Chapter 19, Problem 77QRT

(a)

Interpretation Introduction

Interpretation:

Given equation has to be balanced.

  ${\text{MnO}}_{\text{2}}\text{(s)}\text{+}{\text{Br}}^{\text{-}}\text{(aq)}\stackrel{}{\to }{\text{Mn}}^{\text{2+}}\text{(aq)}\text{+}{\text{Br}}_{\text{2}}\text{(l)}$

Explanation of Solution

Given equation is,

    ${\text{MnO}}_{\text{2}}\text{(s)}\text{+}{\text{Br}}^{\text{-}}\text{(aq)}\stackrel{}{\to }{\text{Mn}}^{\text{2+}}\text{(aq)}\text{+}{\text{Br}}_{\text{2}}\text{(l)}$

The oxidation state of manganese has changed from $+4$ to $+2$.  Oxidation state of bromine has changed from $-1$ to zero.  Therefore, this is a redox reaction.

Balancing the equation:

Balanced chemical equation can be written for the given reaction by the method of half-reactions.  Summing up the half-reactions obtained gives the resultant reaction.

  $\begin{array}{l}\underset{\_}{\begin{array}{l}{\text{2Br}}^{-}\text{(aq)}\stackrel{}{\to }{\text{Br}}_{\text{2}}\text{(l)}+2e^{-}\\ {\text{2e}}^{-}\text{+}{\text{4H}}_{\text{3}}{\text{O}}^{\text{+}}\text{(aq)}\text{+}{\text{MnO}}_{\text{2}}\text{(s)}\stackrel{}{\to }{\text{Mn}}^{\text{2+}}\text{(aq)}+{\text{6H}}_{\text{2}}\text{O(l)}\end{array}}\\ {\text{2Br}}^{-}\text{(aq)}+{\text{4H}}_{\text{3}}{\text{O}}^{\text{+}}\text{(aq)}+{\text{MnO}}_{\text{2}}\text{(s)}\stackrel{}{\to }{\text{Mn}}^{\text{2+}}\text{(aq)}+{\text{6H}}_{\text{2}}\text{O(l)+}{\text{Br}}_{\text{2}}\text{(l)}\end{array}$

Therefore, the balanced chemical equation is,

    ${\text{2Br}}^{-}\text{(aq)}+{\text{4H}}_{\text{3}}{\text{O}}^{\text{+}}\text{(aq)}+{\text{MnO}}_{\text{2}}\text{(s)}\stackrel{}{\to }{\text{Mn}}^{\text{2+}}\text{(aq)}+{\text{6H}}_{\text{2}}\text{O(l)+}{\text{Br}}_{\text{2}}\text{(l)}$

(b)

Interpretation Introduction

Interpretation:

Amount of bromide ions that is required to produce $\text{6}\text{.50}\text{g}$ of bromine has to be calculated.

Explanation of Solution

In the problem statement it is given that the yield is $\text{100}\text{\%}$.  Therefore, stoichiometry can be used to calculate the amount of bromide ions required.

    $\begin{array}{l}\text{6}\text{.50}\text{g}\text{Br2}\text{×}\frac{\text{1}\text{mol}{\text{Br}}_{\text{2}}}{\text{159}\text{.808}\text{g}{\text{Br}}_{\text{2}}}\text{×}\frac{\text{2}\text{mol}{\text{Br}}^{\text{-}}}{\text{1}\text{mol}{\text{Br}}_{\text{2}}}=\frac{\text{13}\text{mol}{\text{Br}}^{\text{-}}}{\text{159}\text{.808}}\\ =\text{0}\text{.08134}\text{mol}{\text{Br}}^{-}\end{array}$

Therefore, the amount of bromide ions required is $\text{0}\text{.08134}\text{mol}$.

(c)

Interpretation Introduction

Interpretation:

The mass of manganese dioxide that is required to produce $\text{6}\text{.50}\text{g}$ of bromine has to be calculated.

Explanation of Solution

In the problem statement it is given that the yield is $\text{100}\text{\%}$.  Therefore, stoichiometry can be used to calculate the mass of manganese dioxide required.

    $\begin{array}{l}\text{6}\text{.50}\text{g}\text{Br2}\text{×}\frac{\text{1}\text{mol}{\text{Br}}_{\text{2}}}{\text{159}\text{.808}\text{g}{\text{Br}}_{\text{2}}}\text{×}\frac{\text{1}\text{mol}{\text{MnO}}_{\text{2}}}{\text{1}\text{mol}{\text{Br}}_{\text{2}}}\text{×}\frac{\text{86}\text{.9368}\text{g}{\text{MnO}}_{\text{2}}}{\text{1}\text{mol}{\text{MnO}}_{\text{2}}}\text{=}\frac{\text{565}\text{.0892}\text{g}{\text{MnO}}_{\text{2}}}{\text{159}\text{.808}}\\ \text{=}\text{3}\text{.536}\text{g}{\text{MnO}}_{\text{2}}\end{array}$

Therefore, the amount of manganese dioxide required is $\text{3}\text{.536}\text{g}$.