Chapter 6, Problem 48QP

Interpretation Introduction

Interpretation:

The molecules to be reacted and the final composition of the mixture in the given reaction are to be identified.

Concept Introduction:

A chemical equation is balanced when the number of atoms of each element and the charges on the atoms for both the reactants and products are the same.

Explanation of Solution

Given Information: A reaction is given in the question.

Initially, $0.130\text{ g}$ of ${\text{C}}_4{\text{H}}_{10}$ is mixed. So, the mass of oxygen that would be reacted can be calculated as follows.

$\begin{array}{c}{\text{Mass of O}}_2=0.130\cancel{{\text{ g C}}_4{\text{H}}_{10}}\times \frac{\text{1 }\cancel{{\text{mol C}}_4{\text{H}}_{10}}}{\text{58}\text{.12}\cancel{{\text{ g of C}}_4{\text{H}}_{10}}}\times \frac{\text{13 }\cancel{{\text{mol O}}_2}}{2\cancel{{\text{mol C}}_2{\text{H}}_6}}\times \frac{\text{32}{\text{.00 g O}}_2}{\text{1 }\cancel{{\text{mol O}}_2}}\\ =0.465{\text{ g O}}_2\end{array}$

$\text{0}{\text{.580 g O}}_2$ is initially present in the reaction, but the reaction consumes only $0.465{\text{ g O}}_2$ . Thus, ${\text{C}}_4{\text{H}}_{10}$ is a limiting reactant. The final composition of the mass of ${\text{CO}}_2$ is calculated with the help of the limiting reactant, ${\text{C}}_4{\text{H}}_{10}$ , which is as follows.

$\begin{array}{c}{\text{Mass of CO}}_2=0.130\cancel{{\text{ g C}}_4{\text{H}}_{10}}\times \frac{\text{1 }\cancel{{\text{mol C}}_4{\text{H}}_{10}}}{\text{58}\text{.12}\cancel{{\text{ g of C}}_4{\text{H}}_{10}}}\times \frac{\text{8 }\cancel{{\text{mol CO}}_2}}{2\cancel{{\text{mol C}}_2{\text{H}}_6}}\times \frac{\text{44}{\text{.01 g CO}}_2}{\text{1 }\cancel{{\text{mol CO}}_2}}\\ =0.394{\text{ g O}}_2\end{array}$

The final composition of the mass of ${\text{H}}_2\text{O}$ is calculated with the help of the limiting reactant, ${\text{C}}_4{\text{H}}_{10}$ , which is as follows.

$\begin{array}{c}{\text{Mass of H}}_2\text{O}=0.130\cancel{{\text{ g C}}_4{\text{H}}_{10}}\times \frac{\text{1 }\cancel{{\text{mol C}}_4{\text{H}}_{10}}}{\text{58}\text{.12}\cancel{{\text{ g of C}}_4{\text{H}}_{10}}}\times \frac{\text{10 }\cancel{{\text{mol H}}_2\text{O}}}{2\cancel{{\text{mol C}}_4{\text{H}}_6}}\times \frac{\text{18}{\text{.02 g H}}_2\text{O}}{\text{1 }\cancel{{\text{mol H}}_2\text{O}}}\\ =0.202{\text{ g H}}_2\text{O}\end{array}$

The number of molecules of butane and oxygen that react to form the desirable amount of carbon dioxide and water is as follows:

$\begin{array}{l}2{\text{C}}_4{\text{H}}_{10}\left(g\right)+13{\text{O}}_2\left(g\right)\to 8{\text{CO}}_2\left(g\right)+10{\text{H}}_2\text{O}\left(g\right)\\ \begin{array}{lllll}\hfill & {\text{C}}_2{\text{H}}_6\hfill & 7{\text{O}}_2\hfill & {\text{CO}}_2\hfill & {\text{H}}_2\text{O}\hfill \\ \text{Intially mixed}\hfill & 0.130\text{ g}\hfill & 0.580\text{ g}\hfill & 0.00\text{ g}\hfill & 0.00\text{ g}\hfill \\ \text{How much react}\hfill & 0.130\text{ g}\hfill & 0.465\text{ g}\hfill & -\hfill & -\hfill \\ \text{Composition of final mixture}\hfill & 0.000\text{ g}\hfill & 0.115\text{ g}\hfill & 0.394\text{ g}\hfill & 0.202\text{ g}\hfill \end{array}\end{array}$

Conclusion

The molecules to be reacted and the final composition of the mixture in the given reaction are shown in the table above.