Chapter 12, Problem 41A

Interpretation Introduction

Interpretation : The mass of the reactants with the mass of the products for the equation is to be calculated and compared. The balanced equation obeys the law of conservation of mass is to be shown.

Concept Introduction : Reactants, products, and an arrow indicating the reaction's direction make up a chemical equation. For a low-energy thermodynamic process, the reactant and product masses must match up in accordance with the law of conservation of mass.

Answer to Problem 41A

The balanced equation $2{\text{KClO}}_{\text{3}}\left(s\right) \to 2\text{KCl}\left(s\right)+3{\text{O}}_{\text{2}}\left(g\right)$ obeys the law of conservation of mass.

The balanced equation $4{\text{NH}}_{\text{3}}\left(g\right)+6\text{NO}\left(g\right) \to 5{\text{N}}_{\text{2}}\left(g\right)+6{\text{H}}_{\text{2}}\text{O}\left(g\right)$ obeys the law of conservation of mass.

The balanced equation $4\text{K}\left(s\right)+{\text{O}}_{\text{2}}\left(g\right) \to 2{\text{K}}_{\text{2}}\text{O}\left(s\right)$ obeys the law of conservation of mass.

Explanation of Solution

One mole of a material has a mass of molar mass.

The standard atomic masses of the constituent atoms are added up to determine the molar mass of the molecule (in g/mol).

Consider the given chemical equation as follows:

  $2{\text{KClO}}_{\text{3}}\left(s\right) \to 2\text{KCl}\left(s\right)+3{\text{O}}_{\text{2}}\left(g\right)$

The molar mass of the reactants and products is to be calculated.

The reactant is $2{\text{KClO}}_{\text{3}}$ and the products are $2\text{KCl}$ and $3{\text{O}}_{\text{2}}$ .

The molar mass of $\text{K}$ is $\text{39}\text{.10}\text{g/mol}$ .

The molar mass of $\text{Cl}$ is $\text{35}\text{.45}\text{g/mol}$ .

The molar mass of $\text{O}$ is $\text{16}\text{.00 g/mol}$ .

Mass of ${\text{KClO}}_{\text{3}}$ is given as follows:

  $\begin{array}{c}\text{Mass of}{\text{KClO}}_{\text{3}}=\text{M}\left(\text{K}\right)\text{+M}\left(\text{Cl}\right)\text{+3M}\left(\text{O}\right)\\ =\left(39.10+35.45+\left(3\times 16.00\right)\right)\text{g/mol}\\ \text{=122}\text{.55}\text{g/mol}\end{array}$

Mass of $\text{KCl}$ is given as follows:

  $\begin{array}{c}\text{Mass of}\text{KCl}=\text{M}\left(\text{K}\right)\text{+M}\left(\text{Cl}\right)\\ =\left(39.10+35.45\right)\text{g/mol}\\ \text{= 74}\text{.55}\text{g/mol}\end{array}$

Mass of ${\text{O}}_2$ is given as follows:

  $\begin{array}{c}\text{Mass of}{\text{O}}_2=2\text{M}\left(\text{O}\right)\\ =\left(2\times 16.00\right)\text{g/mol}\\ \text{=32}\text{.00}\text{g/mol}\end{array}$

The mass of the reactant is given as follows:

  $\begin{array}{c}\text{Mass of}2{\text{KClO}}_{\text{3}}=2\times \text{M}\left({\text{KClO}}_3\right)\\ =\left(\text{2}\times \text{122}\text{.55}\right)\text{g/mol}\\ \text{= 245}\text{.1}\text{g/mol}\end{array}$

The mass of the products is given as follows:

  $\begin{array}{c}\text{Mass of}2\text{KCl}+3{\text{O}}_{\text{2}}=2\text{M}\left(\text{KCl}\right)+3\text{M}\left({\text{O}}_2\right)\\ =\left(\left(\text{2}\times \text{74}\text{.55}\right)+\left(3\times 32.00\right)\right)\text{g/mol}\\ \text{= 245}\text{.1}\text{g/mol}\end{array}$

The mass of the reactants is equal to the mass of the products; thus, the equation obeys the law of conservation of mass.

Consider the given chemical equation as follows:

  $4{\text{NH}}_{\text{3}}\left(g\right)+6\text{NO}\left(g\right) \to 5{\text{N}}_{\text{2}}\left(g\right)+6{\text{H}}_{\text{2}}\text{O}\left(g\right)$

The molar mass of the reactants and products is to be calculated.

The reactants are $4{\text{NH}}_{\text{3}}$ and $6\text{NO}$ and the products are $2\text{KCl}$ and $3{\text{O}}_{\text{2}}$ .

The molar mass of $\text{O}$ is $\text{16}\text{.00 g/mol}$ .

