Chapter 9, Problem 12PE

(a)

Interpretation Introduction

Interpretation:

The number of moles of water are needed to react with $\text{100}\text{g}$ moles of ${\text{Al}}_{\text{4}}{\text{C}}_{\text{3}}$ 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 12PE

The number of moles of water are needed to react with $\text{100}\text{g}$ moles of ${\text{Al}}_{\text{4}}{\text{C}}_{\text{3}}$ is $8.33\text{mol}\text{.}$

Explanation of Solution

Given,

The mass of ${\text{Al}}_{\text{4}}{\text{C}}_{\text{3}}$ is $\text{100}\text{g}\text{.}$

The molecular weight of ${\text{Al}}_{\text{4}}{\text{C}}_{\text{3}}$ is $\text{144}\text{g}\text{.}$

The balanced reaction is,

  ${\text{Al}}_{\text{4}}{\text{C}}_{\text{3}}{\text{+12 H}}_{\text{2}}\text{O}\stackrel{}{\to }\text{4}\text{Al}{\left(\text{OH}\right)}_{\text{3}}{\text{+3CH}}_{\text{4}}$

In the mole ratio, the coefficients of the balanced equation are used.  Therefore the mole ratio is $\frac{\text{12}\text{mol}{\text{H}}_{\text{2}}\text{O}}{\text{1}\text{mol}{\text{Al}}_{\text{4}}{\text{C}}_{\text{3}}}\text{.}$

The number of moles can be calculated as,

  $\text{100}\text{g}{\text{Al}}_{\text{4}}{\text{C}}_{\text{3}}\left(\frac{\text{1}\text{mol}}{\text{144}\text{.0}\text{g}}\right)\left(\frac{\text{12}\text{mol}{\text{H}}_{\text{2}}\text{O}}{\text{1}\text{mol}{\text{Al}}_{\text{4}}{\text{C}}_{\text{3}}}\right)\text{=}\text{8}\text{.33}\text{mol}{\text{H}}_{\text{2}}\text{O}$

The number of moles of water are needed to react with $\text{100}\text{g}$ moles of ${\text{Al}}_{\text{4}}{\text{C}}_{\text{3}}$ is $8.33\text{mol}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The number of moles of $\text{Al}{\left(\text{OH}\right)}_{\text{3}}$ produced from $0.600$ moles of ${\text{CH}}_{\text{4}}$ have to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 12PE

The number of moles of $\text{Al}{\left(\text{OH}\right)}_{\text{3}}$ produced from $0.600$ moles of ${\text{CH}}_{\text{4}}$ is $0.800\text{mol}\text{.}$

Explanation of Solution

The balanced reaction is,

  ${\text{Al}}_{\text{4}}{\text{C}}_{\text{3}}{\text{+12 H}}_{\text{2}}\text{O}\stackrel{}{\to }\text{4}\text{Al}{\left(\text{OH}\right)}_{\text{3}}{\text{+3CH}}_{\text{4}}$

In the mole ratio, the coefficients of the balanced equation are used.  Therefore the mole ratio is $\frac{\text{4}\text{mol}\text{Al}{\left(\text{OH}\right)}_{\text{3}}}{\text{3}\text{mol}{\text{CH}}_{\text{4}}}\text{.}$

The number of moles can be calculated as,

  $\text{0}\text{.600}\text{mol}{\text{CH}}_{\text{4}}\left(\frac{\text{4}\text{mol}\text{Al}{\left(\text{OH}\right)}_{\text{3}}}{\text{3}\text{mol}{\text{CH}}_{\text{4}}}\right)\text{=}\text{0}\text{.800}\text{mol}\text{Al}{\left(\text{OH}\right)}_{\text{3}}$

The number of moles of $\text{Al}{\left(\text{OH}\right)}_{\text{3}}$ produced from $0.600$ moles of ${\text{CH}}_{\text{4}}$ is $0.800\text{mol}\text{.}$

(c)

Interpretation Introduction

Interpretation:

The number of moles of ${\text{CH}}_{\text{4}}$ formed by the reaction of $\text{275}\text{g}$ moles of ${\text{Al}}_{\text{4}}{\text{C}}_{\text{3}}$ has to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 12PE

