Chapter 8, Problem 8.37QP

Nitrous oxide (N₂O) is also called “laughing gas.” It can be prepared by the thermal decomposition of ammonium nitrate (NH₄NO₃). The other product is H₂O. (a) Write a balanced equation for this reaction, (b) How many grams of N₂O are formed if 0.46 mole of NH₄NO₃ is used in the reaction?

Interpretation Introduction

Interpretation: Balanced equation of ammonium nitrate decomposition and the amount of ${\text{N}}_{\text{2}}\text{O}$ produced from $\text{0}\text{.46}$ mole of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ has to be determined.

Concept introduction:

Balanced chemical equation: The mass of products should be equal to mass of its starting materials this is governed by law of conservation of mass.

$\begin{array}{l}\text{Mass}\text{of}\text{starting}\text{materials}\left(\text{Reactants}\right)\text{=}\text{Mass}\text{of}\text{products}\\ {\text{N}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}\to {\text{NH}}_{\text{3}}\end{array}$

In the above reaction 2 nitrogen atoms are there in reactant side but 1 nitrogen in product side, likewise 2 hydrogen’s in reactant side but 3 hydrogen’s in product side so this is unbalanced equation by putting coefficients for each elements in both sides we can balance this reaction as follows:

${\text{N}}_{\text{2}}\text{+}{\text{3H}}_{\text{2}}\to {\text{2NH}}_{\text{3}}$

To write: Balanced chemical equation of $N{H}_{4}N{O}_3$ decomposition

Answer to Problem 8.37QP

Balanced equation for ammonium nitrate decomposition is given below

${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}\to {\text{N}}_{\text{2}}\text{O}\text{+}{\text{H}}_{\text{2}}\text{O}$

Explanation of Solution

The mass of products should be equal to mass of its starting materials this is governed by law of conservation of mass this equation is known as balanced chemical equation.

Decomposition of ammonium nitrate:

Starting material: ammonium nitrate ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$

Products: ${\text{N}}_{\text{2}}\text{O}$ and ${\text{H}}_{\text{2}}\text{O}$

We can write this chemical reaction as follows:

${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}\to {\text{N}}_{\text{2}}\text{O}\text{+}{\text{H}}_{\text{2}}\text{O}$

Above one is an unbalanced equation because the number of hydrogen and oxygen in starting material and products are not equal. So this equation balanced by putting coefficients to corresponding elements as follows:

${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}\to {\text{N}}_{\text{2}}\text{O}\text{+}{\text{2H}}_{\text{2}}\text{O}$

Interpretation Introduction

Interpretation: Balanced equation of ammonium nitrate decomposition and the amount of ${\text{N}}_{\text{2}}\text{O}$ produced from $\text{0}\text{.46}$ mole of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ has to be determined.

Concept introduction:

Balanced chemical equation: The mass of products should be equal to mass of its starting materials this is governed by law of conservation of mass.

$\begin{array}{l}\text{Mass}\text{of}\text{starting}\text{materials}\left(\text{Reactants}\right)\text{=}\text{Mass}\text{of}\text{products}\\ {\text{N}}_{\text{2}}\text{+}{\text{H}}_{\text{2}}\to {\text{NH}}_{\text{3}}\end{array}$

In the above reaction 2 nitrogen atoms are there in reactant side but 1 nitrogen in product side, likewise 2 hydrogen’s in reactant side but 3 hydrogen’s in product side so this is unbalanced equation by putting coefficients for each elements in both sides we can balance this reaction as follows:

${\text{N}}_{\text{2}}\text{+}{\text{3H}}_{\text{2}}\to {\text{2NH}}_{\text{3}}$

To calculate: The amount of ${\text{N}}_{\text{2}}\text{O}$ produced from $\text{0}\text{.46}$ mole of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$

Answer to Problem 8.37QP

The amount of ${\text{N}}_{\text{2}}\text{O}$ produced from $\text{0}\text{.46}$ mole of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ is $\text{20}\text{.2457 g}$