The molar mass of $\text{N}$ is $\text{14}\text{.01 g/mol}$ .

The molar mass of $\text{H}$ is $\text{1}\text{.008 g/mol}$ .

Mass of ${\text{NH}}_{\text{3}}$ is given as follows:

  $\begin{array}{c}\text{Mass of}{\text{NH}}_{\text{3}}=\text{M}\left(\text{N}\right)\text{+3M}\left(\text{H}\right)\\ =\left(14.01+\left(3\times 1.008\right)\right)\text{g/mol}\\ \text{=17}\text{.034}\text{g/mol}\end{array}$

Mass of $\text{NO}$ is given as follows:

  $\begin{array}{c}\text{Mass of}\text{NO}=\text{M}\left(\text{N}\right)\text{+M}\left(\text{O}\right)\\ =\left(14.01+16.00\right)\text{g/mol}\\ \text{= 30}\text{.01}\text{g/mol}\end{array}$

Mass of ${\text{N}}_2$ is given as follows:

  $\begin{array}{c}\text{Mass of}{\text{N}}_2=2\text{M}\left(\text{N}\right)\\ =\left(2\times 14.01\right)\text{g/mol}\\ \text{=28}\text{.02}\text{g/mol}\end{array}$

Mass of ${\text{H}}_2\text{O}$ is given as follows:

  $\begin{array}{c}\text{Mass of}{\text{H}}_2\text{O}=2\text{M}\left(\text{H}\right)+\text{M}\left(\text{O}\right)\\ =\left(\left(2\times 1.008\right)\text{+16}\text{.00}\right)\text{g/mol}\\ \text{=18}\text{.016}\text{g/mol}\end{array}$

The mass of the reactant is given as follows:

  $\begin{array}{c}\text{Mass of}4{\text{NH}}_{\text{3}}+6\text{NO}=4\text{M}\left({\text{NH}}_3\right)+6\text{M}\left(\text{NO}\right)\\ =\left(\left(4\times \text{17}\text{.034}\right)+\left(6\times 30.01\right)\right)\text{g/mol}\\ \text{= 248}\text{.196}\text{g/mol}\end{array}$

The mass of the products is given as follows:

  $\begin{array}{c}\text{Mass of}5{\text{N}}_2+6{\text{H}}_2\text{O}=5\text{M}\left({\text{N}}_2\right)+6\text{M}\left({\text{H}}_2\text{O}\right)\\ =\left(\left(5\times 28.02\right)+\left(6\times 18.016\right)\right)\text{g/mol}\\ \text{= 248}\text{.196}\text{g/mol}\end{array}$

The mass of the reactants is equal to the mass of the products thus the equation obeys the law of conservation of mass.

Consider the given chemical equation as follows:

  $4\text{K}\left(s\right)+{\text{O}}_{\text{2}}\left(g\right) \to 2{\text{K}}_{\text{2}}\text{O}\left(s\right)$

The molar mass of the reactants and products is to be calculated.

The reactants are $4\text{K}$ and ${\text{O}}_{\text{2}}$ and the product is $2{\text{K}}_{\text{2}}\text{O}$ .

The molar mass of $\text{K}$ is $\text{39}\text{.10}\text{g/mol}$ .

The molar mass of $\text{O}$ is $\text{16}\text{.00 g/mol}$ .

Mass of ${\text{O}}_{\text{2}}$ is given as follows:

  $\begin{array}{c}\text{Mass of}{\text{O}}_2=\text{2M}\left(\text{O}\right)\\ =\left(2\times 16.00\right)\text{g/mol}\\ \text{=32}\text{.00}\text{g/mol}\end{array}$

Mass of ${\text{K}}_2\text{O}$ is given as follows:

  $\begin{array}{c}\text{Mass of}{\text{K}}_2\text{O}=2\text{M}\left(\text{K}\right)\text{+M}\left(\text{O}\right)\\ =\left(\left(2\times 39.10\right)+16.00\right)\text{g/mol}\\ \text{= 94}\text{.20}\text{g/mol}\end{array}$

The mass of the reactant is given as follows:

  $\begin{array}{c}\text{Mass of}4\text{K}+{\text{O}}_2=4\text{M}\left(\text{K}\right)+\text{M}\left({\text{O}}_2\right)\\ =\left(\left(4\times 39.10\right)+32.00\right)\text{g/mol}\\ \text{= 188}\text{.4}\text{g/mol}\end{array}$

The mass of the products is given as follows:

  $\begin{array}{c}\text{Mass of}2{\text{K}}_2\text{O}=2\text{M}\left({\text{K}}_2\text{O}\right)\\ =\left(2\times 94.20\right)\text{g/mol}\\ \text{= 188}\text{.4}\text{g/mol}\end{array}$

The mass of the reactants is equal to the mass of the products thus the equation obeys the law of conservation of mass.