Chapter 8, Problem 8.49QP

Interpretation Introduction

Interpretation:

The amount of ${\text{NO}}_{\text{2}}$ and number of moles of excess reactant has to calculate and identify the limiting reactant from the reaction.

Concept introduction:

• Limiting reactant is defined as a reactant which is totally utilized in a chemical reaction and decides when the reaction stops.
• $\text{Number}\text{of moles =}\frac{\text{mass}\text{of}\text{substance}}{\text{molecular}\text{weight}}$

Answer to Problem 8.49QP

• $O_3$ is the limiting reactant.
• $\text{Amount}\text{of}{\text{NO}}_{\text{2}}\text{=}\text{0}\text{.709}\text{g}\text{of}{\text{NO}}_{\text{2}}$
• $\text{unreacted mol NO = 0}\text{.0069 mol NO}$

Explanation of Solution

To calculate the number of moles of ${\text{O}}_{\text{3}}\text{and}\text{NO}$

$\begin{array}{l}\text{Number}{\text{of moles of NO}}_{\text{2}}\text{=}\text{0}\text{.740g}{\text{O}}_{\text{3}}\text{×}\frac{{\text{1 mol O}}_{\text{3}}}{\text{48g/mol}}\text{×}\frac{{\text{1mol NO}}_{\text{2}}}{{\text{1 mol O}}_{\text{3}}}\\ \text{=}\text{0}\text{.0154}\text{moles}\text{of}{\text{NO}}_{\text{2}}\\ \text{Number}{\text{of moles of NO}}_{\text{2}}\text{=}\text{0}\text{.670g}\text{NO}\text{×}\frac{\text{1 mol NO}}{\text{30}\text{.01g/mol}}\text{×}\frac{{\text{1mol NO}}_{\text{2}}}{\text{1 mol NO}}\\ \text{=}\text{0}\text{.0223}\text{moles}\text{of}{\text{NO}}_{\text{2}}\end{array}$

The number of moles of ${\text{O}}_{\text{3}}\text{and}\text{NO}$ can be aclculated using the corresponding given weights and molecular weight.  From the number of moles of ${\text{O}}_{\text{3}}\text{and}\text{NO}$, we can calculate the number of moles of ${\text{NO}}_{\text{2}}$ can be produced.  Here number of moles of ${\text{NO}}_{\text{2}}$ produced by ${\text{O}}_{\text{3}}$ is less than $\text{NO}$.  Hence, $O_3$is the limiting reactant.

Calculate the amount of ${\text{NO}}_{\text{2}}$ produced in the reaction.

$\begin{array}{l}\text{Amount}\text{of}{\text{NO}}_{\text{2}}\text{=}\text{0}\text{.0154}\text{mol}\text{of}{\text{O}}_{\text{3}}\text{×}\frac{\text{1}\text{mol}\text{of}{\text{NO}}_{\text{2}}}{\text{1}\text{mol}\text{of}{\text{O}}_{\text{3}}}\text{×}\frac{\text{46}\text{.01g}\text{of}{\text{NO}}_{\text{2}}}{\text{1}\text{mol}\text{of}{\text{NO}}_{\text{2}}}\\ \text{=}\text{0}\text{.0154}\text{mol}\text{of}{\text{O}}_{\text{3}}\text{×}\text{46}\text{.01g}\text{of}{\text{NO}}_{\text{2}}\\ \text{=}\text{0}\text{.709}\text{g}\text{of}{\text{NO}}_{\text{2}}\end{array}$

From the calculated number of moles of ${\text{O}}_{\text{3}}$ and using stoichiometry the amount of ${\text{NO}}_{\text{2}}$ can be calculated by simple mathematical operations.

Calculate the number of moles of excess reagent remains unreact.

$\begin{array}{l}\text{Number}\text{of}\text{moles}\text{of}\text{NO}\text{after}\text{reaction}\text{=}\text{0}\text{.0154}\text{mol}\text{of}{\text{O}}_{\text{3}}\text{×}\frac{\text{1}\text{mol}\text{NO}}{\text{1}\text{mol}{\text{O}}_{\text{3}}}\\ \text{=}\text{0}\text{.0154}\text{mol}\\ \text{Number}\text{of}\text{moles}\text{of}\text{NO}\text{before}\text{reaction}\text{=}\text{0}\text{.670}\text{g}\text{of}\text{NO}\text{×}\frac{\text{1}\text{mol}\text{NO}}{\text{30}\text{mol}\text{NO}}\\ \text{=}\text{0}\text{.0223}\text{mol}\\ \text{unreacted mol NO = 0}\text{.0223 mol NO - 0}\text{.0154 mol NO}\end{array}$

$unreactedmolNO=0.0069molNO$

The number of moles of unreacted $\text{NO}$ can be calculated by subtracting number of moles of $\text{NO}$ before reaction and number of moles of $\text{NO}$ after reaction.

Conclusion

• $O_3$ is the limiting reactant.
• $\text{Amount}\text{of}{\text{NO}}_{\text{2}}\text{=}\text{0}\text{.709}\text{g}$
• $\text{unreacted mol NO = 0}\text{.0069 mol NO}$