Chapter 5, Problem 87P

Interpretation Introduction

Interpretation:

The following table should be completed using the given information. The limiting reactant and the reactant used in excess should be predicted.

  

Concept Introduction:

Mole is the amount of the substance that contains the same number of particles or atoms or molecules. Molar mass is defined as average mass of atoms present in the chemical formula. It is the sum of the atomic masses of all the atoms present in the chemical formula of any compound.

Answer to Problem 87P

The limiting reactant is ${\text{O}}_{\text{2}}$ and reactant that is present in excess is ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}$ . The complete table is mentioned below:

  $\begin{array}{l}\text{Reactants}\text{Products}\\ \begin{array}{cccccccc}\text{Equation}& {\text{C}}_{\text{2}}{\text{H}}_{\text{4}}& +& {\text{3O}}_{\text{2}}& \to & {\text{2CO}}_{\text{2}}& +& {\text{2H}}_{\text{2}}\text{O}\\ \text{Initial}\text{quantities}\left(\text{molecules}\right)& 8& & 15& & 0& & 0\\ \text{Molecules}\text{used}\text{or}\text{formed}& 5& & 15& & 10& & 10\\ \text{Molecules}\text{remaining}& 3& & 0& & 10& & 10\end{array}\end{array}$

Explanation of Solution

The given reaction is ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}+{\text{3O}}_{\text{2}}\to {\text{2CO}}_{\text{2}}+{\text{2H}}_{\text{2}}\text{O}$ . The given molecules of ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}$ is 8, ${\text{O}}_{\text{2}}$ is 15, ${\text{CO}}_{\text{2}}$ is 0 and ${\text{H}}_{\text{2}}\text{O}$ is 0. According to the reaction conditions, 1 molecule of ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}$ reacts with 3 molecule of ${\text{O}}_{\text{2}}$ to produce 2 moles of ${\text{CO}}_{\text{2}}$ and 2 moles of ${\text{H}}_{\text{2}}\text{O}$.

Therefore, 8 moles of ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}$ react with 24 moles of ${\text{O}}_{\text{2}}$ ; but there are only 15 molecules of ${\text{O}}_{\text{2}}$ . The substance that is totally consumed when the reaction is completed is known as limiting reactant. Therefore, the limiting reactant is ${\text{O}}_{\text{2}}$ and reactant that is present in excess is ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}$.

The amount of product would be formed according to the ${\text{O}}_{\text{2}}$ . The amount of ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}$ is calculated as follows:

  $\begin{array}{c}\text{3}\text{moles}\text{of}{\text{O}}_{\text{2}}=\text{1}\text{mole}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\\ 1\text{mole}\text{of}{\text{O}}_{\text{2}}=\frac{\text{1}}{3}\text{mole}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\\ 15\text{mole}\text{of}{\text{O}}_{\text{2}}=\left(\frac{\text{1}}{3}\times 15\right)\text{moles}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\\ =5\text{moles}\text{of}{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\end{array}$

The amount of ${\text{CO}}_{\text{2}}$ is calculated as follows:

  $\begin{array}{c}\text{3}\text{moles}\text{of}{\text{O}}_{\text{2}}=2\text{moles}\text{of}{\text{CO}}_{\text{2}}\\ 1\text{mole}\text{of}{\text{O}}_{\text{2}}=\frac{2}{3}\text{mole}\text{of}{\text{CO}}_{\text{2}}\\ 15\text{mole}\text{of}{\text{O}}_{\text{2}}=\left(\frac{2}{3}\times 15\right)\text{moles}\text{of}{\text{CO}}_{\text{2}}\\ =10\text{moles}\text{of}{\text{CO}}_{\text{2}}\end{array}$

The amount of ${\text{H}}_{\text{2}}\text{O}$ is calculated as follows:

  $\begin{array}{c}\text{3}\text{moles}\text{of}{\text{O}}_{\text{2}}=2\text{moles}\text{of}{\text{H}}_{\text{2}}\text{O}\\ 1\text{mole}\text{of}{\text{O}}_{\text{2}}=\frac{2}{3}\text{mole}\text{of}{\text{H}}_{\text{2}}\text{O}\\ 15\text{mole}\text{of}{\text{O}}_{\text{2}}=\left(\frac{2}{3}\times 15\right)\text{moles}\text{of}{\text{H}}_{\text{2}}\text{O}\\ =10\text{moles}\text{of}{\text{H}}_{\text{2}}\text{O}\end{array}$

Therefore, the limiting reactant is ${\text{O}}_{\text{2}}$ and reactant that is present in excess is ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}$ . The complete table is mentioned below:

  $\begin{array}{l}\text{Reactants}\text{Products}\\ \begin{array}{cccccccc}\text{Equation}& {\text{C}}_{\text{2}}{\text{H}}_{\text{4}}& +& {\text{3O}}_{\text{2}}& \to & {\text{2CO}}_{\text{2}}& +& {\text{2H}}_{\text{2}}\text{O}\\ \text{Initial}\text{quantities}\left(\text{molecules}\right)& 8& & 15& & 0& & 0\\ \text{Molecules}\text{used}\text{or}\text{formed}& 5& & 15& & 10& & 10\\ \text{Molecules}\text{remaining}& 3& & 0& & 10& & 10\end{array}\end{array}$

Conclusion

The limiting reactant is ${\text{O}}_{\text{2}}$ and reactant that is present in excess is ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}$ . The complete table is mentioned below:

  $\begin{array}{l}\text{Reactants}\text{Products}\\ \begin{array}{cccccccc}\text{Equation}& {\text{C}}_{\text{2}}{\text{H}}_{\text{4}}& +& {\text{3O}}_{\text{2}}& \to & {\text{2CO}}_{\text{2}}& +& {\text{2H}}_{\text{2}}\text{O}\\ \text{Initial}\text{quantities}\left(\text{molecules}\right)& 8& & 15& & 0& & 0\\ \text{Molecules}\text{used}\text{or}\text{formed}& 5& & 15& & 10& & 10\\ \text{Molecules}\text{remaining}& 3& & 0& & 10& & 10\end{array}\end{array}$