Chapter 9, Problem 17PE

(a)

Interpretation Introduction

Interpretation:

The number moles of water are needed to react $\text{10}\text{mol}$ of oxygen has to be given.

Concept Introduction:

Mole ratio:

A mole ratio is a ratio between the numbers of moles of any two species involved in a chemical reaction.

Example,

In the reaction ${\text{2H}}_{\text{2}}{\text{+O}}_{\text{2}}\stackrel{}{\to }{\text{2H}}_{\text{2}}\text{O}$, the mole ratio can be written as $\frac{\text{2}\text{mol}{\text{H}}_{\text{2}}}{\text{1}\text{mol}{\text{O}}_{\text{2}}}.$

Answer to Problem 17PE

The number moles of water are needed to react $\text{10}\text{mol}$ of oxygen is $0.87\text{mol}\text{.}$

Explanation of Solution

Given,

The number of moles of oxygen is $\text{10}\text{mol}\text{.}$

The balanced reaction is,

  ${\text{2C}}_{\text{12}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{6}}\text{+23}{\text{O}}_{\text{2}}{\text{+2H}}_{\text{2}}\text{O}\stackrel{}{\to }\text{24}{\text{CO}}_{\text{2}}\text{+12HCl}$

In the mole ratio, the coefficients of the balanced equation are used.  Therefore the mole ratio is $\frac{\text{2}\text{mol}{\text{H}}_{\text{2}}\text{O}}{\text{23}\text{mol}{\text{O}}_{\text{2}}}\text{.}$

The number of moles can be calculated as,

  $\text{10}\text{mol}{\text{O}}_{\text{2}}\left(\frac{\text{2}\text{mol}{\text{H}}_{\text{2}}\text{O}}{\text{23}\text{mol}{\text{O}}_{\text{2}}}\right)\text{=}\text{0}\text{.87}\text{mol}{\text{H}}_{\text{2}}\text{O}$

The number moles of water are needed to react $\text{10}\text{mol}$ of oxygen is $0.87\text{mol}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The number of grams of $\text{HCl}$ has to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 17PE

The number of grams of $\text{HCl}$ is $3325\text{g}\text{.}$

Explanation of Solution

Given,

The number of moles of water is $\text{15}\text{.2}\text{mol}\text{.}$

The balanced reaction is,

  ${\text{2C}}_{\text{12}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{6}}\text{+23}{\text{O}}_{\text{2}}{\text{+2H}}_{\text{2}}\text{O}\stackrel{}{\to }\text{24}{\text{CO}}_{\text{2}}\text{+12HCl}$

In the mole ratio, the coefficients of the balanced equation are used.  Therefore the mole ratio is $\frac{\text{12}\text{mol}\text{HCl}}{\text{2}\text{mol}{\text{H}}_{\text{2}}\text{O}}\text{.}$

The number of moles can be calculated as,

  $\text{15}\text{.2}\text{mol}{\text{H}}_{\text{2}}\text{O}\left(\frac{\text{12}\text{mol}\text{HCl}}{\text{2}\text{mol}{\text{H}}_{\text{2}}\text{O}}\right)\text{=91}\text{.2}\text{mol}\text{HCl}$

Conversion of moles to grams:

The molar mass of $\text{HCl}$ is $36.46\text{g/mol}\text{.}$

  $\frac{\text{36}\text{.46}\text{g}}{\text{1}\text{mol}}\text{×91}\text{.2}\text{mol}\text{HCl=3325}\text{g}\text{HCl}$

The number of grams of $\text{HCl}$ is $3325\text{g}\text{.}$

(c)

Interpretation Introduction

Interpretation:

The number of moles of carbon dioxide are produced when $\text{76}\text{.5}\text{g}\text{HCl}$ produced has to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 17PE

The number of moles of carbon dioxide is $\text{4}\text{.19}\text{mol}\text{.}$

Explanation of Solution

Given,

The molecular weight of hydrochloric acid is $\text{36}\text{.46}\text{g}\text{.}$

The molecular weight of carbon dioxide is $\text{44}\text{.01}\text{g}\text{.}$

The given mass of hydrochloric acid is $\text{76}\text{.5}\text{g}\text{.}$

The balanced reaction is,

  ${\text{2C}}_{\text{12}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{6}}\text{+23}{\text{O}}_{\text{2}}{\text{+2H}}_{\text{2}}\text{O}\stackrel{}{\to }\text{24}{\text{CO}}_{\text{2}}\text{+12HCl}$

