Chapter 3, Problem 3.58P

(a)

Interpretation Introduction

Interpretation:

A balanced equation for the following reaction is to be written.

$\underset{\_}{}\text{Cu}\left(s\right)+\underset{\_}{}{\text{S}}_8\left(s\right)\to \underset{\_}{}{\text{Cu}}_2\text{S}\left(s\right)$

Concept introduction:

In a balanced chemical equation, the total mass of reactants and products are equal in a balanced chemical equation, thus, it obeyed the law of conservation of mass.

Following are the steps to write a balanced chemical equation.

Step 1: Identify the most complex substance and choose an element such that the element must be present only in one reactant and one product. Place the stoichiometric coefficient before the element(s) such that the number of atoms of that element(s) is the same on both sides.

Step 2: Balance the remaining atoms by placing the stoichiometric coefficients before the element(s) such that the number of atoms of that element(s) is the same on both sides. Identify the least complex substance and end with it. Generally, oxygen atoms are balanced in last.

Step 3: In a balanced chemical reaction, the smallest whole number coefficients are most preferred. Hence, adjusting the coefficients in such a way that the smallest whole number coefficients are obtained for each element. Generally, oxygen atoms are balanced in last.

Step 4: Check whether the chemical equation is balanced or not by counting the number of atoms of each element on both sides.

Answer to Problem 3.58P

The balanced chemical reaction is as follows:

$16\text{Cu}\left(s\right)+{\text{S}}_8\left(s\right)\to 8{\text{Cu}}_2\text{S}\left(s\right)$

Explanation of Solution

The equation implies that $8\text{ S}$ atoms are present on the reactant side while only $1\text{ S}$ atom is present on the product side. Balance sulfur $\left(\text{S}\right)$ atoms first. To balance $\text{S}$, place the coefficient8 in front of ${\text{Cu}}_2\text{S}$. The equation becomes,

$\underset{\_}{}\text{Cu}\left(s\right)+\underset{\_}{}{\text{S}}_8\left(s\right)\to \underset{\_}{8}{\text{Cu}}_2\text{S}\left(s\right)$

Sulfur atoms are balanced, thus, place a 1 in front of ${\text{S}}_8$. The equation becomes,

$\underset{\_}{}\text{Cu}\left(s\right)+\underset{\_}{1}{\text{S}}_8\left(s\right)\to \underset{\_}{8}{\text{Cu}}_2\text{S}\left(s\right)$

Balance copper $\left(\text{Cu}\right)$ atoms $\text{1 Cu}$ atom is present on the reactant side while $\text{16 Cu}$ atoms are present on the product side. To balance copper $\left(\text{Cu}\right)$ atoms, place the coefficient 16 in front of $\text{Cu}$. The equation becomes,

$\underset{\_}{16}\text{Cu}\left(s\right)+\underset{\_}{1}{\text{S}}_8\left(s\right)\to \underset{\_}{8}{\text{Cu}}_2\text{S}\left(s\right)$

Check whether the equation is balanced or not as follows:

$\text{Reactants}\left(16\text{ Cu, 8 S}\right)\to \text{Products}\left(16\text{ Cu, 8 S}\right)$

Atoms of each element are the same on both sides, thus, the chemical reaction is balanced.

Conclusion

The balanced chemical reaction is as follows:

$16\text{Cu}\left(s\right)+{\text{S}}_8\left(s\right)\to 8{\text{Cu}}_2\text{S}\left(s\right)$

(b)

Interpretation Introduction

Interpretation:

A balanced equation for the following reaction is to be written.

$\underset{\_}{}{\text{P}}_4{\text{O}}_{10}\left(s\right)+\underset{\_}{}{\text{H}}_2\text{O}\left(l\right)\to \underset{\_}{}{\text{H}}_3{\text{PO}}_4\left(l\right)$

Concept introduction:

In a balanced chemical equation, the total mass of reactants and products are equal in a balanced chemical equation, thus, it obeyed the law of conservation of mass.

Following are the steps to write a balanced chemical equation.

Step 1: Identify the most complex substance and choose an element such that the element must be present only in one reactant and one product. Place the stoichiometric coefficient before the element(s) such that the number of atoms of that element(s) is the same on both sides.

Step 2: Balance the remaining atoms by placing the stoichiometric coefficients before the element(s) such that the number of atoms of that element(s) is the same on both sides. Identify the least complex substance and end with it. Generally, oxygen atoms are balanced in last.