Explanation of Solution

${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}\to {\text{N}}_{\text{2}}\text{O}\text{+}{\text{2H}}_{\text{2}}\text{O}$

The above balanced equation shows 1 mole of ${\text{N}}_{\text{2}}\text{O}$ produced from 1 mole ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$

$\frac{\text{1}\text{mole}\text{of}{\text{N}}_{\text{2}}\text{O}}{\text{1}\text{mole}\text{of}{\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}}$

Molar mass of $N{H}_{4}N{O}_3$ calculation is given below:

$\begin{array}{l}\text{Molar}\text{mass}\text{of}{\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}\text{=}\left(\left(\text{2×mass}\text{of}\text{N}\right)\text{+}\left(\text{4×mass}\text{of}\text{H}\right)\text{+}\left(\text{3×mass}\text{of}\text{O}\right)\right)\text{grams}\\ \text{=}\left(\left(\text{2×14}\text{.0067}\right)\text{+}\left(\text{4×1}\text{.00794}\right)\text{+}\left(\text{3×15}\text{.9994}\right)\right)\text{grams}\\ \text{=}\left(\text{28}\text{.0134}\text{+}\text{4}\text{.03176}\text{+}\text{47}\text{.9982}\right)\text{grams}\\ \text{=}\text{80}\text{.04336}\text{grams}\end{array}$

Molar mass of $N_{2}O$ calculation is given below:

$\begin{array}{l}\text{Molar}\text{mass}\text{of}{\text{N}}_{\text{2}}\text{O}\text{=}\left(\left(\text{2×mass}\text{of}\text{N}\right)\text{+}\text{mass}\text{of}\text{O}\right)\text{grams}\\ \text{=}\left(\left(\text{2×14}\text{.0067}\right)\text{+}\text{15}\text{.9994}\right)\text{grams}\\ \text{=}\left(\text{28}\text{.0134}\text{+}\text{15}\text{.9994}\right)\text{grams}\\ \text{=}\text{44}\text{.0128}\text{grams}\end{array}$

Moles to gram conversion of $N{H}_{4}N{O}_3$

$\begin{array}{l}\text{moles}\text{of}{\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}\text{×}\text{molecular}\text{mass}\text{of}\text{NH}\text{N}{\text{O}}_{\text{3}}\text{=}\text{weight}\text{of}{\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}\text{in}\text{grams}\\ \text{0}\text{.46}\text{×}\text{80}\text{.04336}\text{=}\text{36}\text{.8199}\text{grams}\end{array}$

Production of $N_{2}O$calculation is given below:

$\text{Weight}\text{of}{\text{N}}_{\text{2}}\text{O}\text{produced}\text{in}\text{grams}\text{=}\text{Mass}\text{of}{\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}\text{ in grams×}\frac{\text{Molar}\text{mass}\text{of}{\text{N}}_{\text{2}}\text{O}\text{in}\text{grams}}{\text{Molar}\text{mass}\text{of}{\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}\text{in}\text{grams}}$ $\text{36}\text{.8199g}\text{×}\frac{\text{44}\text{.0128g}}{\text{80}\text{.04336g}}\text{=}\text{20}\text{.2457g}$

The amount of ${\text{N}}_{\text{2}}\text{O}$ produced from $\text{0}\text{.46}$ mole of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ is $\text{20}\text{.2457 g}$

Conclusion

Balanced equation of ammonium nitrate decomposition and the amount of ${\text{N}}_{\text{2}}\text{O}$ produced from $\text{0}\text{.46}$ moles of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ were determined.

(a) Balanced equation for ammonium nitrate decomposition is given below

${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}\to {\text{N}}_{\text{2}}\text{O}\text{+}{\text{H}}_{\text{2}}\text{O}$

(b) The amount of ${\text{N}}_{\text{2}}\text{O}$ produced from $\text{0}\text{.46}$ mole of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ is $\text{20}\text{.2457 g}$