Chapter 5, Problem 88P

Completer the followin table using the given balanced equation and the initial quantities of reactants. Label the limiting reactant and the reactant used in excess.

	Reactants	Products
Equation	${\text{2C}}_{\text{4}}{\text{H}}_{\text{10}}+{\text{13O}}_{\text{2}}\to 8{\text{CO}}_{\text{2}}{\text{+ 10H}}_{\text{2}}\text{O}$
Initial quantities	4.20 mol 6.50 molo.00 molo.oo mol
Molecules used or formed
Molecules remaining

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 88P

The limiting reactant is ${\text{O}}_{\text{2}}$ and reactant that is present in excess is ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}$ . The complete table is mentioned below:

  $\begin{array}{l}\text{Reactants}\text{Products}\\ \begin{array}{cccccccc}\text{Equation}& {\text{2C}}_{\text{4}}{\text{H}}_{\text{10}}& +& {\text{13O}}_{\text{2}}& \to & {\text{8CO}}_{\text{2}}& +& {\text{10H}}_{\text{2}}\text{O}\\ \text{Initial}\text{quantities}\left(\text{mole}\right)& 4.20& & 6.50& & 0& & 0\\ \text{Molecules}\text{used}\text{or}\text{formed}& 1& & 6.50& & 4& & 5\\ \text{Molecules}\text{remaining}& 3.20& & 0& & 4& & 5\end{array}\end{array}$

Explanation of Solution

The given reaction is ${\text{2C}}_{\text{4}}{\text{H}}_{\text{10}}+1{\text{3O}}_{\text{2}}\to 8{\text{CO}}_{\text{2}}+10{\text{H}}_{\text{2}}\text{O}$ . The given moles of ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}$ is 4.20, ${\text{O}}_{\text{2}}$ is 6.50, ${\text{CO}}_{\text{2}}$ is 0 and ${\text{H}}_{\text{2}}\text{O}$ is 0. According to the reaction conditions, 2 moles of ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}$ reacts with 13 moles of ${\text{O}}_{\text{2}}$ to produce 8 moles of ${\text{CO}}_{\text{2}}$ and 10 moles of ${\text{H}}_{\text{2}}\text{O}$ . The mole of ${\text{O}}_{\text{2}}$ would react with 1 mole of ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}$ is calculated as follows:

  $\begin{array}{c}2\text{moles}\text{of}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}=\text{13}\text{mole}\text{of}{\text{O}}_{\text{2}}\\ 1\text{mole}\text{of}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}=\frac{\text{13}}{2}\text{mole}\text{of}{\text{O}}_{\text{2}}\end{array}$

Therefore, the moles of ${\text{O}}_{\text{2}}$ would react with 4.20 moles of ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}$ is calculated as follows:

  $\begin{array}{c}4.20\text{mole}\text{of}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}=\left(\frac{\text{13}}{2}\times 4.20\right)\text{moles}\text{of}{\text{O}}_{\text{2}}\\ =27.3\text{moles}\text{of}{\text{O}}_{\text{2}}\end{array}$

Therefore, 4.20 moles of ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}$ reacts with 27.3 moles of ${\text{O}}_{\text{2}}$ ; but there are only 6.50 moles of ${\text{O}}_{\text{2}}$ present. 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 ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}$.

The amount of product would be formed according to the ${\text{O}}_{\text{2}}$ . The amount of ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}$ is calculated as follows:

  $\begin{array}{c}\text{13}\text{moles}\text{of}{\text{O}}_{\text{2}}=2\text{mole}\text{of}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\\ 1\text{mole}\text{of}{\text{O}}_{\text{2}}=\frac{2}{13}\text{mole}\text{of}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\\ 6.50\text{mole}\text{of}{\text{O}}_{\text{2}}=\left(\frac{2}{13}\times 6.50\right)\text{moles}\text{of}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\\ =1\text{mole}\text{of}{\text{C}}_{\text{4}}{\text{H}}_{\text{10}}\end{array}$

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

  $\begin{array}{c}\text{13}\text{moles}\text{of}{\text{O}}_{\text{2}}=8\text{moles}\text{of}{\text{CO}}_{\text{2}}\\ 1\text{mole}\text{of}{\text{O}}_{\text{2}}=\frac{8}{13}\text{mole}\text{of}{\text{CO}}_{\text{2}}\\ 6.50\text{moles}\text{of}{\text{O}}_{\text{2}}=\left(\frac{8}{13}\times 6.50\right)\text{moles}\text{of}{\text{CO}}_{\text{2}}\\ =4\text{moles}\text{of}{\text{CO}}_{\text{2}}\end{array}$

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

  $\begin{array}{c}\text{13}\text{moles}\text{of}{\text{O}}_{\text{2}}=10\text{moles}\text{of}{\text{H}}_{\text{2}}\text{O}\\ 1\text{mole}\text{of}{\text{O}}_{\text{2}}=\frac{10}{13}\text{mole}\text{of}{\text{H}}_{\text{2}}\text{O}\\ 6.50\text{moles}\text{of}{\text{O}}_{\text{2}}=\left(\frac{10}{13}\times 6.50\right)\text{moles}\text{of}{\text{H}}_{\text{2}}\text{O}\\ =5\text{moles}\text{of}{\text{H}}_{\text{2}}\text{O}\end{array}$

Therefore, the limiting reactant is ${\text{O}}_{\text{2}}$ and reactant that is present in excess is ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}$ . The complete table is mentioned below:

  $\begin{array}{l}\text{Reactants}\text{Products}\\ \begin{array}{cccccccc}\text{Equation}& {\text{2C}}_{\text{4}}{\text{H}}_{\text{10}}& +& {\text{13O}}_{\text{2}}& \to & {\text{8CO}}_{\text{2}}& +& {\text{10H}}_{\text{2}}\text{O}\\ \text{Initial}\text{quantities}\left(\text{mole}\right)& 4.20& & 6.50& & 0& & 0\\ \text{Molecules}\text{used}\text{or}\text{formed}& 1& & 6.50& & 4& & 5\\ \text{Molecules}\text{remaining}& 3.20& & 0& & 4& & 5\end{array}\end{array}$

Conclusion

The limiting reactant is ${\text{O}}_{\text{2}}$ and reactant that is present in excess is ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}$ . The complete table is mentioned below:

  $\begin{array}{l}\text{Reactants}\text{Products}\\ \begin{array}{cccccccc}\text{Equation}& {\text{2C}}_{\text{4}}{\text{H}}_{\text{10}}& +& {\text{13O}}_{\text{2}}& \to & {\text{8CO}}_{\text{2}}& +& {\text{10H}}_{\text{2}}\text{O}\\ \text{Initial}\text{quantities}\left(\text{mole}\right)& 4.20& & 6.50& & 0& & 0\\ \text{Molecules}\text{used}\text{or}\text{formed}& 1& & 6.50& & 4& & 5\\ \text{Molecules}\text{remaining}& 3.20& & 0& & 4& & 5\end{array}\end{array}$