Chapter 4, Problem 4.161QP

The following “cycle of copper” experiment is performed in some general chemistry laboratories. The series of reactions starts with copper and ends with metallic copper. The steps are as follows: (1) A piece of copper wire of known mass is allowed to react with concentrated nitric acid [the products are copper(II) nitrate, nitrogen dioxide, and water]. (2) The copper(II) nitrate is treated with a sodium hydroxide solution to form copper(II) hydroxide precipitate. (3) On heating, copper(II) hydroxide decomposes to yield copper(II) oxide. (4) The copper(II) oxide is reacted with concentrated sulfuric acid to yield copper(II) sulfate. (5) Copper(II) sulfate is treated with an excess of zinc metal to form metallic copper. (6) The remaining zinc metal is removed by treatment with hydrochloric acid, and metallic copper is filtered, dried, and weighed. (a) Write a balanced equation for each step and classify the reactions. (b) Assuming that a student started with 65.6 g of copper, calculate the theoretical yield at each step. (c) Considering the nature of the steps, comment on why it is possible to recover most of the copper used at the start.

Interpretation Introduction

Interpretation:

The balanced equations for the given reaction are should be written and type of reactions are should be classified and theoretical yield of the each steps should be calculated then possibility of Copper recovered from given steps should be commended.

Concept introduction:

Balanced equation:

• Balanced equation is equation for the chemical reaction, which is written by the number of atoms for each element and total charges of ions are equal in the reaction for in both reactants and the products sides.

Redox reaction:

• The oxidation number of a molecule is varied by involving in reaction is called redox reaction.  The oxidation and reduction of chemical compounds takes place while they are involving in the reaction is called redox reaction.

Acid -Base reaction:

• The reaction between acid and base to produce a salt is called acid-base reaction.

Precipitation reaction:

• The precipitate is formed, when the two solutions mixed together is called precipitation reaction.

Decomposition reaction:

• In the reaction, reactants are decomposed to give a product is known as decomposition reaction.
• To write the balanced equations for given reactions and explain the type of reaction.

Answer to Problem 4.161QP

Copper wire metal is treated with concentrated Nitric acid to gives Copper (II) nitrate and it is a redox reaction.

${\text{Cu}}_{\text{(s)}}{\text{ + 4HNO}}_{\text{3(aq)}}\to {\text{Cu(NO}}_{\text{3}}{\text{)}}_{\text{2(aq)}}{\text{ + 2NO}}_{\text{2(g)}}\uparrow {\text{ + 2H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

Copper (II) nitrate is treated with Sodium hydroxide to gives Copper (II) hydroxide precipitate and it is a precipitation reaction.

${\text{Cu(NO}}_{\text{3}}{\text{)}}_{\text{2(aq)}}{\text{ + 2NaOH}}_{\text{(aq) }}\to {\text{Cu(OH)}}_{\text{2(s)}}\downarrow {\text{+2NaNO}}_{\text{3(aq)}}$

On heating of Copper (II) hydroxide to gives Copper (II) oxide and it is a decomposition reaction.

${\text{Cu(OH)}}_{\text{2(s)}}\stackrel{\Delta }{\to }{\text{CuO}}_{\text{(s)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(g)}}$

Copper (II) oxide is treated with con. Sulfuric acid to gives Copper sulphate and it is an acid-base reaction.

${\text{CuO}}_{\text{(s)}}{\text{+H}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}\to {\text{CuSO}}_{\text{4(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

Copper sulphate is treated with Zinc metal to gives Zinc (II) sulphate and it is a redox reaction.

${\text{CuSO}}_{\text{4(aq)}}{\text{+Zn}}_{\text{(s)}}\to {\text{Cu}}_{\text{(s)}}{\text{+ZnSO}}_{\text{4(aq)}}$

Zinc (II) sulphate is treated with $\text{HCl}$ to gives and it is a redox reaction.

${\text{Zn}}_{\text{(s)}}{\text{+ 2HCl}}_{\text{(aq)}}\to {\text{ZnCl}}_{\text{2(aq) }}{\text{+ H}}_{\text{2(g)}}$

Explanation of Solution

Copper wire metal is treated with concentrated Nitric acid to gives Copper (II) nitrate.

${\text{Cu}}_{\text{(s)}}{\text{ + 4HNO}}_{\text{3(aq)}}\to {\text{Cu(NO}}_{\text{3}}{\text{)}}_{\text{2(aq)}}{\text{ + 2NO}}_{\text{2(g)}}\uparrow {\text{ + 2H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

The above reaction is a redox reaction.

Copper (II) nitrate is treated with Sodium hydroxide to gives Copper (II) hydroxide precipitate.

${\text{Cu(NO}}_{\text{3}}{\text{)}}_{\text{2(aq)}}{\text{ + 2NaOH}}_{\text{(aq) }}\to {\text{Cu(OH)}}_{\text{2(s)}}\downarrow {\text{+2NaNO}}_{\text{3(aq)}}$

The above reaction is a precipitation reaction.

On heating of Copper (II) hydroxide to gives Copper (II) oxide

${\text{Cu(OH)}}_{\text{2(s)}}\stackrel{\Delta }{\to }{\text{CuO}}_{\text{(s)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(g)}}$

The above reaction is a decomposition reaction.

