Chapter 4, Problem 5MCP

(a)

Interpretation Introduction

Interpretation:

A balanced equation for the reaction of atmospheric oxygen with ethane has to be written.  Also, when $\text{1 mol}$ of ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$ reacts, the number of moles of water produced has to be given.

Explanation of Solution

Given:

Atmospheric oxygen reacts with ethane to produce water and carbon dioxide as product.

Balanced equation:

The balanced chemical equation for the given reaction is written as,

  ${\text{2C}}_{\text{2}}{\text{H}}_{\text{6}}{\text{+ 7O}}_{\text{2}}\stackrel{}{\to }{\text{4CO}}_{\text{2}}{\text{+ 6H}}_{\text{2}}\text{O}$

The given moles of ethane is ${\text{1 mol C}}_{\text{2}}{\text{H}}_{\text{6}}$.

From the balanced equation, it is known that two moles of ethane reacts and gives six moles of water.  The molar ratio is ${\text{2 mol C}}_{\text{2}}{\text{H}}_{\text{6}}{\text{ : 6 mol H}}_{\text{2}}\text{O}$.

Calculate the moles of water produced from one mole of ethane as follows,

  ${\text{1 mol C}}_{\text{2}}{\text{H}}_{\text{6}}\times \frac{6{\text{ mol H}}_{\text{2}}\text{O}}{{\text{2 mol C}}_{\text{2}}{\text{H}}_{\text{6}}}{\text{ = 3 mol H}}_{\text{2}}\text{O}$

Therefore, the number of moles of water produced is ${\text{3 mol H}}_{\text{2}}\text{O}$

(b)

Interpretation Introduction

Interpretation:

A balanced equation for the reaction of atmospheric oxygen with ethylene has to be written.  Also, when $\text{1 mol}$ of ${\text{C}}_{\text{2}}{\text{H}}_4$ reacts, the number of moles of water produced has to be given.

Explanation of Solution

Given:

Atmospheric oxygen reacts with ethylene to produce water and carbon dioxide as product.

Balanced equation:

The balanced chemical equation for the given reaction is written as,

  ${\text{C}}_{\text{2}}{\text{H}}_4{\text{+ 3O}}_{\text{2}}\stackrel{}{\to }{\text{ 2CO}}_{\text{2}}{\text{+ 2H}}_{\text{2}}\text{O}$

The given moles of ethylene is ${\text{1 mol C}}_{\text{2}}{\text{H}}_4$.

From the balanced equation, it is known that one mole of ethylene reacts and gives two moles of water.  The molar ratio is ${\text{1 mol C}}_{\text{2}}{\text{H}}_4{\text{ : 2 mol H}}_{\text{2}}\text{O}$.

Calculate the moles of water produced from one mole of ethylene as follows,

  ${\text{1 mol C}}_{\text{2}}{\text{H}}_4\times \frac{{\text{2 mol H}}_{\text{2}}\text{O}}{{\text{1 mol C}}_{\text{2}}{\text{H}}_4}{\text{ = 2 mol H}}_{\text{2}}\text{O}$

Therefore, the number of moles of water produced is ${\text{2 mol H}}_{\text{2}}\text{O}$

(c)

Interpretation Introduction

Interpretation:

The number of moles of ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}$ used in reaction of $0.10\text{ g}$ of ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}$ has to be calculated.

Concept Introduction:

Moles:

Mole of the substance is found by dividing the mass of the substance by its molar mass.

  $\text{No}\text{. of moles (n) = }\frac{\text{mass}}{\text{Molar mass}}$

Explanation of Solution

Given:

Atmospheric oxygen reacts with ethylene to produce water and carbon dioxide as product.

The amount of ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}$ reacts is $\text{0}\text{.10 g}$.

The balanced chemical equation for the given reaction is written as,

  ${\text{C}}_{\text{2}}{\text{H}}_4{\text{+ 3O}}_{\text{2}}\stackrel{}{\to }{\text{ 2CO}}_{\text{2}}{\text{+ 2H}}_{\text{2}}\text{O}$

The molar mass of ethylene is $\text{28}\text{.05 g/mol}$

Determine the moles of ethylene present in given amount of ethylene as follows,

  $\begin{array}{l}\text{moles}\text{= }\frac{\text{mass}}{\text{Molar mass}}\\ =\frac{\text{0}\text{.10 g}}{\text{28}\text{.05 g/mol}}\\ =0.0036\text{ mol}\end{array}$

The number of moles of ethylene used in reaction is $0.0036\text{ moles}\text{.}$

(d)

Interpretation Introduction

Interpretation:

The number of molecules of ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}$ used in reaction of $0.10\text{ g}$ of ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}$ has to be calculated.