Step 3: In a balanced chemical reaction, the smallest whole number coefficients are most preferred. Hence, adjusting the coefficients in such a way that the smallest whole number coefficients are obtained for each element.

Step 4: Check whether the chemical equation is balanced or not by counting the number of atoms of each element on both sides.

Answer to Problem 3.58P

The balanced chemical reaction is as follows:

${\text{P}}_4{\text{O}}_{10}\left(s\right)+6{\text{H}}_2\text{O}\left(l\right)\to 4{\text{H}}_3{\text{PO}}_4\left(l\right)$

Explanation of Solution

The equation implies that $4\text{ P}$ atoms are present on the reactant side while only $1\text{ P}$ atom is present on the product side. Balance phosphorous $\left(\text{P}\right)$ atoms first. To balance $\text{P}$, place the coefficient4 in front of ${\text{H}}_3{\text{PO}}_4$. The equation becomes,

$\underset{\_}{}{\text{P}}_4{\text{O}}_{10}\left(s\right)+\underset{\_}{}{\text{H}}_2\text{O}\left(l\right)\to \underset{\_}{4}{\text{H}}_3{\text{PO}}_4\left(l\right)$

Phosphorous atoms are balanced, thus, place the coefficient 1 in front of ${\text{P}}_4{\text{O}}_{10}$. The equation becomes,

$\underset{\_}{1}{\text{P}}_4{\text{O}}_{10}\left(s\right)+\underset{\_}{}{\text{H}}_2\text{O}\left(l\right)\to \underset{\_}{4}{\text{H}}_3{\text{PO}}_4\left(l\right)$

Next, balance hydrogen $\left(\text{H}\right)$ atoms. $\text{2 H}$ atoms are present on the reactant side while $\text{12 H}$ atoms are present on the product side. To balance $\text{H}$ atoms, place the coefficient 6 in front of ${\text{H}}_2\text{O}$. The equation becomes,

$\underset{\_}{1}{\text{P}}_4{\text{O}}_{10}\left(s\right)+\underset{\_}{6}{\text{H}}_2\text{O}\left(l\right)\to \underset{\_}{4}{\text{H}}_3{\text{PO}}_4\left(l\right)$

Check whether the equation is balanced or not as follows:

$\text{Reactants}\left(4\text{ P, 12 H, 16 O}\right)\to \text{Products}\left(4\text{ P, 12 H, 16 O}\right)$

Atoms of each element are the same on both sides, thus, the chemical reaction is balanced.

Conclusion

The balanced chemical reaction is as follows:

${\text{P}}_4{\text{O}}_{10}\left(s\right)+6{\text{H}}_2\text{O}\left(l\right)\to 4{\text{H}}_3{\text{PO}}_4\left(l\right)$

(c)

Interpretation Introduction

Interpretation:

A balanced equation for the following reaction is to be written.

$\underset{\_}{}{\text{B}}_2{\text{O}}_3\left(s\right)+\underset{\_}{}\text{NaOH}\left(aq\right)\to \underset{\_}{}{\text{Na}}_3{\text{BO}}_3\left(aq\right)+\underset{\_}{}{\text{H}}_2\text{O}\left(l\right)$

Concept introduction:

In a balanced chemical equation, the total mass of reactants and products are equal in a balanced chemical equation, thus, it obeyed the law of conservation of mass.

Following are the steps to write a balanced chemical equation.

Step 1: Identify the most complex substance and choose an element such that the element must be present only in one reactant and one product. Place the stoichiometric coefficient before the element(s) such that the number of atoms of that element(s) is the same on both sides.

Step 2: Balance the remaining atoms by placing the stoichiometric coefficients before the element(s) such that the number of atoms of that element(s) is the same on both sides. Identify the least complex substance and end with it. Generally, oxygen atoms are balanced in last.

Step 3: In a balanced chemical reaction, the smallest whole number coefficients are most preferred. Hence, adjusting the coefficients in such a way that the smallest whole number coefficients are obtained for each element.

Step 4: Check whether the chemical equation is balanced or not by counting the number of atoms of each element on both sides.

