Chapter 8, Problem 41A

(a)

Interpretation Introduction

Interpretation:

The given decomposition reaction needs to be balanced.

  ${\text{NI}}_{\text{3}}\left(\text{s}\right)\to {\text{N}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{I}}_{\text{2}}\left(\text{s}\right)$

Concept Introduction:

The reaction in which a single break down into two or more elements is known as decomposition reaction.

Explanation of Solution

The unbalanced reaction is shown below:

  ${\text{NI}}_{\text{3}}\left(\text{s}\right)\to {\text{N}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{I}}_{\text{2}}\left(\text{s}\right)$

In reactant side, there is one nitrogen atom and in product side there are 2 nitrogen atoms. So, multiply with 2 in reactant side and to balance the Iodine atom, multiply 3 with product side.

The decomposition reaction is shown below:

  ${\text{2NI}}_{\text{3}}\left(\text{s}\right)\to {\text{N}}_{\text{2}}\left(\text{g}\right)\text{+}{\text{3I}}_{\text{2}}\left(\text{s}\right)$

Nitrogen triiodide is an inorganic compound which decomposed into nitrogen gas and iodine solid.

(b)

Interpretation Introduction

Interpretation:

The given decomposition reaction needs to be balanced.

  ${\text{BaCO}}_{\text{3}}\left(\text{s}\right)\to \text{BaO}\left(\text{s}\right)\text{+}{\text{CO}}_{\text{2}}\left(\text{g}\right)$

Concept Introduction:

The reaction in which a single break down into two or more elements is known as decomposition reaction.

Explanation of Solution

The unbalanced reaction is shown below:

  ${\text{BaCO}}_{\text{3}}\left(\text{s}\right)\to \text{BaO}\left(\text{s}\right)\text{+}{\text{CO}}_{\text{2}}\left(\text{g}\right)$

The decomposition reaction is shown below:

  ${\text{BaCO}}_{\text{3}}\left(\text{s}\right)\to \text{BaO}\left(\text{s}\right)\text{+}{\text{CO}}_{\text{2}}\left(\text{g}\right)$

Metal carbonate decomposed into solid of metal oxide and produce carbon dioxide gas.

(c)

Interpretation Introduction

Interpretation:

The given decomposition reaction needs to be balanced.

  ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(\text{s}\right)\to \text{C}\left(\text{s}\right)\text{+}{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$

Concept Introduction:

The reaction in which a single break down into two or more elements is known as decomposition reaction.

Explanation of Solution

The unbalanced reaction is shown below:

  ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(\text{s}\right)\to \text{C}\left(\text{s}\right)\text{+}{\text{H}}_{\text{2}}\text{O}\left(\text{g}\right)$

In reactant side, there are 6 carbon atom and 12 hydrogen atom and in product side there is one carbon atom and 1 H₂O molecule. Therefore, multiply with 6 for C atom and H₂O molecule to balance the reaction.

The decomposition reaction is shown below:

  ${\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\left(\text{s}\right)\to \text{6C}\left(\text{s}\right)\text{+}{\text{6H}}_{\text{2}}\text{O}\left(\text{g}\right)$

When glucose decomposed it will form carbon solid and water molecules.

(d)

Interpretation Introduction

Interpretation:

The given decomposition reaction needs to be balanced.

  $\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}{\text{SO}}_{\text{4}}\left(\text{s}\right)\to {\text{CuSO}}_{\text{4}}\left(\text{s}\right)\text{+}{\text{NH}}_{\text{3}}\left(\text{g}\right)$

Concept Introduction:

The reaction in which a single break down into two or more elements is known as decomposition reaction.

Explanation of Solution

The unbalanced reaction is shown below:

  $\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}{\text{SO}}_{\text{4}}\left(\text{s}\right)\to {\text{CuSO}}_{\text{4}}\left(\text{s}\right)\text{+}{\text{NH}}_{\text{3}}\left(\text{g}\right)$

In the reactant side, there are 4 NH₃ molecule and in product side there is one NH₃ molecule. Therefore; multiply with 4 to balance the reaction.

The decomposition reaction is shown below:

  $\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}{\text{SO}}_{\text{4}}\left(\text{s}\right)\to {\text{CuSO}}_{\text{4}}\left(\text{s}\right)\text{+}4{\text{NH}}_{\text{3}}\left(\text{g}\right)$

Tetrammine copper(II) sulphate decompose into copper sulphate and produce ammonia gas.

(e)

Interpretation Introduction

Interpretation:

The given decomposition reaction needs to be balanced.

Concept Introduction:

The reaction in which a single break down into two or more elements is known as decomposition reaction.

Explanation of Solution

The unbalanced reaction is shown below:

  ${\text{NaN}}_{\text{3}}\left(\text{s}\right)\to {\text{Na}}_{\text{3}}\text{N}\left(\text{s}\right)\text{+}{\text{N}}_{\text{2}}\left(\text{g}\right)$

In the reactant side; there are 1 sodium atom and in the product side there are 3 sodium atoms. So, multiply with 3 in the reactant side to balance the sodium atom and to balance the nitrogen atom multiply with 5/2 in the product side.

The decomposition reaction is shown below:

  ${\text{3NaN}}_{\text{3}}\left(\text{s}\right)\to {\text{Na}}_{\text{3}}\text{N}\left(\text{s}\right)\text{+}\frac{5}{2}{\text{N}}_{\text{2}}\left(\text{g}\right)$

When sodium azide decompose it will form solid of sodium nitride and nitrogen gas.