Chapter 22, Problem 22.96QP

(a)

Interpretation Introduction

Interpretation:

The preparation of given set of substances from $D_{2}O$ has to be given.

To give: The preparation of $NaOD$ from $D_{2}O$

Answer to Problem 22.96QP

$2N{a}_{(s)}+2D_2{O}_{(l)}\stackrel{}{\to }2NaO{D}_{(aq)}+D_{2(g)}$

Explanation of Solution

Preparation of $NaOD$ from $D_{2}O$

$2N{a}_{(s)}+2D_2{O}_{(l)}\stackrel{}{\to }2NaO{D}_{(aq)}+D_{2(g)}$

Metallic sodium reacts with deuterated water and gives deuterated sodium hydroxide $NaOD$.

(b)

Interpretation Introduction

Interpretation:

The preparation of given set of substances from $D_{2}O$ has to be given.

To give: The preparation of $DCl$ from $D_{2}O$

Answer to Problem 22.96QP

$\begin{array}{l}2D_2{O}_{(l)}\stackrel{electrolysis}{\to }2D_{2\left(g\right)}+O_{2\left(g\right)}\\ D_{2(g)}+C{l}_{2(g)}\stackrel{}{\to }2DC{l}_{(g)}\end{array}$

Explanation of Solution

Preparation of $DCl$ from $D_{2}O$

$\begin{array}{l}2D_2{O}_{(l)}\stackrel{electrolysis}{\to }2D_{2\left(g\right)}+O_{2\left(g\right)}\\ D_{2(g)}+C{l}_{2(g)}\stackrel{}{\to }2DC{l}_{(g)}\end{array}$

The preparation of $DCl$ from $D_{2}O$ is a two step process.

Electrolysis of $D_{2}O$ will gives $D_2$ and $O_2$.  Formed $D_2$ reacts with molecular chlorine and produces $DCl$.

(c)

Interpretation Introduction

Interpretation:

The preparation of given set of substances from $D_{2}O$ has to be given.

To give: The preparation of $N{D}_3$ from $D_{2}O$

Answer to Problem 22.96QP

$M{g}_3{N}_{2(s)}+6D_2{O}_{(l)}\stackrel{}{\to }3Mg{(OD)}_{2(s)}+2N{D}_3{}_{(g)}$

Explanation of Solution

Preparation of $N{D}_3$ from $D_{2}O$

$M{g}_3{N}_{2(s)}+6D_2{O}_{(l)}\stackrel{}{\to }3Mg{(OD)}_{2(s)}+2N{D}_3{}_{(g)}$

Reaction of $M{g}_3{N}_2$ with $D_{2}O$ gives $N{D}_3$ and $Mg{(OD)}_2$.

(d)

Interpretation Introduction

Interpretation:

The preparation of given set of substances from $D_{2}O$ has to be given.

To give: The preparation of $C_2{D}_2$ from $D_{2}O$

Answer to Problem 22.96QP

$Ca{C}_{2(s)}+2D_2{O}_{(l)}\stackrel{}{\to }{C}_2{D}_{2(g)}+Ca{\left(OD\right)}_2{}_{(aq)}$

Explanation of Solution

Preparation of $C_2{D}_2$ from $D_{2}O$

$Ca{C}_{2(s)}+2D_2{O}_{(l)}\stackrel{}{\to }{C}_2{D}_{2(g)}+Ca{\left(OD\right)}_2{}_{(aq)}$

Reaction of $Ca{C}_2$ with $D_{2}O$ gives $C_2{D}_2$ and $Ca{(OD)}_2$.

(e)

Interpretation Introduction

Interpretation:

The preparation of given set of substances from $D_{2}O$ has to be given.

To give: The preparation of $C{D}_4$ from $D_{2}O$

Answer to Problem 22.96QP

$B{e}_2{C}_{(s)}+4D_2{O}_{(l)}\stackrel{}{\to }2Be{(OD)}_{2(aq)}+C{D}_{4_{(aq)}}$

Explanation of Solution

Preparation of $C{D}_4$ from $D_{2}O$

$B{e}_2{C}_{(s)}+4D_2{O}_{(l)}\stackrel{}{\to }2Be{(OD)}_{2(aq)}+C{D}_{4_{(aq)}}$

Reaction of beryllium carbide with $D_{2}O$ gives $C{D}_4$ and $Be{(OD)}_2$.

(f)

Interpretation Introduction

Interpretation:

The preparation of given set of substances from $D_{2}O$ has to be given.

To give: Preparation of $D_{2}S{O}_4$ from $D_{2}O$

Answer to Problem 22.96QP

$S{O}_{3(g)}+D_2{O}_{(l)}\stackrel{}{\to }{D}_{2}S{O}_{4(aq)}$

Explanation of Solution

Preparation of $D_{2}S{O}_4$ from $D_{2}O$

$S{O}_{3(g)}+D_2{O}_{(l)}\stackrel{}{\to }{D}_{2}S{O}_{4(aq)}$

Reaction of sulfur trioxide with deuterated water $D_{2}O$ gives deuterated sulfuric acid $D_{2}S{O}_4$