moles of molybdenum needed using its molar mass.
• Then, using the balanced equations, we find out how many moles of \( \text{MoS}_2 \) and \( \text{O}_2 \) are required.

These stoichiometric calculations are essential for optimizing the use of resources when producing substances like molybdenum.

Ideal Gas Law

The Ideal Gas Law is a vital tool in determining the behavior of gases. It is expressed as \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the amount of substance in moles, \( R \) is the gas constant (0.0821 L atm/mol K), and \( T \) is temperature in Kelvin.

In the context of molybdenum production, the Ideal Gas Law helps us calculate the volumes of oxygen and hydrogen gases required under certain conditions. After determining the moles of gases needed through stoichiometry:

• We plug these values along with the given temperature (converted to Kelvin) and pressure into the equation.
• The equation then provides the necessary volume for each gas for the reaction to proceed efficiently.

The Ideal Gas Law connects the conditions of temperature, pressure, and volume, allowing precise measurement and use of gases in chemical processes.

Molybdenum Production

Molybdenum is a valuable metal with various industrial applications, including alloys and catalysts. Its production from the mineral molybdenite (\( \text{MoS}_2 \)) involves a two-step process:

• Oxidation: \( \text{MoS}_2 \) is reacted with oxygen to form molybdenum trioxide (\( \text{MoO}_3 \)) and sulfur dioxide (\( \text{SO}_2 \)).This step converts the mineral into a more reactive oxide form.
• Reduction: The molybdenum trioxide is then treated with hydrogen gas, reducing it to pure metallic molybdenum and producing water as a byproduct.

This process utilizes the principles of stoichiometry and balancing chemical equations to ensure that every step efficiently uses the reactants to yield the maximum amount of molybdenum, highlighting the importance of precise quantitative analysis in industrial chemistry.

Balanced Chemical Equations

A balanced chemical equation is a chemical sentence that represents a chemical reaction. It shows the same number of atoms of each element on both sides, adhering to the law of conservation of mass. Balancing equations is crucial as it ensures that the reaction adheres to the chemical laws and reflects reality.

In the molybdenum production process:

• The equation for oxidizing \( \text{MoS}_2 \) into \( \text{MoO}_3 \) and \( \text{SO}_2 \) was balanced to show that \( \text{MoS}_2 \) reacts with a certain amount of \( \text{O}_2 \) to form products.
• Similarly, the reduction of \( \text{MoO}_3 \) with hydrogen is balanced, showing the exact proportions needed to form pure molybdenum.

Balanced equations are indispensable for calculating the precise amounts of reactants and products involved, ensuring efficient and predictable reaction outcomes.