Chapter 3, Problem 3.157QE

(a)

Interpretation Introduction

Interpretation:

The mass of ${\text{K[PtCl}}_{\text{3}}{\text{(C}}_{\text{2}}{\text{H}}_{\text{4}}\text{)]}$ that can be prepared from $\text{45}{\text{.8g K}}_2{\text{[PtCl}}_4\text{]}$ and $\text{12}\text{.5 g}$ ethylene has to be calculated.

Explanation of Solution

The reaction is:

    ${\text{K}}_2{\text{[PtCl}}_4{\text{] + C}}_{\text{2}}{\text{H}}_{\text{4}}\stackrel{}{\to }{\text{ K[PtCl}}_{\text{3}}{\text{(C}}_{\text{2}}{\text{H}}_{\text{4}}\text{)]}$.

Molar mass of ${\text{K}}_2{\text{[PtCl}}_4\text{]}$ is $\text{415}\text{.09g/mol}$.

Molar mass of ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}$ is $\text{28}\text{.05g/mol}$.

Then,

  ${\text{Amount K}}_2{\text{[PtCl}}_4\text{]=45}\text{.8 g×}\left(\frac{{\text{1mol K}}_2{\text{[PtCl}}_4\text{]}}{\text{(415}{\text{.09)g K}}_2{\text{[PtCl}}_4\text{]}}\right)\text{=0}\text{.11mol}$.

  ${\text{Amount C}}_{\text{2}}{\text{H}}_{\text{4}}\text{= 12}\text{.5 g×}\left(\frac{{\text{1mol C}}_{\text{2}}{\text{H}}_{\text{4}}}{\text{(28}{\text{.05)g C}}_{\text{2}}{\text{H}}_{\text{4}}}\right)\text{=0}\text{.44mol}$

${\text{Amount K[PtCl}}_{\text{3}}{\text{(C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{)]based on K}}_2{\text{[PtCl}}_4\text{]=0}{\text{.11mol K}}_2{\text{[PtCl}}_4\text{]×}\left(\frac{{\text{1mol K[PtCl}}_{\text{3}}{\text{(C}}_{\text{2}}{\text{H}}_{\text{4}}\text{)]}}{{\text{1 mol K}}_2{\text{[PtCl}}_4\text{]}}\right)\text{=0}{\text{.11mol K[PtCl}}_{\text{3}}{\text{(C}}_{\text{2}}{\text{H}}_{\text{4}}\text{)]}$

${\text{Amount K[PtCl}}_{\text{3}}{\text{(C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{)]based on C}}_{\text{2}}{\text{H}}_{\text{4}}\text{=0}{\text{.44mol C}}_{\text{2}}{\text{H}}_{\text{4}}\text{×}\left(\frac{{\text{1mol K[PtCl}}_{\text{3}}{\text{(C}}_{\text{2}}{\text{H}}_{\text{4}}\text{)]}}{{\text{1 mol C}}_{\text{2}}{\text{H}}_{\text{4}}}\right)\text{=0}{\text{.44mol K[PtCl}}_{\text{3}}{\text{(C}}_{\text{2}}{\text{H}}_{\text{4}}\text{)]}$

Therefore ${\text{K}}_2{\text{[PtCl}}_4\text{]}$ is the limiting reagent.

Mass of ${\text{K[PtCl}}_{\text{3}}{\text{(C}}_{\text{2}}{\text{H}}_{\text{4}}\text{)]}$ can be prepared is:

    ${\text{Mass of K[PtCl}}_{\text{3}}{\text{(C}}_{\text{2}}{\text{H}}_{\text{4}}\text{)]=}\left[\frac{\text{45}\text{.8×368}\text{.5945}}{\text{415}\text{.09}}\right]\text{=40}\text{.669g}$

(b)

Interpretation Introduction

Interpretation:

The reactant which is present in excess has to be identified and the mass of it that remains at the end of the reaction has to be calculated.

Explanation of Solution

The reaction is:

    ${\text{K}}_2{\text{[PtCl}}_4{\text{] + C}}_{\text{2}}{\text{H}}_{\text{4}}\stackrel{}{\to }{\text{ K[PtCl}}_{\text{3}}{\text{(C}}_{\text{2}}{\text{H}}_{\text{4}}\text{)]}$.

Molar mass of ${\text{K}}_2{\text{[PtCl}}_4\text{]}$ is $\text{415}\text{.09g/mol}$.

Molar mass of ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}$ is $\text{28}\text{.05g/mol}$.

Then,

  ${\text{Amount K}}_2{\text{[PtCl}}_4\text{]=45}\text{.8 g×}\left(\frac{{\text{1mol K}}_2{\text{[PtCl}}_4\text{]}}{\text{(415}{\text{.09)g K}}_2{\text{[PtCl}}_4\text{]}}\right)\text{=0}\text{.11mol}$.

  ${\text{Amount C}}_{\text{2}}{\text{H}}_{\text{4}}\text{= 12}\text{.5 g×}\left(\frac{{\text{1mol C}}_{\text{2}}{\text{H}}_{\text{4}}}{\text{(28}{\text{.05)g C}}_{\text{2}}{\text{H}}_{\text{4}}}\right)\text{=0}\text{.44mol}$

Ethylene is present in excess amount.

The mass of ethylene present is:

    ${\text{Mass of C}}_{\text{2}}{\text{H}}_{\text{4}}\text{=0}\text{.44mol×(28}{\text{.05)g C}}_{\text{2}}{\text{H}}_{\text{4}}\text{=12}\text{.342g}$