Chapter 4, Problem 152AP

Interpretation Introduction

Interpretation:

The balanced chemical equations of the reactions are to be represented and thetheoretical yield of copper at each step is to be determined.

Concept introduction:

A chemical reaction always follows law of conservation of mass, according to which, when a chemical reaction occurs, the mass of atoms in products should be equal to the mass of atoms in reactants.

Classification of reactions depends on the reactants and products involved in the reaction.

The number of moles of a reactant is calculated by the formula as follows:

$m=\frac{wt}{MW}$

Here, $m$ is the number of moles of compound, $wt$ is the weight of the compound, and $MW$ is the molecular weight of the compound.

Answer to Problem 152AP

Solution:

(a)

$\text{Cu}\left(s\right)+4{\text{HNO}}_3\left(aq\right)\to \text{Cu}{\left({\text{NO}}_3\right)}_2\left(s\right)+2{\text{NO}}_2\left(g\right)+2{\text{H}}_2\text{O}\left(l\right)$(Redox Reaction)

$\text{Cu}{\left({\text{NO}}_3\right)}_2\left(aq\right)+2\text{NaOH}\left(aq\right)\to \text{Cu}{\left(\text{OH}\right)}_2\left(s\right)+2{\text{NaNO}}_{\text{3}}\left(aq\right)$(Precipitation Reaction)

$\text{Cu}{\left(\text{OH}\right)}_2\left(s\right)\to \text{CuO}\left(s\right)+{\text{H}}_2\text{O}\left(g\right)$ (Decomposition Reaction)

$\text{CuO}\left(s\right)+{\text{H}}_2{\text{SO}}_4\left(aq\right)\to {\text{CuSO}}_4\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)$(Acid-Base Reaction)

${\text{CuSO}}_4\left(aq\right)+\text{Zn}\left(s\right)\to \text{Cu}\left(s\right)+{\text{ZnSO}}_4\left(aq\right)$(Redox Reaction)

$\text{Zn}\left(s\right)+2\text{HCl}\left(aq\right)\to {\text{ZnCl}}_2\left(aq\right)+{\text{H}}_2\left(g\right)$ (Redox Reaction)

(b)

$\text{Yield of copper}=194\text{ g Cu}{({\text{NO}}_3)}_2$

$\text{Yield of copper}=101\text{ g Cu}({\text{OH}}_2)$

$\text{Yield of copper}=82.1\text{ g CuO}$

$\text{Yield of copper}=165{\text{ g CuSO}}_4$

$\text{Yield of copper}=65.6\text{ g Cu}$

(c) All the steps in the reaction are quantitative.

Explanation of Solution

a)Balanced equation for each step and classify reactions

The equation for the reaction is as follows:

$\text{Cu}\left(s\right)+4{\text{HNO}}_3\left(aq\right)\to \text{Cu}{\left({\text{NO}}_3\right)}_2\left(s\right)+2{\text{NO}}_2\left(g\right)+2{\text{H}}_2\text{O}\left(l\right)$

This is a redox reaction as here, copper oxidizes to form ${\text{Cu}}^{2+}$ ions and nitrogen reduces from $+5$ in ${\text{NO}}_3{}^{-}$ to $+4$ in ${\text{NO}}_2$.

2)

The equation for the reaction is as follows:

$\text{Cu}{\left({\text{NO}}_3\right)}_2\left(aq\right)+2\text{NaOH}\left(aq\right)\to \text{Cu}{\left(\text{OH}\right)}_2\left(s\right)+2{\text{NaNO}}_{\text{3}}\left(aq\right)$

This is a precipitation reaction. In this reaction, $\text{Cu}{\left(\text{OH}\right)}_2$ is obtained as a precipitate.

3)

The equation for the reaction is as follows:

$\text{Cu}{\left(\text{OH}\right)}_2\left(s\right)\to \text{CuO}\left(s\right)+{\text{H}}_2\text{O}\left(g\right)$

This is a decomposition reaction as in this reaction, $\text{Cu}{\left(\text{OH}\right)}_2$ decomposes to form products.

