Chapter 9, Problem 25PE

(a)

Interpretation Introduction

Interpretation:

The limiting reactant and the number of moles aluminum sulfate produced have to be given.

Concept Introduction:

The limiting reactant of the reaction is the reactant that is completely used during the reaction.  Using the mole ratio and starting amounts of the reactants limiting reactant can be determined.

Example:

Consider a reaction starts with $\text{30}{\text{g CaCO}}_{\text{3}}\text{\&}\text{11}\text{g}\text{HCl}\text{.}$ The values in grams has to be converted to moles by dividing with their molecular weights.  According to the mole ratio $\text{0}\text{.3}\text{g}$ of calcium carbonate require $\text{0}\text{.6}\text{g}$ of $\text{HCl}$ completely.  Therefore HCl is the limiting agent.

Explanation of Solution

Given,

The mass of $\text{Al}{\left(\text{OH}\right)}_{\text{3}}$ is $35\text{g}\text{.}$

The mass of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $35\text{g}\text{.}$

The molecular weight of $\text{Al}{\left(\text{OH}\right)}_{\text{3}}$ is $78\text{g/mol}\text{.}$

The molecular weight of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\text{98}\text{.07}\text{g/mol}\text{.}$

The molecular weight of ${\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}$ is $\text{342}\text{.15}\text{g/mol}\text{.}$

The given reaction is,

  $\text{2Al}{\left(\text{OH}\right)}_{\text{3}}{\text{+3H}}_{\text{2}}{\text{SO}}_{\text{4}}\stackrel{}{\to }{\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}{\text{+6H}}_{\text{2}}\text{O}$

The limiting reactant can be given by finding the mass of the product.

The moles of ${\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}$ from $\text{Al}{\left(\text{OH}\right)}_{\text{3}}$ can be calculated as,

  $\text{35}\text{g}\text{Al}{\left(\text{OH}\right)}_{\text{3}}\left(\frac{\text{1}\text{mol}\text{Al}{\left(\text{OH}\right)}_{\text{3}}}{\text{78}\text{g}\text{Al}{\left(\text{OH}\right)}_{\text{3}}}\right)\left(\frac{\text{1}\text{mol}{\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}}{\text{2}\text{mol}\text{Al}{\left(\text{OH}\right)}_{\text{3}}}\right)\text{=0}\text{.2243}\text{mol}{\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}$

The moles of ${\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}$ from ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ can be calculated as,

  $\text{35}\text{g}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\frac{\text{1}\text{mol}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{98}\text{.07}\text{g}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}\right)\left(\frac{\text{1}\text{mol}{\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}}{\text{3}\text{mol}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}\right)\text{=0}\text{.1189}\text{mol}{\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}$

The moles of ${\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}$ from aluminum hydroxide is $\text{0}\text{.2243}\text{mol}\text{.}$

The moles of ${\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}$ from sulfuric acid is $\text{0}\text{.1189}\text{mol}\text{.}$

Since ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ produces less ${\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}$, ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is the limiting reactant.

(b)

Interpretation Introduction

Interpretation:

The grams of aluminum sulfate produced and the excess reactant has to be given.

Explanation of Solution

Given,

The mass of $\text{Al}{\left(\text{OH}\right)}_{\text{3}}$ is $25\text{g}\text{.}$

The mass of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $45\text{g}\text{.}$

The molecular weight of $\text{Al}{\left(\text{OH}\right)}_{\text{3}}$ is $78\text{g/mol}\text{.}$

The molecular weight of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\text{98}\text{.07}\text{g/mol}\text{.}$

The molecular weight of ${\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}$ is $\text{342}\text{.15}\text{g/mol}\text{.}$

The given reaction is,

  $\text{2Al}{\left(\text{OH}\right)}_{\text{3}}{\text{+3H}}_{\text{2}}{\text{SO}}_{\text{4}}\stackrel{}{\to }{\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}{\text{+6H}}_{\text{2}}\text{O}$

The excess reactant can be given by finding the mass of the product.

The mass of ${\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}$ from $\text{Al}{\left(\text{OH}\right)}_{\text{3}}$ can be calculated as,

  $\text{25}\text{g}\text{Al}{\left(\text{OH}\right)}_{\text{3}}\left(\frac{\text{1}\text{mol}\text{Al}{\left(\text{OH}\right)}_{\text{3}}}{\text{78}\text{g}\text{Al}{\left(\text{OH}\right)}_{\text{3}}}\right)\left(\frac{\text{1}\text{mol}{\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}}{\text{2}\text{mol}\text{Al}{\left(\text{OH}\right)}_{\text{3}}}\right)\left(\frac{\text{342}\text{.15}\text{g}{\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}}{\text{1}\text{mol}}\right)\text{=54}\text{.83}\text{g}{\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}$

The mass of ${\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}$ from ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ can be calculated as,

  $\text{45}\text{g}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\frac{\text{1}\text{mol}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{98}\text{.07}\text{g}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}\right)\left(\frac{\text{1}\text{mol}{\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}}{\text{3}\text{mol}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}\right)\left(\frac{\text{342}\text{.15}\text{g}{\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}}{\text{1}\text{mol}}\right)\text{=52}\text{.3}\text{g}{\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}$

The mass of ${\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}$ from aluminum hydroxide is $\text{54}\text{.83}\text{g}\text{.}$

The mass of ${\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}$ from sulfuric acid is $\text{52}\text{.3}\text{g}\text{.}$

Since ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ produces less ${\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}$, ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is the limiting reactant and $\text{Al}{\left(\text{OH}\right)}_{\text{3}}$ is the excess reactant.

(c)

Interpretation Introduction

Interpretation:

The number of moles of substance present in the container after the completion of reaction and the substance that will be present in the container has to be given.

Explanation of Solution

Given,

The number of moles of $\text{Al}{\left(\text{OH}\right)}_{\text{3}}$ is $\text{2}\text{.5}\text{mol}\text{.}$

The number of moles of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\text{5}\text{.5}\text{mol}\text{.}$

The given reaction is,

  $\text{2Al}{\left(\text{OH}\right)}_{\text{3}}{\text{+3H}}_{\text{2}}{\text{SO}}_{\text{4}}\stackrel{}{\to }{\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}{\text{+6H}}_{\text{2}}\text{O}$

The moles of ${\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}$ from $\text{Al}{\left(\text{OH}\right)}_{\text{3}}$ can be calculated as,

  $\text{2}\text{.5}\text{mol}\text{Al}{\left(\text{OH}\right)}_{\text{3}}\left(\frac{\text{1}\text{mol}{\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}}{\text{2}\text{mol}\text{Al}{\left(\text{OH}\right)}_{\text{3}}}\right)\text{=1}\text{.25 mol}{\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}$

The number of moles of unreacted ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\text{5}\text{.5-1}\text{.25=4}\text{.25}\text{mol}$

When the reaction is completed $1.25\text{mol}$ of $\text{Al}{\left(\text{OH}\right)}_{\text{3}}$ is produced and $4.25\text{mol}$ of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is unreacted and it will be present inside the container.