Chapter 6, Problem 47QP

Interpretation Introduction

Interpretation:

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

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, $0.269\text{ g}$ of ${\text{C}}_2{\text{H}}_6$ is mixed. So, the mass of oxygen that would be reacted can be calculated as follows.

$\begin{array}{c}{\text{Mass of O}}_2=0.260\cancel{{\text{ g C}}_2{\text{H}}_6}\times \frac{\text{1 }\cancel{{\text{mol C}}_2{\text{H}}_6}}{\text{30}\text{.07}\cancel{{\text{ g of C}}_2{\text{H}}_6}}\times \frac{\text{7 }\cancel{{\text{mol O}}_2}}{2\cancel{{\text{mol C}}_2{\text{H}}_6}}\times \frac{\text{32}{\text{.00 g O}}_2}{\text{1 }\cancel{{\text{mol O}}_2}}\\ =0.968{\text{ g O}}_2\end{array}$

$\text{1}{\text{.00 g O}}_2$ is initially mixed during the reaction, but the reaction only consumes $0.968{\text{ g O}}_2$ . Thus, ${\text{C}}_2{\text{H}}_6$ is a limiting reactant.

Similarly, the amount of carbon dioxide is calculated with the help of the limiting reactant, ${\text{C}}_2{\text{H}}_6$ is as follows.

$\begin{array}{c}{\text{Mass of CO}}_2=0.260\cancel{{\text{ g C}}_2{\text{H}}_6}\times \frac{\text{1 }\cancel{{\text{mol C}}_2{\text{H}}_6}}{\text{30}\text{.07}\cancel{{\text{ g of C}}_2{\text{H}}_6}}\times \frac{\text{4 }\cancel{{\text{mol CO}}_2}}{2\cancel{{\text{mol C}}_2{\text{H}}_6}}\times \frac{\text{44}{\text{.01 g CO}}_2}{\text{1 }\cancel{{\text{mol CO}}_2}}\\ =0.761{\text{ g CO}}_2\end{array}$

Similarly, the amount of water is calculated with the help of the limiting reactant, ${\text{C}}_2{\text{H}}_6$ is as follows.

$\begin{array}{c}{\text{Mass of H}}_2\text{O}=0.260\cancel{{\text{ g C}}_2{\text{H}}_6}\times \frac{\text{1 }\cancel{{\text{mol C}}_2{\text{H}}_6}}{\text{30}\text{.07}\cancel{{\text{ g of C}}_2{\text{H}}_6}}\times \frac{\text{6 }\cancel{{\text{mol H}}_2\text{O}}}{2\cancel{{\text{mol C}}_2{\text{H}}_6}}\times \frac{\text{18}{\text{.02 g H}}_2\text{O}}{\text{1 }\cancel{{\text{mol H}}_2\text{O}}}\\ =0.467{\text{ g H}}_2\text{O}\end{array}$

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

$\begin{array}{l}2{\text{C}}_2{\text{H}}_6\left(g\right)+7{\text{O}}_2\left(g\right)\to 2{\text{CO}}_2\left(g\right)+6{\text{H}}_2\text{O}\left(g\right)\\ \begin{array}{lllll}\hfill & {\text{C}}_2{\text{H}}_6\hfill & 7{\text{O}}_2\hfill & {\text{CO}}_2\hfill & {\text{H}}_2\text{O}\hfill \\ \text{Intially mixed}\hfill & 0.260\text{ g}\hfill & 1.00\text{ g}\hfill & 0.00\text{ g}\hfill & 0.00\text{ g}\hfill \\ \text{How much react}\hfill & 0.260\text{ g}\hfill & 0.968\text{ g}\hfill & -\hfill & -\hfill \\ \text{Composition of final mixture}\hfill & 0.000\text{ g}\hfill & 0.03\text{ g}\hfill & 0.761\text{ g}\hfill & 0.467\text{ g}\hfill \end{array}\end{array}$

Conclusion

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