Chapter 9, Problem 11PE

(a)

Interpretation Introduction

Interpretation:

The moles of $\text{HCl}$ reacted with $1.05$ moles of ${\text{MnO}}_{\text{2}}$ 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 11PE

The mole of $\text{HCl}$ react with $1.05$ moles of ${\text{MnO}}_{\text{2}}$ is $4.20\text{mol}\text{.}$

Explanation of Solution

The balanced reaction is,

  ${\text{MnO}}_{\text{2}}\left(\text{s}\right)\text{+4}\text{HCl}\left(\text{aq}\right)\stackrel{}{\to }{\text{Cl}}_{\text{2}}\left(\text{g}\right){\text{+MnCl}}_{\text{2}}\left(\text{aq}\right){\text{+2H}}_{\text{2}}\text{O}$

In the mole ratio, the coefficients of the balanced equation are used.  Therefore the mole ratio is $\frac{\text{4}\text{mol}\text{HCl}}{\text{1}\text{mol}{\text{MnO}}_{\text{2}}}\text{.}$

The number of moles can be calculated as,

  $\text{1}\text{.05}\text{mol}{\text{MnO}}_{\text{2}}\left(\frac{\text{4}\text{mol}\text{HCl}}{\text{1}\text{mol}{\text{MnO}}_{\text{2}}}\right)\text{=}\text{4}\text{.20}\text{mol}\text{NaOH}$

The mole of $\text{HCl}$ react with $1.05$ moles of ${\text{MnO}}_{\text{2}}$ is $4.20\text{mol}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The moles of ${\text{MnCl}}_{\text{2}}$ produced from $1.25$ moles of ${\text{H}}_{\text{2}}\text{O}$ have to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 11PE

The moles of ${\text{MnCl}}_{\text{2}}$ produced from $1.25$ moles of ${\text{H}}_{\text{2}}\text{O}$ is $0.625\text{mol}\text{.}$

Explanation of Solution

The balanced reaction is,

  ${\text{MnO}}_{\text{2}}\left(\text{s}\right)\text{+4}\text{HCl}\left(\text{aq}\right)\stackrel{}{\to }{\text{Cl}}_{\text{2}}\left(\text{g}\right){\text{+MnCl}}_{\text{2}}\left(\text{aq}\right){\text{+2H}}_{\text{2}}\text{O}$

In the mole ratio, the coefficients of the balanced equation are used.  Therefore the mole ratio is $\frac{\text{1}\text{mol}{\text{MnCl}}_{\text{2}}}{\text{2}\text{mol}{\text{H}}_{\text{2}}\text{O}}\text{.}$

The number of moles can be calculated as,

  $\text{1}\text{.25}\text{mol}{\text{H}}_{\text{2}}\text{O}\left(\frac{\text{1}\text{mol}{\text{MnCl}}_{\text{2}}}{\text{2}\text{mol}{\text{H}}_{\text{2}}\text{O}}\right)\text{=}\text{0}\text{.625}\text{mol}{\text{MnCl}}_{\text{2}}$

The moles of ${\text{MnCl}}_{\text{2}}$  produced from $1.25$ moles of ${\text{H}}_{\text{2}}\text{O}$ is $0.625\text{mol}\text{.}$

(c)

Interpretation Introduction

Interpretation:

The grams of ${\text{Cl}}_{\text{2}}$ will produce when reacts with $3.28$ moles of ${\text{MnO}}_{\text{2}}$ have to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 11PE

The grams of ${\text{Cl}}_{\text{2}}$ will produce when reacts with $3.28$ moles of ${\text{MnO}}_{\text{2}}$ is $116\text{g}\text{.}$

Explanation of Solution

The balanced reaction is,

  ${\text{MnO}}_{\text{2}}\left(\text{s}\right)\text{+4}\text{HCl}\left(\text{aq}\right)\stackrel{}{\to }{\text{Cl}}_{\text{2}}\left(\text{g}\right){\text{+MnCl}}_{\text{2}}\left(\text{aq}\right){\text{+2H}}_{\text{2}}\text{O}$

In the mole ratio, the coefficients of the balanced equation are used.  Therefore the mole ratio is $\frac{\text{1}\text{mol}{\text{Cl}}_{\text{2}}}{\text{1}\text{mol}{\text{MnO}}_{\text{2}}}\text{.}$

The number of moles can be calculated as,

  $\text{3}\text{.28}\text{mol}{\text{MnO}}_{\text{2}}\left(\frac{\text{1}\text{mol}{\text{Cl}}_{\text{2}}}{\text{1}\text{mol}{\text{MnO}}_{\text{2}}}\right)\text{=3}\text{.28}\text{mol}{\text{Cl}}_{\text{2}}$

Conversion of moles to grams:

The molar mass of chlorine is $35.45\text{g/mol}\text{.}$

  $\frac{\text{35}\text{.45}\text{g}}{\text{1}\text{mol}}\text{×3}\text{.28}\text{mol}{\text{Cl}}_{\text{2}}\text{=116}\text{g}{\text{Cl}}_{\text{2}}$

The grams of ${\text{Cl}}_{\text{2}}$ will produce when reacts with $3.28$ moles of ${\text{MnO}}_{\text{2}}$ is $116\text{g}\text{.}$

(d)

Interpretation Introduction

Interpretation:

The moles of $\text{HCl}$ required to produce from $\text{15}\text{.0}\text{kg}$ of ${\text{MnCl}}_{\text{2}}$ have to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 11PE

The moles of $\text{HCl}$ required to produce from $\text{15}\text{.0}\text{kg}$ of ${\text{MnCl}}_{\text{2}}$ is $476\text{mol}\text{.}$

Explanation of Solution

The mass of ${\text{MnCl}}_{\text{2}}$ is $\text{15}\text{kg}\text{.}$

The molecular weight of ${\text{MnCl}}_{\text{2}}$$126\text{g/mol}\text{.}$

The mass in kilogram has to converted to gram,

  $\text{1}\text{kg=1000}\text{g}$

  $\text{15}\text{kg=15000}\text{g}$

The number of moles can be calculated as,

  $\text{Number}\text{of}\text{moles=}\frac{\text{Mass}}{\text{Molecular}\text{weight}}$

  $\text{Number}\text{of}\text{moles=}\frac{\text{15000}\text{g}}{\text{126}\text{g/mol}}$

  $\text{Number}\text{of}\text{moles=119}\text{mol}$

The balanced reaction is,

  ${\text{MnO}}_{\text{2}}\left(\text{s}\right)\text{+4}\text{HCl}\left(\text{aq}\right)\stackrel{}{\to }{\text{Cl}}_{\text{2}}\left(\text{g}\right){\text{+MnCl}}_{\text{2}}\left(\text{aq}\right){\text{+2H}}_{\text{2}}\text{O}$

In the mole ratio, the coefficients of the balanced equation are used.  Therefore the mole ratio is $\frac{\text{4}\text{mol}\text{HCl}}{\text{1}\text{mol}{\text{MnCl}}_{\text{2}}}\text{.}$

The number of moles can be calculated as,

  $\text{119}\text{mol}{\text{MnCl}}_{\text{2}}\left(\frac{\text{4}\text{mol}\text{HCl}}{\text{1}\text{mol}{\text{MnCl}}_{\text{2}}}\right)\text{=}\text{476}\text{mol}{\text{MnCl}}_{\text{2}}$

The moles of $\text{HCl}$ required to produce from $\text{15}\text{.0}\text{kg}$ of ${\text{MnCl}}_{\text{2}}$ is $476\text{mol}\text{.}$