Answer to Problem 3.58P

The balanced chemical reaction is as follows:

${\text{B}}_2{\text{O}}_3\left(s\right)+6\text{NaOH}\left(aq\right)\to 2{\text{Na}}_3{\text{BO}}_3\left(aq\right)+3{\text{H}}_2\text{O}\left(l\right)$

Explanation of Solution

The equation implies that $4\text{ B}$ atoms are present on the reactant side while only $1\text{ B}$ atom is present on the product side. Balance boron $\left(\text{B}\right)$ atoms first and oxygen $\left(\text{O}\right)$ atoms in last. To balance $\text{B}$, place the coefficient 2 in front of ${\text{Na}}_3{\text{BO}}_3$. The equation becomes,

$\underset{\_}{}{\text{B}}_2{\text{O}}_3\left(s\right)+\underset{\_}{}\text{NaOH}\left(aq\right)\to \underset{\_}{2}{\text{Na}}_3{\text{BO}}_3\left(aq\right)+\underset{\_}{}{\text{H}}_2\text{O}\left(l\right)$

Boron atoms are balanced, next, balance sodium $\left(\text{Na}\right)$ atoms. $\text{1 Na}$ atom is present on the reactant side while $\text{6 Na}$ atoms are present on the product side. To balance $\text{Na}$ atoms, place the coefficient 6 in front of $\text{NaOH}$. The equation becomes,

$\underset{\_}{}{\text{B}}_2{\text{O}}_3\left(s\right)+\underset{\_}{6}\text{NaOH}\left(aq\right)\to \underset{\_}{2}{\text{Na}}_3{\text{BO}}_3\left(aq\right)+\underset{\_}{}{\text{H}}_2\text{O}\left(l\right)$

Sodium atoms are balanced, next, balance hydrogen $\left(\text{H}\right)$ atoms. $\text{6 H}$ atoms are present on the reactant side while $\text{2 H}$ atoms are present on the product side. To balance $\text{H}$ atoms, place the coefficient 3 in front of ${\text{H}}_2\text{O}$. The equation becomes,

$\underset{\_}{}{\text{B}}_2{\text{O}}_3\left(s\right)+\underset{\_}{6}\text{NaOH}\left(aq\right)\to \underset{\_}{2}{\text{Na}}_3{\text{BO}}_3\left(aq\right)+\underset{\_}{3}{\text{H}}_2\text{O}\left(l\right)$

Hydrogen atoms are balanced, next, balance oxygen $\left(\text{O}\right)$ atoms. $\text{9 O}$ atoms are present on both sides. Hence, oxygen atoms are balanced. Place the coefficient 1 in front of ${\text{B}}_2{\text{O}}_3$. The equation becomes,

$\underset{\_}{1}{\text{B}}_2{\text{O}}_3\left(s\right)+\underset{\_}{6}\text{NaOH}\left(aq\right)\to \underset{\_}{2}{\text{Na}}_3{\text{BO}}_3\left(aq\right)+\underset{\_}{3}{\text{H}}_2\text{O}\left(l\right)$

Check whether the equation is balanced or not as follows:

$\text{Reactants}\left(2\text{ B, }6\text{ Na, 6 H, 9 O}\right)\to \text{Products}\left(2\text{ B, }6\text{ Na, 6 H, 9 O}\right)$

Atoms of each element are the same on both sides, thus, the chemical reaction is balanced.

Conclusion

The balanced chemical reaction is as follows:

${\text{B}}_2{\text{O}}_3\left(s\right)+6\text{NaOH}\left(aq\right)\to 2{\text{Na}}_3{\text{BO}}_3\left(aq\right)+3{\text{H}}_2\text{O}\left(l\right)$

(d)

Interpretation Introduction

Interpretation:

A balanced equation for the following reaction is to be written.

$\underset{\_}{}{\text{CH}}_3{\text{NH}}_2\left(g\right)+\underset{\_}{}{\text{O}}_2\left(g\right)\to \underset{\_}{}{\text{CO}}_2\left(g\right)+\underset{\_}{}{\text{H}}_2\text{O}\left(g\right)+\underset{\_}{}{\text{N}}_2\left(g\right)$

Concept introduction:

In a balanced chemical equation, the total mass of reactants and products are equal in a balanced chemical equation, thus, it obeyed the law of conservation of mass.

Following are the steps to write a balanced chemical equation.

Step 1: Identify the most complex substance and choose an element such that the element must be present only in one reactant and one product. Place the stoichiometric coefficient before the element(s) such that the number of atoms of that element(s) is the same on both sides.

Step 2: Balance the remaining atoms by placing the stoichiometric coefficients before the element(s) such that the number of atoms of that element(s) is the same on both sides. Identify the least complex substance and end with it. Generally, oxygen atoms are balanced in last.

Step 3: In a balanced chemical reaction, the smallest whole number coefficients are most preferred. Hence, adjusting the coefficients in such a way that the smallest whole number coefficients are obtained for each element.

Step 4: Check whether the chemical equation is balanced or not by counting the number of atoms of each element on both sides.

