Chapter 8, Problem 135AE

Metallic molybdenum can be produced from the mineral molybdenite, MoS₂. 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

$\begin{array}{l}{\text{MoS}}_{\text{2}}\left(s\right)+\frac{7}{2}{\text{O}}_{\text{2}}\left(g\right)\to {\text{MoO}}_{\text{3}}\left(s\right)+{\text{ 2SO}}_{\text{2}}\left(g\right)\hfill \\ {\text{MoO}}_{\text{3}}\left(s\right)+{\text{ 3H}}_{\text{2}}\left(g\right)\to \text{ Mo}\left(s\right)+{\text{ 3H}}_{\text{2}}\text{O}\left(l\right)\hfill \end{array}$

Calculate the volumes of air and hydrogen gas at 17°C and 1.00 atm that are necessary to produce 1.00 × 10³ kg pure molybdenum from MoS₂. Assume air contains 21% oxygen by volume, and assume 100% yield for each reaction.

Interpretation Introduction

Interpretation:

The volume of air and volume of ${\text{H}}_{\text{2}}$ required to producing $\text{1}\text{×}{\text{10}}^{\text{3}}\text{Kg}$ of molybdenum from ${\text{MoS}}_{\text{2}}$ in the given reactions at ${17}^{\text{o}}\text{C}$ and $\text{1}\text{atm}$ are to be calculated.

Concept introduction:

• Number of moles of a substance from its given mass is,

  $\text{Number of moles}\text{=}\frac{\text{Given}\text{mass}}{\text{Molecular}\text{mass}}$

• Mole ratios between the reactants of a reaction are depends upon the coefficients of respective reactant in a balanced chemical equation.
• According to ideal gas equation for finding volume of a gas,

$\text{Volume}\text{=}\frac{\text{Number}\text{of}\text{moles}\times \text{R}\times \text{Temperature}}{\text{pressure}}$

Answer to Problem 135AE

Answer

The volume of air required to producing $\text{1}\text{×}{\text{10}}^{\text{3}}\text{Kg}$ of molybdenum is $\text{4}\text{.1}\text{×}{\text{10}}^{\text{6}}\text{L}$.

The volume of ${\text{H}}_{\text{2}}$ required to producing $\text{1}\text{×}{\text{10}}^{\text{3}}\text{Kg}$ of molybdenum is $\text{7}\text{.42}\text{×}{\text{10}}^5\text{L}$.

Explanation of Solution

Explanation

• To find: the number of moles of producing $\text{1}\text{×}{\text{10}}^{\text{3}}\text{Kg}$ of molybdenum.

The number of moles of molybdenum produced is $\text{1}\text{.04}\text{×}{\text{10}}^{\text{4}}\text{mol}$.

The given produced mass of molybdenum is $\text{1}\text{×}{\text{10}}^{\text{3}}\text{Kg}\text{=}\text{1}\text{×}{\text{10}}^{\text{6}}\text{g}$

So the number of moles of molybdenum produced is,

The number of moles of a substance from its given mass is,

$\text{Number of moles}\text{=}\frac{\text{Given}\text{mass}}{\text{Molecular}\text{mass}}$

Therefore, the number of moles of molybdenum produced is,

$\text{Number of moles}\text{of sulfur}\text{=}\text{1}\text{×}{\text{10}}^{\text{6}}\text{g}\text{×}\frac{\text{1}\text{mole}}{\text{95}\text{.94}\text{g}}\text{=}\text{1}\text{.04}\text{×}{\text{10}}^{\text{4}}\text{mol}$

Hence, the number of moles of molybdenum produced is $\text{1}\text{.04}\text{×}{\text{10}}^{\text{4}}\text{mol}$.

• To find: the molar ratio between molybdenum to ${\text{H}}_{\text{2}}$, ${\text{MoO}}_{\text{3}}$ and ${\text{O}}_{\text{2}}$ in the given reactions.

Molar ratio between molybdenum to ${\text{H}}_{\text{2}}$ is $1:3$.

Molar ratio between molybdenum to ${\text{MoO}}_{\text{3}}$ is $1:1$.

Molar ratio between molybdenum to ${\text{O}}_{\text{2}}$ is $1:\frac{7}{2}$.

