Chapter 8, Problem 8.39QP

Interpretation Introduction

Interpretation: The amount of ${\text{O}}_{\text{2}}$ gas produced in grams from decomposition of $\text{46}\text{.0}\text{g}{\text{KClO}}_{\text{3}}$ 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 ${\text{O}}_{\text{2}}$ gas produced in grams from $\text{46}\text{.0}\text{g}{\text{KClO}}_{\text{3}}$

Answer to Problem 8.39QP

The amount of ${\text{O}}_{\text{2}}$ gas produced in grams from $\text{46}\text{.0}\text{g}{\text{KClO}}_{\text{3}}$ is $\text{18}\text{.1085}\text{g}$.

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.

Balanced equation of potassium chlorate decomposition is given below:

${\text{2KClO}}_{\text{3}}\to \text{2KCl}\text{+}{\text{3O}}_{\text{2}}$

The above balanced equation shows 3 moles of ${\text{O}}_{\text{2}}$ produced from 2 moles of ${\text{KClO}}_{\text{3}}$

$\frac{\text{3}\text{moles}\text{of}{\text{O}}_{\text{2}}}{\text{2}\text{moles}\text{of}{\text{KClO}}_{\text{3}}}$

Molecular mass of $KCl{O}_3$ calculation is given below:

$\begin{array}{l}\text{Molecular}\text{mass}\text{of}{\text{KClO}}_{\text{3}}\text{=}\left(\text{mass}\text{of}\text{K+}\text{mass}\text{of}\text{Cl+}\left(\text{3×mass}\text{of}\text{O}\right)\right)\text{grams}\\ \text{=}\left(\text{39}\text{.0983+}\text{35}\text{.453}\text{+}\left(\text{3×15}\text{.9994}\right)\right)\text{grams}\\ \text{=}\left(\text{39}\text{.0983+}\text{35}\text{.453}\text{+47}\text{.9982}\right)\text{grams}\\ \text{=}\text{122}\text{.5495}\text{grams}\end{array}$

Molecular mass of $O_2$ calculation is given below:

$\begin{array}{l}\text{Molecular}\text{mass}\text{of}{\text{O}}_{\text{2}}\text{=}\text{2×mass}\text{of}\text{O}\text{in}\text{grams}\\ \text{=}\text{2×15}\text{.9994}\text{grams}\\ \text{=}\text{31}\text{.9988}\text{grams}\end{array}$

Grams to moles conversion of $KCl{O}_3$

$\begin{array}{l}\frac{\text{Weight}\text{of}{\text{KClO}}_{\text{3}}}{\text{Molecular}\text{weight}\text{of}{\text{KClO}}_{\text{3}}}\text{=}\text{moles}\text{of}{\text{KClO}}_{\text{3}}\\ \frac{\text{46}}{\text{122}\text{.5495}}\text{=}\text{0}\text{.3754}\text{moles}\text{of}{\text{KClO}}_{\text{3}}\end{array}$

The amount of $O_2$ production in moles is given below:

$\begin{array}{l}\text{Moles}\text{of}{\text{KClO}}_{\text{3}}\text{×}\frac{\text{3}\text{moles}\text{of}{\text{O}}_{\text{2}}}{\text{2}\text{moles}\text{of}{\text{KClO}}_{\text{3}}}\text{=}\text{moles}\text{of}{\text{O}}_{\text{2}}\\ \text{0}\text{.3754}\text{×1}\text{.5}\text{moles}\text{of}{\text{O}}_{\text{2}}\text{=}\text{0}\text{.5631}\text{moles}\text{of}{\text{O}}_{\text{2}}\end{array}$

The amount of $O_2$ production in grams is given below:

$\begin{array}{l}\text{Moles}\text{of}{\text{O}}_{\text{2}}\text{×}\text{molar mass}\text{of}{\text{O}}_{\text{2}}\text{=}\text{weight}\text{of}{\text{O}}_{\text{2}}\text{in}\text{grams}\\ \text{0}\text{.5631}\text{×31}\text{.9988=18}\text{.1085}\text{grams}\end{array}$

The amount of ${\text{O}}_{\text{2}}$ gas produced in grams from $\text{46}\text{.0}\text{g}{\text{KClO}}_{\text{3}}$ is $\text{18}\text{.1085}\text{g}$.

Conclusion

The amount of ${\text{O}}_{\text{2}}$ gas produced in grams from $\text{46}\text{.0}\text{g}{\text{KClO}}_{\text{3}}$ was determined.

The amount of ${\text{O}}_{\text{2}}$ gas produced in grams from $\text{46}\text{.0}\text{g}{\text{KClO}}_{\text{3}}$ is $\text{18}\text{.1085}\text{g}$.