4)

The equation for the reaction is as follows:

$\text{CuO}\left(s\right)+{\text{H}}_2{\text{SO}}_4\left(aq\right)\to {\text{CuSO}}_4\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)$

This is an acid-base reaction. Here, CuO acts as a baseand ${\text{H}}_2{\text{SO}}_4$ as an acid. Thus, CuO accepts protons from ${\text{H}}_2{\text{SO}}_4$ to form ${\text{H}}_2\text{O}$ as the product.

5)

The equation for the reaction is as follows:

${\text{CuSO}}_4\left(aq\right)+\text{Zn}\left(s\right)\to \text{Cu}\left(s\right)+{\text{ZnSO}}_4\left(aq\right)$

This reaction is a redox reaction as here, Cu reducesfrom ${\text{Cu}}^{+2}$ to Cu metal and Zn oxidizes from Zn metal to ${\text{Zn}}^{2+}$.

6)

The equation for the reaction is as follows:

$\text{Zn}\left(s\right)+2\text{HCl}\left(aq\right)\to {\text{ZnCl}}_2\left(aq\right)+{\text{H}}_2\left(g\right)$

This reaction is a redox reaction as here, hydrogen reducesfrom ${\text{H}}^{+}$ to ${\text{H}}_2$ and Zn oxidizesfrom Zn to ${\text{Zn}}^{2+}$.

b) Theoretical Yield at each step

Initial mass ofcopper metal is $65.6\text{ g}$.

Initially, the number of moles of copper in the reaction is calculated as follows:

$\begin{array}{c}{m}_{\text{Cu}}=\left(\frac{65.6}{63.55}\right)\\ =1.032\text{ moles}\end{array}$

From the reactions given above, it can be observed that the molar ratio between reactant and product is 1:1.

Using the initial number of moles of copper available, the yield of copper in all the reactions is determined as follows:

1) The mass of $\text{Cu}{\left({\text{NO}}_3\right)}_2$ is as follows:

$1.032=\left(\frac{w{t}_{\text{Cu}{\left({\text{NO}}_3\right)}_2}}{187.57}\right)$

Rearranging the equation:

$\begin{array}{c}w{t}_{\text{Cu}{\left({\text{NO}}_3\right)}_2}=\left(1.032\right)\left(187.57\right)\\ =194\text{ g Cu}{\left({\text{NO}}_3\right)}_2\end{array}$

2) The mass of $\text{Cu}{\left(\text{OH}\right)}_2$ is as follows:

$1.032=\left(\frac{w{t}_{\text{Cu}{\left(\text{O}H\right)}_2}}{97.566}\right)$

Rearranging the equation:

$\begin{array}{c}w{t}_{\text{Cu}{\left(\text{OH}\right)}_2}=\left(1.032\right)\left(97.566\right)\\ =101\text{ g Cu}{\left(\text{OH}\right)}_2\end{array}$

3) The mass of CuO is as follows:

$1.032=\left(\frac{w{t}_{\text{CuO}}}{79.55}\right)$

Rearranging the equation:

$\begin{array}{c}w{t}_{\text{Cu}{\left(\text{OH}\right)}_2}=\left(1.032\right)\left(79.55\right)\\ =82.1\text{ g CuO}\end{array}$

4) The mass of CuO is as follows:

$1.032=\left(\frac{w{t}_{{\text{CuSO}}_{\text{4}}}}{159.62}\right)$

Rearranging the equation:

$\begin{array}{c}w{t}_{{\text{CuSO}}_{\text{4}}}=\left(1.032\right)\left(159.62\right)\\ =165{\text{ g CuSO}}_{\text{4}}\end{array}$

5) The mass of Cu is as follows:

$1.032=\left(\frac{w{t}_{\text{Cu}}}{63.55}\right)$

Rearranging the equation:

$\begin{array}{c}w{t}_{\text{Cu}}=\left(1.032\right)\left(63.55\right)\\ =65.6\text{ g Cu}\end{array}$

c)Comment on why it is possible to recover most of the copper used at the start of reaction.

The recovery of most of the copper is possible in these reactions because fewer impurities are involved in the reactants. Thus, the reactions are clean and almost quantitative, therefore the recovery yield is high.