The balance equation of given reactions are,

${\text{MoS}}_{\text{2}}\text{+}\frac{\text{7}}{\text{2}}{\text{O}}_{\text{2}}\to {\text{MoO}}_{\text{3}}\text{+}{\text{2SO}}_{\text{2}}$

${\text{MoO}}_{\text{3}}\text{+}{\text{3H}}_{\text{2}}\to \text{Mo}\text{+}{\text{3H}}_2\text{O}$

Here, in second reaction,

1 mole of ${\text{MoO}}_{\text{3}}$ is reacted with 3 mole of ${\text{H}}_{\text{2}}$ to produce 1 mole of molybdenum in the reaction.

Therefore,

The molar ratio between molybdenum to ${\text{MoO}}_{\text{3}}$ is $1:1$.

The molar ratio between molybdenum to ${\text{H}}_{\text{2}}$ is $1:3$.

In first reaction,

1 mole of ${\text{MoS}}_{\text{2}}$ is reacted with $\frac{\text{7}}{\text{2}}\text{mol}$ ${\text{O}}_{\text{2}}$ to produce 1 mole of ${\text{MoO}}_{\text{3}}$ in the reaction.

Therefore,

The mole ratio between ${\text{MoO}}_{\text{3}}$ to ${\text{O}}_{\text{2}}$ is $1:7$.

Since, the molar ratio between molybdenum to ${\text{MoO}}_{\text{3}}$ is $1:1$.

Hence, the molar ratio between molybdenum to ${\text{O}}_{\text{2}}$ is $1:\frac{7}{2}$.

• To find: the number of moles of ${\text{H}}_{\text{2}}$, ${\text{O}}_{\text{2}}$ in the given reactions.

The number of moles of ${\text{H}}_{\text{2}}$ is $\text{3}\text{.12}\text{×}{\text{10}}^{\text{4}}\text{mol}$.

The number of moles of ${\text{O}}_{\text{2}}$ is $\text{3}\text{.64}\text{×}{\text{10}}^{\text{4}}\text{mol}$

The molar ratio between molybdenum to ${\text{H}}_{\text{2}}$ is calculated as $1:3$.

The molar ratio between molybdenum to ${\text{O}}_{\text{2}}$ is calculated as $1:\frac{7}{2}$.

The number of moles of molybdenum is $\text{1}\text{.04}\text{×}{\text{10}}^{\text{4}}\text{mol}$.

Therefore,

• The number of moles of ${\text{H}}_{\text{2}}$ is,

$\text{1}\text{.04}\text{×}{\text{10}}^{\text{4}}\text{mol}\text{Mo}\text{×}\frac{\text{3}\text{mol}{\text{H}}_{\text{2}}}{\text{1}\text{mol}\text{Mo}}\text{=}\text{3}\text{.12}\text{×}{\text{10}}^{\text{4}}\text{mol}{\text{H}}_{\text{2}}$

• The number of moles of ${\text{O}}_{\text{2}}$ is,

$\text{1}\text{.04}\text{×}{\text{10}}^{\text{4}}\text{mol}\text{Mo}\text{×}\frac{\frac{\text{7}}{\text{2}}\text{mol}{\text{O}}_{\text{2}}}{\text{1}\text{mol}\text{Mo}}\text{=}\text{3}\text{.36}\text{×}{\text{10}}^{\text{4}}\text{mol}{\text{O}}_{\text{2}}$

Hence,

The number of moles of ${\text{H}}_{\text{2}}$ in the given reactions is $\text{3}\text{.12}\text{×}{\text{10}}^{\text{4}}\text{mol}$.

The number of moles of ${\text{O}}_{\text{2}}$ in the given reactions is $\text{3}\text{.64}\text{×}{\text{10}}^{\text{4}}\text{mol}$

• To determine: the volume of ${\text{H}}_{\text{2}}$ required to producing $\text{1}\text{×}{\text{10}}^{\text{3}}\text{Kg}$ of molybdenum from ${\text{MoS}}_{\text{2}}$ in the given reactions

The volume of ${\text{H}}_{\text{2}}$ required to producing $\text{1}\text{×}{\text{10}}^{\text{3}}\text{Kg}$ of molybdenum is $\text{7}\text{.42}\text{×}{\text{10}}^5\text{L}$.

