Chapter 3, Problem 3.162QE

Although copper does not usually react with acids, it does react with concentrated nitric acid. The reaction is complicated, but one outcome is

$\text{Cu}(\text{s})+{\text{HNO}}_3(\text{conc})\to \text{Cu}{({\text{NO}}_3)}_2(\text{aq})+{\text{NO}}_2(\text{g})+{\text{H}}_2\text{O}(\mathcal{l})$

1. (a) Name all of the reactants and products.
2. (b) Balance the reaction.
3. (c) Assign oxidation numbers to the atoms. Is this a redox reaction?
4. (d) Pre-1983 pennies were made of pure copper. If such a penny had a mass of 3.10 g, how many moles of Cu are in one penny? How many atoms of copper are in one penny?
5. (e) What mass of HNO₃ would be needed to completely react with a pre-1983 penny?

Interpretation Introduction

Interpretation:

The reactants and products in the given reaction has to be named.

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

Explanation of Solution

The given reaction is:

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

The name of the reactants is:

    $\begin{array}{l}{\text{Cu}}_{\text{(s)}}\text{-Copper metal}\\ {\text{HNO}}_{\text{3}}\text{- Nitric acid}\end{array}$

The name of the products is:

    $\begin{array}{l}{\text{Cu(NO}}_{\text{3}}{\text{)}}_{\text{2(aq)}}\text{-Copper nitrate}\\ {\text{NO}}_{\text{2}}\text{- Nitrogen dioxide}\\ {\text{H}}_{\text{2}}\text{O-Water}\end{array}$

Interpretation Introduction

Interpretation:

The given reaction has to be balanced.

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

Explanation of Solution

The given reaction is:

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

Add 2 as coefficient in front of nitrogen dioxide.

Then the equation will be:

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

Now balance the nitrogen atoms, for that place 4 in front of nitric acid.

Equation will be:

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

Finally balance the hydrogen and oxygen atoms, for that place a coefficient of 2 in front of water.

Then the balanced equation is:

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

Interpretation Introduction

Interpretation:

The oxidation numbers of the atoms present in the equation has to be given.  Also, whether the reaction is a redox reaction or not has to be given.

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

Explanation of Solution

The given equation is:

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

The oxidation number of metallic copper is 0.

Oxidation number of hydrogen in nitric acid is +1.

Oxidation number of nitrogen in nitric acid is +5.

Oxidation number of oxygen in nitric acid is -6.

Oxidation number of copper in copper nitrate is +2.

Nitrate is a polyatomic ion and its oxidation state is-1.

Oxidation number of nitrogen in nitrogen dioxide is +4.

Oxidation number of oxygen in nitrogen dioxide is -2.

Oxidation state of hydrogen in water is +1.

Oxidation number of oxygen in water is 2.

Interpretation Introduction

Interpretation:

$\text{pre-1983}$ pennies were made of pure copper.  If such penny had a mass of $\text{3}\text{.10g}$, then the number of moles and copper atoms present in one penny has to be determined.

Explanation of Solution

  $\text{Number of moles =}\frac{\text{Mass in gram of copper}}{\text{Molar mass of copper}}\text{=}\frac{\text{3}\text{.10g}}{\text{63}\text{.5g/mol}}\text{=0}\text{.048moles}$.

  $\begin{array}{l}\text{1mole Cu=6}{\text{.023×10}}^{\text{23}}\text{atoms}\\ \text{0}\text{.048 mole Cu =0}\text{.048}\times \text{6}{\text{.023×10}}^{\text{23}}\text{atoms}=2.94{\text{×10}}^{\text{22}}\text{atoms}\end{array}$

Interpretation Introduction

Interpretation:

The mass of nitric acid would be needed to completely react with a $\text{pre-1983}$ penny has to be determined.

Explanation of Solution

The equation is:

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

Molar mass of nitric acid is $\text{63}\text{.01g/mol}$.

Molar mass of copper is $\text{63}\text{.5g/mol}$.

Therefore,

Nitric acid required is:

    $\text{=}\frac{\text{63}\text{.01g/mol×3}\text{.10g}}{\text{63}\text{.5g/mol}}\text{=3}\text{.07g}$