Chapter 3, Problem 3.37P

Write balanced equations for each of the following by inserting the correct coefficients in the blanks:

1. $\text{\_\_Cu}\left(s\right)+\_\_{\text{S}}_8\left(s\right)\to \_\_{\text{Cu}}_2\text{S}\left(s\right)$ ${\text{\_\_P}}_4{\text{O}}_{10}\left(s\right)+\_\_{\text{H}}_2\text{O}\left(\text{l}\right)\to \_\_{\text{H}}_3{\text{PO}}_4\left(\text{l}\right)$ ${\text{\_\_B}}_2{\text{O}}_3\left(s\right)+\_\_\text{NaOH}\left(aq\right)\to \_\_{\text{Na}}_3{\text{BO}}_3\left(aq\right)+\_\_{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$ ${\text{\_\_CH}}_3{\text{NH}}_2\left(g\right)+\_\_{\text{O}}_2\left(g\right)\to \_\_{\text{CO}}_{\text{2}}\left(g\right)+\_\_{\text{H}}_{\text{2}}\text{O}\left(g\right)+\_\_{\text{N}}_2\left(g\right)$

Interpretation Introduction

Interpretation:The correct coefficients should be inserted in the indicated equation to form the balanced equation.

  $\underset{\_}{}\text{Cu}\left(s\right)+\underset{\_}{}{\text{S}}_{\text{8}}\to \underset{\_}{}{\text{Cu}}_{\text{2}}\text{S}$

Concept introduction:In accordance with law of conservation of mass it is suggested total amount of reactants consumed is equal to the total amount of products formed in any balanced equation.

In order to balance an equation, one starts with the most complex chemical species and then place the appropriate coefficient so as to make the stoichiometric coefficient equal on each side.

After proper adjustment of coefficient, the atoms are checked for an equal number on each side and finally, the physical states of matter are written in parenthesis.

Answer to Problem 3.37P

Correct coefficients in the blank are written to form the balanced equation as follows:

  $\underset{\_}{\text{ 16 }}\text{Cu}\left(s\right)+{\text{S}}_{\text{8}}\to \underset{\_}{\text{ 8 }}{\text{Cu}}_{\text{2}}\text{S}$

Explanation of Solution

The unbalanced equation that needs to be balanced is as follows:

  $\underset{\_}{}\text{Cu}\left(s\right)+\underset{\_}{}{\text{S}}_{\text{8}}\to \underset{\_}{}{\text{Cu}}_{\text{2}}\text{S}$

Since left side has 8 $\text{S}$ while the right side has 1 $\text{S}$ atom so first the stoichiometric coefficient of $\text{S}$ on the right is changed to 8. This is done when 8 is placed as a coefficient of ${\text{Cu}}_{\text{2}}\text{S}$ .

  $\underset{\_}{}\text{Cu}\left(s\right)+\underset{\_}{}{\text{S}}_{\text{8}}\to \underset{\_}{\text{ 8 }}{\text{Cu}}_{\text{2}}\text{S}$

Place 16 as the coefficient of $\text{Cu}$ so as to make equal $\text{Cu}$ on each side and coefficient of $\text{S}$ will remain 1.

  $\underset{\_}{\text{ 16 }}\text{Cu}\left(s\right)+{\text{S}}_{\text{8}}\to \underset{\_}{\text{ 8 }}{\text{Cu}}_{\text{2}}\text{S}$

Since above equation has equal copper and sulfur atoms on each side so the equation is now balanced.

Interpretation Introduction

Interpretation: The correct coefficient should be inserted in the indicated equation to form the balanced equation.

  $\underset{\_}{}{\text{P}}_{\text{4}}{\text{O}}_{\text{10}}\left(s\right)+\underset{\_}{}{\text{H}}_{\text{2}}\text{O}\left(l\right)\to \underset{\_}{}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\left(l\right)$

Concept introduction:In accordance with law of conservation of mass it is suggested total amount of reactants consumed is equal to the total amount of products formed in any balanced equation.

In order to balance an equation, one starts with the most complex chemical species and then place the appropriate coefficient so as to make the stoichiometric coefficient equal on each side.

After proper adjustment of coefficient, the atoms are checked for an equal number on each side and finally, the physical states of matter are written in parenthesis.

