Chapter 9, Problem 31PE

(a)

Interpretation Introduction

Interpretation:

The theoretical yield of $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ produced has to be given.

Answer to Problem 31PE

The theoretical yield of $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ from Copper is $\text{27}\text{.04}\text{g}\text{.}$

The theoretical yield of $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ from ${\text{HNO}}_{\text{3}}$ is $43.40\text{g}\text{.}$

Explanation of Solution

Given,

The mass of copper is $27.5\text{g}\text{.}$

The mass of ${\text{HNO}}_{\text{3}}$ is $\text{125}\text{g}\text{.}$

The molecular weight of copper is $\text{63}\text{.546}\text{g/mol}\text{.}$

The molecular weight of $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is $187.5\text{g/mol}\text{.}$

The molecular weight of ${\text{HNO}}_{\text{3}}$ is $\text{63g/mol}\text{.}$

The balanced equation is,

  $\text{3}\text{Cu+8}{\text{HNO}}_{\text{3}}\stackrel{}{\to }\text{3}\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}{\text{+4H}}_{\text{2}}\text{O+2NO}$

In the reaction there is three mole of $\text{Cu}\text{.}$  Therefore the mass concentration of $\text{Cu}$ is $\text{3×63}\text{.546g}\text{=190}\text{.638}\text{g}\text{.}$  In the reaction there is eight mole of ${\text{HNO}}_{\text{3}}\text{.}$  Therefore the mass concentration of ${\text{HNO}}_{\text{3}}$ is $\text{8×63g}\text{=504}\text{g}\text{.}$

The theoretical yield of $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ from Copper can be calculated as,

  Grams of $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\text{=}\frac{\text{187}\text{.5}\text{gCu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\text{×27}\text{.5}\text{g}\text{Cu }}{\text{190}\text{.638}\text{g}\text{Cu}}$

  Grams of $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\text{=27}\text{.04}\text{g}$

The theoretical yield of $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ from Copper is $\text{27}\text{.04}\text{g}\text{.}$

The theoretical yield of $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ from ${\text{HNO}}_{\text{3}}$ can be calculated as,

  Grams of $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\text{=}\frac{\text{187}\text{.5}\text{gCu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\text{×125}\text{g}{\text{HNO}}_{\text{3}}}{\text{540}\text{g}{\text{HNO}}_{\text{3}}}$

  Grams of $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\text{=43}\text{.40}\text{g}$

The theoretical yield of $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ from ${\text{HNO}}_{\text{3}}$ is $43.40\text{g}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The grams of excess reactant remain have to be given.

Answer to Problem 31PE

The grams of excess reactant remain is $86.12\text{g}\text{.}$

Explanation of Solution

The theoretical yield of $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ from Copper is $\text{27}\text{.04}\text{g}\text{.}$

The theoretical yield of $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ from ${\text{HNO}}_{\text{3}}$ is $43.40\text{g}\text{.}$

The molecular weight of $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is $187.5\text{g/mol}\text{.}$

The molecular weight of ${\text{HNO}}_{\text{3}}$ is $\text{63g/mol}\text{.}$

Since Copper produces less $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ copper is the limiting reactant and Nitric acid is the excess reactant.

The balanced equation is,

  $\text{3}\text{Cu+8}{\text{HNO}}_{\text{3}}\stackrel{}{\to }\text{3}\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}{\text{+4H}}_{\text{2}}\text{O+2NO}$

The mole ratio is $\left(\frac{\text{8}{\text{HNO}}_{\text{3}}}{\text{3}\text{mol}\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}\right).$

The grams of excess reactant reacted can be calculated as,

  $\text{43}\text{.40}\text{g}\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\left(\frac{\text{1}\text{mol}}{\text{187}\text{.5}\text{g}\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}\right)\left(\frac{\text{8}\text{mol}{\text{HNO}}_{\text{3}}}{\text{3}\text{mol}\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}\right)\left(\frac{\text{63}\text{g}}{\text{1}\text{mol}}\right)\text{=38}\text{.88}\text{g}$

The grams of excess reactant remain is $\text{125}\text{g}\left(\text{Initial}\right)\text{-38}\text{.88}\text{g}\left(\text{Final}\right)\text{=86}\text{.12}\text{g}$

The grams of excess reactant remain is $86.12\text{g}\text{.}$

(c)

Interpretation Introduction

Interpretation:

The Actual yield of $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ has to be given.

Answer to Problem 31PE

The actual yield of $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ from Copper is $\text{14}\text{.85}\text{g}\text{.}$

The actual yield of $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ from ${\text{HNO}}_{\text{3}}$ is $23.83\text{g}\text{.}$

Explanation of Solution

Given,

The percentage yield of $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is $\text{87}\text{.3\%}\text{.}$

The theoretical yield of $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ from Copper is $\text{17}\text{.02}\text{g}\text{.}$

The theoretical yield of $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ from ${\text{HNO}}_{\text{3}}$ is $27.31\text{g}\text{.}$

The actual yield of $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ from Copper can be calculated as,

  The percentage yield=$\frac{\text{Actual}\text{yield}}{\text{theoretical}\text{yield}}\text{×100}$

  $\text{Actual}\text{Yield=}\frac{\text{Percentage}\text{yield}\text{×Theoretical}\text{yield}}{\text{100}}$

  $\text{Actual}\text{Yield=}\frac{87.3\text{×17}\text{.02}\text{g}}{\text{100}}$

  $\text{Actual}\text{Yield=14}\text{.85}\text{g}$

The actual yield of $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ from Copper is $\text{14}\text{.85}\text{g}\text{.}$

The actual yield of $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ from ${\text{HNO}}_{\text{3}}$ can be calculated as,

  $\text{Actual}\text{Yield=}\frac{87.3\text{×27}\text{.31}\text{g}}{\text{100}}$

  $\text{Actual}\text{Yield=23}\text{.83}\text{g}$

The actual yield of $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ from ${\text{HNO}}_{\text{3}}$ is $23.83\text{g}\text{.}$