The number of number of moles of ${\text{H}}_{\text{2}}$ reacted in the given reactions is $\text{3}\text{.12}\text{×}{\text{10}}^{\text{4}}\text{mol}$.

The pressure is given as $1\text{atm}$.

The temperature is given as ${\text{17}}^{\text{o}}\text{C}\text{=}\text{290}\text{K}$.

The equation for finding volume of a gas from its number of moles is,

According to ideal gas equation for finding volume of a gas,

$\text{Volume}\text{=}\frac{\text{Number}\text{of}\text{moles}\times \text{R}\times \text{Temperature}}{\text{pressure}}$

Therefore,

$\begin{array}{l}\text{Volume}{\text{of H}}_{\text{2}}\text{=}\frac{\text{3}\text{.12}\text{×}{\text{10}}^{\text{4}}\text{mol}\times \text{0}\text{.08206}\text{L}\text{atm/K}\text{mol}\times 290\text{K}}{1\text{atm}}\\ =\text{7}\text{.42}\text{×}{\text{10}}^5\text{L}\end{array}$

Hence, the volume of ${\text{H}}_{\text{2}}$ required to producing $\text{1}\text{×}{\text{10}}^{\text{3}}\text{Kg}$ of molybdenum is $\text{7}\text{.42}\text{×}{\text{10}}^5\text{L}$.

• To determine: volume of air required to producing $\text{1}\text{×}{\text{10}}^{\text{3}}\text{Kg}$ of molybdenum from ${\text{MoS}}_{\text{2}}$ in the given reactions

The volume of air required to producing $\text{1}\text{×}{\text{10}}^{\text{3}}\text{Kg}$ of molybdenum is $\text{4}\text{.1}\text{×}{\text{10}}^{\text{6}}\text{L}$.

The number of moles of ${\text{O}}_{\text{2}}$ in the given reactions is calculated as $\text{3}\text{.64}\text{×}{\text{10}}^{\text{4}}\text{mol}$

The pressure is given as $1\text{atm}$.

The temperature is given as ${\text{17}}^{\text{o}}\text{C}\text{=}\text{290}\text{K}$.

The equation for finding volume of a gas from its number of moles is,

According to ideal gas equation for finding volume of a gas,

$\text{Volume}\text{=}\frac{\text{Number}\text{of}\text{moles}\times \text{R}\times \text{Temperature}}{\text{pressure}}$

$\begin{array}{l}\text{Volume}{\text{of O}}_{\text{2}}\text{=}\frac{\text{3}\text{.64}\text{×}{\text{10}}^{\text{4}}\text{mol}\text{×}\text{0}\text{.08206}\text{L}\text{atm/K}\text{mol}\text{×}\text{290K}}{\text{1}\text{atm}}\\ \text{=}\text{8}\text{.66}\text{×}{\text{10}}^{\text{5}}\text{L}\end{array}$

      

The $21\%$ volume of air containing ${\text{O}}_{\text{2}}$ by volume.

That is,

$\text{volume}\text{of}\text{air}\text{×}\frac{\text{21}}{\text{100}}\text{=}\text{volume}\text{of}{\text{O}}_{\text{2}}$

Therefore,

$\begin{array}{l}\text{volume}\text{of}\text{air}=\frac{100\times \text{8}\text{.66}\text{×}{\text{10}}^{\text{5}}\text{L}}{21}\\ =\text{4}\text{.1}\text{×}{\text{10}}^{\text{6}}\text{L}\end{array}$

Hence, the volume of air required to producing $\text{1}\text{×}{\text{10}}^{\text{3}}\text{Kg}$ of molybdenum is $\text{4}\text{.1}\text{×}{\text{10}}^{\text{6}}\text{L}$.

Conclusion

Conclusion

The volume of air and volume of ${\text{H}}_{\text{2}}$ required to producing $\text{1}\text{×}{\text{10}}^{\text{3}}\text{Kg}$ of molybdenum from ${\text{MoS}}_{\text{2}}$ in the given reactions are determined by ideal gas equation. The number of moles of ${\text{H}}_{\text{2}}$ and ${\text{O}}_{\text{2}}$ are calculated by checking their mole ratio between produced molybdenum.