The number of moles of ${\text{CH}}_{\text{4}}$ formed by the reaction of $\text{275}\text{g}$ moles of ${\text{Al}}_{\text{4}}{\text{C}}_{\text{3}}$ is $5.72\text{mol}\text{.}$

Explanation of Solution

Given,

The mass of ${\text{Al}}_{\text{4}}{\text{C}}_{\text{3}}$ is $275\text{g}\text{.}$

The molecular weight of ${\text{Al}}_{\text{4}}{\text{C}}_{\text{3}}$ is $\text{144}\text{g}\text{.}$

The balanced reaction is,

  ${\text{Al}}_{\text{4}}{\text{C}}_{\text{3}}{\text{+12 H}}_{\text{2}}\text{O}\stackrel{}{\to }\text{4}\text{Al}{\left(\text{OH}\right)}_{\text{3}}{\text{+3CH}}_{\text{4}}$

In the mole ratio, the coefficients of the balanced equation are used.  Therefore the mole ratio is $\frac{\text{3}\text{mol}{\text{CH}}_{\text{4}}}{\text{1}\text{mol}{\text{Al}}_{\text{4}}{\text{C}}_{\text{3}}}\text{.}$

The number of moles can be calculated as,

  $\text{275}\text{g}{\text{Al}}_{\text{4}}{\text{C}}_{\text{3}}\left(\frac{\text{1}\text{mol}}{\text{144}\text{.0}\text{g}}\right)\left(\frac{\text{3}\text{mol}{\text{CH}}_{\text{4}}}{\text{1}\text{mol}{\text{Al}}_{\text{4}}{\text{C}}_{\text{3}}}\right)\text{=}\text{5}\text{.72}\text{mol}{\text{CH}}_{\text{4}}$

The number of moles of ${\text{CH}}_{\text{4}}$ formed by the reaction of $\text{275}\text{g}$ moles of ${\text{Al}}_{\text{4}}{\text{C}}_{\text{3}}$ is $5.72\text{mol}\text{.}$

(d)

Interpretation Introduction

Interpretation:

The gram of water is required to produce $\text{4}\text{.22}\text{mol}$ of $\text{Al}{\left(\text{OH}\right)}_{\text{3}}$ have to be given.

Concept Introduction:

Refer to part (a).

Answer to Problem 12PE

The gram of water is required to produce $\text{4}\text{.22}\text{mol}$ of $\text{Al}{\left(\text{OH}\right)}_{\text{3}}$ is $\text{228}\text{g}\text{.}$

Explanation of Solution

The balanced reaction is,

  ${\text{Al}}_{\text{4}}{\text{C}}_{\text{3}}{\text{+12 H}}_{\text{2}}\text{O}\stackrel{}{\to }\text{4}\text{Al}{\left(\text{OH}\right)}_{\text{3}}{\text{+3CH}}_{\text{4}}$

In the mole ratio, the coefficients of the balanced equation are used.  Therefore the mole ratio is $\frac{\text{12}\text{mol}{\text{H}}_{\text{2}}\text{O}}{\text{4}\text{mol}\text{Al}{\left(\text{OH}\right)}_{\text{3}}{}_{\text{4}}}\text{.}$

The number of moles can be calculated as,

  $\text{4}\text{.22}\text{mol}\text{Al}{\left(\text{OH}\right)}_{\text{3}}\left(\frac{\text{12}\text{mol}{\text{H}}_{\text{2}}\text{O}}{\text{4}\text{mol}\text{Al}{\left(\text{OH}\right)}_{\text{3}}{}_{\text{4}}}\right)\text{=}\text{12}\text{.66}\text{mol}{\text{H}}_{\text{2}}\text{O}$

Conversion of moles to grams:

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

  $\frac{\text{18}\text{g}}{\text{1}\text{mol}}\text{×12}\text{.66}\text{mol}{\text{H}}_{\text{2}}\text{O=228}\text{g}{\text{H}}_{\text{2}}\text{O}$

The gram of water is required to produce $\text{4}\text{.22}\text{mol}$ of $\text{Al}{\left(\text{OH}\right)}_{\text{3}}$ is $\text{228}\text{g}\text{.}$