In the mole ratio, the coefficients of the balanced equation are used.  Therefore the mole ratio is $\frac{\text{24}\text{mol}{\text{CO}}_{\text{2}}}{\text{12}\text{mol}\text{HCl}}\text{.}$

The number of moles can be calculated as,

  $\text{76}\text{.5}\text{g}\text{HCl}\left(\frac{\text{1}\text{mol}}{\text{36}\text{.46}\text{g}}\right)\left(\frac{\text{24}\text{mol}{\text{CO}}_{\text{2}}}{\text{12}\text{mol}\text{HCl}}\right)\text{=4}\text{.19 mol}{\text{CO}}_{\text{2}}$

The number of moles of carbon dioxide is $\text{4}\text{.19}\text{mol}\text{.}$

(d)

Interpretation Introduction

Interpretation:

The number of grams of ${\text{C}}_{\text{12}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{6}}$ produced has to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 17PE

The number of grams of ${\text{C}}_{\text{12}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{6}}$ is $\text{1645}\text{g}\text{.}$

Explanation of Solution

Given,

The molecular weight of ${\text{C}}_{\text{12}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{6}}$ is $\text{361}\text{g}\text{.}$

The molecular weight of carbon dioxide is $\text{44}\text{.01}\text{g}\text{.}$

The given mass of carbon di oxide is $100.25\text{g}\text{.}$

The balanced reaction is,

  ${\text{2C}}_{\text{12}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{6}}\text{+23}{\text{O}}_{\text{2}}{\text{+2H}}_{\text{2}}\text{O}\stackrel{}{\to }\text{24}{\text{CO}}_{\text{2}}\text{+12HCl}$

In the mole ratio, the coefficients of the balanced equation are used.  Therefore the mole ratio is $\frac{\text{24}\text{mol}{\text{CO}}_{\text{2}}}{\text{2}\text{mol}{\text{C}}_{\text{12}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{6}}}\text{.}$

The number of grams can be calculated as,

  $\text{100}\text{.25}\text{g}{\text{CO}}_{\text{2}}\left(\frac{\text{1}\text{mol}}{\text{44}\text{.01}\text{g}{\text{CO}}_{\text{2}}}\right)\left(\frac{\text{24}\text{mol}{\text{CO}}_{\text{2}}}{\text{2}\text{mol}{\text{C}}_{\text{12}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{6}}}\right)\left(\frac{\text{361}\text{g}}{\text{1}\text{mol}}\right)\text{=9868}\text{g}$

The number of grams of ${\text{C}}_{\text{12}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{6}}$ is $9868\text{g}\text{.}$

(e)

Interpretation Introduction

Interpretation:

The number of grams of $\text{HCl}$ produced has to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 17PE

The number of grams of $\text{HCl}$ is $\text{1515}\text{g}\text{.}$

Explanation of Solution

Given,

The molecular weight of ${\text{C}}_{\text{12}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{6}}$ is $\text{361}\text{g}\text{.}$

The molecular weight of $\text{HCl}$ is $36.46\text{g}\text{.}$

The given mass of ${\text{C}}_{\text{12}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{6}}$ is $\text{2}\text{.5}\text{kg}\text{.}$

The mass in kilogram has to converted to gram,

  $\text{1}\text{kg=1000}\text{g}$

  $\text{2}\text{.5kg=2500}\text{g}$

The balanced reaction is,

  ${\text{2C}}_{\text{12}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{6}}\text{+23}{\text{O}}_{\text{2}}{\text{+2H}}_{\text{2}}\text{O}\stackrel{}{\to }\text{24}{\text{CO}}_{\text{2}}\text{+12HCl}$

In the mole ratio, the coefficients of the balanced equation are used.  Therefore the mole ratio is $\frac{\text{12}\text{mol}\text{HCl}}{\text{2}\text{mol}{\text{C}}_{\text{12}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{6}}}\text{.}$

The number of grams can be calculated as,

  $2500\text{g}{\text{C}}_{\text{12}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{6}}\left(\frac{\text{1}\text{mol}}{361\text{g}{\text{C}}_{\text{12}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{6}}}\right)\left(\frac{\text{12}\text{mol}\text{HCl}}{\text{2}\text{mol}{\text{C}}_{\text{12}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{6}}}\right)\left(\frac{36.46\text{g}}{\text{1}\text{mol}}\right)\text{=1515}\text{g}$

The number of grams of $\text{HCl}$ is $\text{1515}\text{g}\text{.}$