Concept Introduction:

Steps to find number of molecules from mass:

  $\left(\text{Gram}\right)\underset{\text{g}\times \frac{\text{1 mol}}{\text{g}}=\text{mol}}{\overset{\text{Divide by molar mass}}{\to }}\left(\text{moles}\right)\underset{\text{mol}\times \frac{6.022\times {10}^{23}\text{ particles}}{\text{1 mol}}=\text{particles}}{\overset{\text{multiply by Avogadro's number }}{\to }}\left(\begin{array}{l}\text{Particles (or)}\\ \text{molecules}\end{array}\right)$

Explanation of Solution

Given:

Atmospheric oxygen reacts with ethylene to produce water and carbon dioxide as product.

The amount of ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}$ reacts is $\text{0}\text{.10 g}$.

The balanced chemical equation for the given reaction is written as,

  ${\text{C}}_{\text{2}}{\text{H}}_4{\text{+ 3O}}_{\text{2}}\stackrel{}{\to }{\text{ 2CO}}_{\text{2}}{\text{+ 2H}}_{\text{2}}\text{O}$

The molar mass of ethylene is $\text{28}\text{.05 g/mol}$

Determine the moles of ethylene present in given amount of ethylene as follows,

  $\begin{array}{l}\text{moles}\text{= }\frac{\text{mass}}{\text{Molar mass}}\\ =\frac{\text{0}\text{.10 g}}{\text{28}\text{.05 g/mol}}\\ =0.0036\text{ mol}\end{array}$

The number of moles of ethylene used in reaction is $0.0036\text{ moles}\text{.}$

The number of molecules is found as follows,

  $0.0036\text{ mol}\times \frac{6.022\times {10}^{23}\text{ molecules}}{\text{ 1 mol}}=2.17\times {10}^{21}\text{ molecules}$

Therefore, the number of molecules of ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}$ that are used are $2.17\times {10}^{21}\text{ molecules}$.

(e)

Interpretation Introduction

Interpretation:

The number of moles of each products produced by the reaction of $0.10\text{ g}$ of ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}$ has to be calculated.

Concept Introduction:

Refer subpart (c).

Explanation of Solution

Given:

Atmospheric oxygen reacts with ethylene to produce water and carbon dioxide as product.

The amount of ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}$ reacts is $\text{0}\text{.10 g}$.

The balanced chemical equation for the given reaction is written as,

  ${\text{C}}_{\text{2}}{\text{H}}_4{\text{+ 3O}}_{\text{2}}\stackrel{}{\to }{\text{ 2CO}}_{\text{2}}{\text{+ 2H}}_{\text{2}}\text{O}$

The molar mass of ethylene is $\text{28}\text{.05 g/mol}$

Determine the moles of ethylene present in given amount of ethylene as follows,

  $\begin{array}{l}\text{moles}\text{= }\frac{\text{mass}}{\text{Molar mass}}\\ =\frac{\text{0}\text{.10 g}}{\text{28}\text{.05 g/mol}}\\ =0.0036\text{ mol}\end{array}$

The number of moles of ethylene used in reaction is $0.0036\text{ moles}\text{.}$

From balanced equation, it is known that two moles of water and two moles of carbon dioxide are produced from one mole of ethylene.  The molar ratio is,

  $\begin{array}{l}{\text{1 mol C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{ : 2 mol H}}_{\text{2}}\text{O}\\ {\text{1 mol C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{ : 2 mol CO}}_{\text{2}}\end{array}$

The number of moles of each product produced can be calculated as below.