Copper (II) oxide is treated with con. Sulfuric acid to gives Copper sulphate

${\text{CuO}}_{\text{(s)}}{\text{+H}}_{\text{2}}{\text{SO}}_{\text{4(aq)}}\to {\text{CuSO}}_{\text{4(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

The above reaction is an acid-base reaction.

Copper sulphate is treated with Zinc metal to gives Zinc (II) sulphate.

${\text{CuSO}}_{\text{4(aq)}}{\text{+Zn}}_{\text{(s)}}\to {\text{Cu}}_{\text{(s)}}{\text{+ZnSO}}_{\text{4(aq)}}$

The above reaction is a redox reaction.

Zinc (II) sulphate is treated with $\text{HCl}$ to gives.

${\text{Zn}}_{\text{(s)}}{\text{+ 2HCl}}_{\text{(aq)}}\to {\text{ZnCl}}_{\text{2(aq) }}{\text{+ H}}_{\text{2(g)}}$

The above reaction is a redox reaction.

Conclusion

The balanced equations for the given reaction are written and type of reactions is classified

Interpretation Introduction

Concept introduction:

Molarity:

• The concentration of the solutions is given by the term of molarity and it is given by ratio between numbers of moles of solute present in litter of solution.

$\text{Molarity}\text{=}\frac{\text{No}\text{.mole}}{\text{volume}\text{(L)}}$

Mole:

• The mole of compound is given by multiplication of volume and molarity of the solution.

  $\text{Mole}\text{=}\text{Volume}\text{(L)×}\text{Molarity}\text{(M)}$

Explanation of Solution

Record the given data.

Taken mass of Copper wire = $\text{65}\text{.6}\text{g}$

Taken mass of copper wire is recorded as shown above.ss

Calculate the theoretical yield of the each step involved in the given process.

Molar mass of Copper is $\text{63}\text{.55}\text{g}$

The mole Copper present in of taken gram copper wire is,

$\begin{array}{l}\text{=}\text{65}\text{.6}\text{g×}\frac{\text{1}\text{mol}\text{Cu}}{\text{63}\text{.55}\text{g}\text{Cu}}\\ \text{=}\text{1}\text{.032}\text{mole}\text{Cu}\end{array}$

Theoretical yield in each reaction containing $\text{1}\text{.032}\text{mole}$ of Copper are,

Molar mass of ${\text{Cu(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ is $\text{187}\text{.57}\text{g}$

$\begin{array}{l}\text{=}\text{1}\text{.032}\text{mole}\text{×}\frac{\text{187}\text{.57}\text{g}{\text{Cu(NO}}_{\text{3}}{\text{)}}_{\text{2}}}{\text{1}\text{mole}{\text{Cu(NO}}_{\text{3}}{\text{)}}_{\text{2}}}\\ \text{=}\text{194}\text{.}0\text{g}{\text{Cu(NO}}_{\text{3}}{\text{)}}_{\text{2}}\end{array}$

Molar mass of ${\text{Cu(OH)}}_{\text{2}}$ is $\text{97}\text{.566 g}$

$\begin{array}{l}\text{=}\text{1}\text{.032}\text{mole}\text{×}\frac{97.566\text{g}{\text{Cu(OH)}}_{\text{2}}}{\text{1}\text{mole}{\text{Cu(OH)}}_{\text{2}}}\\ \text{=}101.0\text{g}{\text{Cu(OH)}}_{\text{2}}\end{array}$

Molar mass of $\text{CuO}$ is $\text{79}\text{.55 g}$

$\begin{array}{l}\text{=}\text{1}\text{.032}\text{mole}\text{×}\frac{79.55\text{g}\text{CuO}}{\text{1}\text{mole}\text{CuO}}\\ \text{=}8\text{2}\text{.1}\text{g}\text{CuO}\end{array}$

Molar mass of ${\text{CuSO}}_{\text{4}}$ is $\text{159}\text{.62}\text{g}$

$\begin{array}{l}\text{=}\text{1}\text{.032}\text{mole}\text{×}\frac{159.62\text{g}{\text{CuSO}}_{\text{4}}}{\text{1}\text{mole}{\text{CuSO}}_{\text{4}}}\\ \text{=}165.0\text{g}{\text{CuSO}}_{\text{4}}\end{array}$

Molar mass of $\text{Cu}$ is $\text{63}\text{.55}\text{g}$

$\begin{array}{l}\text{=}\text{1}\text{.032}\text{mole}\text{×}\frac{\text{63}\text{.55}\text{g}\text{Cu}}{\text{1}\text{mole}\text{Cu}}\\ \text{=}\text{65}\text{.}\text{6}\text{g}\text{Cu}\end{array}$

• Theoretical yield in each reaction containing $\text{1}\text{.032}\text{mole}$ of Copper are calculated by the calculated mole of Copper taken in to process and molar masses of products of each steps are plugged in the above equations to gives the Theoretical yield in each reactions in the given process.
• Theoretical yield of varying in VI step because washing technique, which is used in the recovering of Copper in to reaction mixture.

Conclusion

The theoretical yield of the each step involved in the given process was calculated.

To comment the effectiveness of given Copper recovery process.

• Each step in the given process is clean.
•  Each step in the given process nearly quantitative.
•  Therefore yield of the over all process should be high.

The effectiveness of given Copper recovery process was commended.