Answer to Problem 3.58P

The balanced chemical reaction is as follows:

$4{\text{CH}}_3{\text{NH}}_2\left(g\right)+9{\text{O}}_2\left(g\right)\to 4{\text{CO}}_2\left(g\right)+10{\text{H}}_2\text{O}\left(g\right)+2{\text{N}}_2\left(g\right)$

Explanation of Solution

The equation implies that $1\text{ N}$ atom is present on the reactant side while $2\text{ N}$ atoms are present on the product side. Balance nitrogen $\left(\text{N}\right)$ atoms first and oxygen $\left(\text{O}\right)$ atoms in last. To balance $\text{N}$, place the coefficient 2 in front of ${\text{CH}}_3{\text{NH}}_2$. The equation becomes,

$\underset{\_}{2}{\text{CH}}_3{\text{NH}}_2\left(g\right)+\underset{\_}{}{\text{O}}_2\left(g\right)\to \underset{\_}{}{\text{CO}}_2\left(g\right)+\underset{\_}{}{\text{H}}_2\text{O}\left(g\right)+\underset{\_}{}{\text{N}}_2\left(g\right)$

Nitrogen atoms are balanced, hence, place the coefficient 1 in front of ${\text{N}}_2$. Next, balance hydrogen $\left(\text{H}\right)$ atoms. $\text{10 H}$ atoms are present on the reactant side while $\text{2 H}$ atoms are present on the product side. To balance $\text{H}$ atoms, place the coefficient 5 in front of ${\text{H}}_2\text{O}$. The equation becomes,

$\underset{\_}{2}{\text{CH}}_3{\text{NH}}_2\left(g\right)+\underset{\_}{}{\text{O}}_2\left(g\right)\to \underset{\_}{}{\text{CO}}_2\left(g\right)+\underset{\_}{5}{\text{H}}_2\text{O}\left(g\right)+\underset{\_}{1}{\text{N}}_2\left(g\right)$

Hydrogen atoms are balanced. Next, balance carbon $\left(\text{C}\right)$ atoms. $\text{2 C}$ atoms are present on the reactant side while $\text{1 C}$ atom is present on the product side. To balance $\text{C}$ atoms, place the coefficient 2 in front of ${\text{CO}}_2$. The equation becomes,

$\underset{\_}{2}{\text{CH}}_3{\text{NH}}_2\left(g\right)+\underset{\_}{}{\text{O}}_2\left(g\right)\to \underset{\_}{2}{\text{CO}}_2\left(g\right)+\underset{\_}{5}{\text{H}}_2\text{O}\left(g\right)+\underset{\_}{1}{\text{N}}_2\left(g\right)$

Carbon atoms are balanced, next, balance oxygen $\left(\text{O}\right)$ atoms. $\text{9 O}$ atoms are present on the reactant side while $\text{7 O}$ atoms are present on the product side. Hence, place the coefficient $\frac{9}{2}$ in front of ${\text{O}}_2$. The equation becomes,

$\underset{\_}{2}{\text{CH}}_3{\text{NH}}_2\left(g\right)+\underset{\_}{\frac{9}{2}}{\text{O}}_2\left(g\right)\to \underset{\_}{2}{\text{CO}}_2\left(g\right)+\underset{\_}{5}{\text{H}}_2\text{O}\left(g\right)+\underset{\_}{1}{\text{N}}_2\left(g\right)$

Multiply the stoichiometric coefficients by 2 to remove the fraction as follows:

$\underset{\_}{4}{\text{CH}}_3{\text{NH}}_2\left(g\right)+\underset{\_}{9}{\text{O}}_2\left(g\right)\to \underset{\_}{4}{\text{CO}}_2\left(g\right)+\underset{\_}{10}{\text{H}}_2\text{O}\left(g\right)+\underset{\_}{2}{\text{N}}_2\left(g\right)$

Check whether the equation is balanced or not as follows:

$\text{Reactants}\left(4\text{ C, }4\text{ N, 20 H, 18 O}\right)\to \text{Products}\left(4\text{ C, }4\text{ N, 20 H, 18 O}\right)$

Atoms of each element are the same on both sides, thus, the chemical reaction is balanced.

Conclusion

The balanced chemical reaction is as follows:

$4{\text{CH}}_3{\text{NH}}_2\left(g\right)+9{\text{O}}_2\left(g\right)\to 4{\text{CO}}_2\left(g\right)+10{\text{H}}_2\text{O}\left(g\right)+2{\text{N}}_2\left(g\right)$