  $\begin{array}{l}0.0036{\text{ mol C}}_{\text{2}}{\text{H}}_{\text{4}}\times \frac{2{\text{ mol H}}_{\text{2}}\text{O}}{1{\text{ mol C}}_{\text{2}}{\text{H}}_{\text{4}}}=\text{0}{\text{.0072 mol H}}_{\text{2}}\text{O}\\ 0.0036{\text{ mol C}}_{\text{2}}{\text{H}}_{\text{4}}\times \frac{2{\text{ mol CO}}_{\text{2}}}{1{\text{ mol C}}_{\text{2}}{\text{H}}_{\text{4}}}=\text{0}{\text{.0072 mol CO}}_{\text{2}}\end{array}$

Therefore, the number of moles of ${\text{H}}_{\text{2}}\text{O}\text{and}{\text{CO}}_{\text{2}}$ formed are $\text{0}{\text{.0072 mol H}}_{\text{2}}\text{O and 0}{\text{.0072 mol CO}}_{\text{2}}$.

(f)

Interpretation Introduction

Interpretation:

The amount (in $\text{g}$) of each products produced in the given reaction has to be calculated.

Concept Introduction:

Mass:

Mass of the compound is calculated by mole of the compound multiplied with molar mass of the compound.

  $\text{Mass}\text{=}\text{Molar}\text{mass}\text{×}\text{mole}$

Explanation of Solution

Given:

Atmospheric oxygen reacts with ethylene to produce water and carbon dioxide as product.

The amount of ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}$ reacts is $\text{0}\text{.10 g}$.

The balanced chemical equation for the given reaction is written as,

  ${\text{C}}_{\text{2}}{\text{H}}_4{\text{+ 3O}}_{\text{2}}\stackrel{}{\to }{\text{ 2CO}}_{\text{2}}{\text{+ 2H}}_{\text{2}}\text{O}$

From subpart (e), the moles of each product (water and carbon dioxide) formed are found as $\text{0}{\text{.0072 mol H}}_{\text{2}}\text{O and 0}{\text{.0072 mol CO}}_{\text{2}}$

The molar mass of ${\text{H}}_{\text{2}}\text{O}$ is $\text{18}\text{.015 g/mol}$.

Determine the mass of ${\text{H}}_{\text{2}}\text{O}$ formed as follows,

  $\begin{array}{l}\text{Mass}\text{=}\text{ Molar}\text{mass}\text{×}\text{mole}\\ \text{= 18}\text{.015 g/mol}\times \text{0}{\text{.0072 mol H}}_{\text{2}}\text{O}\\ =0.1297\text{ g}\\ =0.1297{\text{ g H}}_{\text{2}}\text{O}\end{array}$

The molar mass of ${\text{CO}}_{\text{2}}$ is $\text{44}\text{.01 g/mol}$.

Determine the mass of ${\text{CO}}_{\text{2}}$ formed as follows,

  $\begin{array}{l}\text{Mass}\text{=}\text{ Molar}\text{mass}\text{×}\text{mole}\\ \text{= 44}\text{.01 g/mol}\times \text{0}{\text{.0072 mol CO}}_{\text{2}}\\ =0.3169\text{ g}\\ =0.3169{\text{ g CO}}_{\text{2}}\end{array}$

Therefore, the mass of products produced are $0.1297{\text{ g H}}_{\text{2}}\text{O}$ and $0.3169{\text{ g CO}}_{\text{2}}$.

(g)

Interpretation Introduction

Interpretation:

The actual yield of the given reaction when it is $95\%$ efficient has to be calculated.

Explanation of Solution

Given:

Atmospheric oxygen reacts with ethylene to produce water and carbon dioxide as product.

The amount of ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}$ reacts is $\text{0}\text{.10 g}$.

The balanced chemical equation for the given reaction is written as,

  ${\text{C}}_{\text{2}}{\text{H}}_4{\text{+ 3O}}_{\text{2}}\stackrel{}{\to }{\text{ 2CO}}_{\text{2}}{\text{+ 2H}}_{\text{2}}\text{O}$

From subpart (f), the mass $\text{(in g)}$ of each products formed is $0.1297{\text{ g H}}_{\text{2}}\text{O}$ and $0.3169{\text{ g CO}}_{\text{2}}$.

Therefore, the total yield of products is $\text{0}\text{.4466 g}$.

The actual yield when the reaction is only $95\%$ efficient is calculated as follows,

  $\text{0}\text{.4466 g}\times \frac{95}{100}=0.4243\text{ g}$

Therefore, the actula yield of the reaction is only $0.4243\text{ g}$.