Answer to Problem 3.37P

The correct coefficient in the blank is written to form the balanced equation as follows:

  ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}\left(s\right)+\underset{\_}{\text{ 6 }}{\text{H}}_{\text{2}}\text{O}\left(l\right)\to \underset{\_}{\text{ 4 }}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\left(l\right)$

Explanation of Solution

The unbalanced equation that needs to be balanced is as follows:

  $\underset{\_}{}{\text{P}}_{\text{4}}{\text{O}}_{\text{10}}\left(s\right)+\underset{\_}{}{\text{H}}_{\text{2}}\text{O}\left(l\right)\to \underset{\_}{}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\left(l\right)$

Since the left side has 4 $\text{P}$ while the right side has 1 $\text{P}$ atom so first the stoichiometric coefficient of $\text{P}$ on the right is changed to 4. This is done when 4 is placed as a coefficient of ${\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$ .

  ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}\left(s\right)+\underset{\_}{}{\text{H}}_{\text{2}}\text{O}\left(l\right)\to \underset{\_}{\text{ 4 }}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\left(l\right)$

Place 6 as coefficient of ${\text{H}}_{\text{2}}\text{O}$ so as to make equal 12 $\text{H}$ on each side and coefficient of ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$

  $\text{S}$ will remain 1.

  ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}\left(s\right)+\underset{\_}{\text{ 6 }}{\text{H}}_{\text{2}}\text{O}\left(l\right)\to \underset{\_}{\text{ 4 }}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}\left(l\right)$

Since above equation has an equal copper and sulphur atom on each side so the equation is now balanced.

Interpretation Introduction

Interpretation: The correct coefficient should be inserted in the indicated equation to form the balanced equation.

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

Concept introduction:In accordance with law of conservation of mass it is suggested total amount of reactants consumed is equal to the total amount of products formed in any balanced equation.

In order to balance an equation, one starts with the most complex chemical species and then place the appropriate coefficient so as to make the stoichiometric coefficient equal on each side.

After proper adjustment of coefficient, the atoms are checked for an equal number on each side and finally, the physical states of matter are written in parenthesis.

Answer to Problem 3.37P

The correct coefficient in the blank is written to form the balanced equation as follows:

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

Explanation of Solution

The unbalanced equation that needs to be balanced is as follows:

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

Since the left side has 2 $\text{B}$ while the right side has 1 $\text{B}$ atom so first the stoichiometric coefficient of $\text{B}$ on the right is changed to 2. This is done when 2 is placed as a coefficient of ${\text{Na}}_{\text{3}}{\text{BO}}_3$ .

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

Place 6 as the coefficient of $\text{NaOH}$ so as to make 6 $\text{Na}$ on each side and coefficient of ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$ will remain 1.

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

Place 3 as the coefficient of ${\text{H}}_{\text{2}}\text{O}$ so as to make 6 $\text{H}$ on each side and coefficient of ${\text{B}}_2{\text{O}}_3$ will remain 1.

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

Since above equation has9 oxygen atoms on each side so the equation is now balanced.

Interpretation Introduction

Interpretation: The correct coefficient should be inserted in the indicated equation to form the balanced equation.

  $\underset{\_}{}{\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}\left(s\right)+\underset{\_}{}{\text{O}}_2\left(l\right)\to \underset{\_}{}{\text{CO}}_{\text{2}}\left(l\right)+\underset{\_}{}{\text{H}}_{\text{2}}\text{O}\left(g\right)+\underset{\_}{}{\text{N}}_{\text{2}}\left(g\right)$

Concept introduction:In accordance with law of conservation of mass it is suggested total amount of reactants consumed is equal to the total amount of products formed in any balanced equation.

In order to balance an equation, one starts with the most complex chemical species and then place the appropriate coefficient so as to make the stoichiometric coefficient equal on each side.

After proper adjustment of coefficient, the atoms are checked for an equal number on each side and finally, the physical states of matter are written in parenthesis.

Answer to Problem 3.37P

The correct coefficient in the blank is written to form the balanced equation as follows:

  $\underset{\_}{\text{ 4 }}{\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}\left(g\right)+\underset{\_}{\text{ 9 }}{\text{O}}_2\left(g\right)\to \underset{\_}{\text{ 4 }}{\text{CO}}_{\text{2}}\left(g\right)+\underset{\_}{\text{ 10 }}{\text{H}}_{\text{2}}\text{O}\left(g\right)+\underset{\_}{\text{ 2 }}{\text{N}}_{\text{2}}\left(g\right)$

Explanation of Solution

The unbalanced equation that needs to be balanced is as follows:

  $\underset{\_}{}{\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}\left(g\right)+\underset{\_}{}{\text{O}}_2\left(g\right)\to \underset{\_}{}{\text{CO}}_{\text{2}}\left(g\right)+\underset{\_}{}{\text{H}}_{\text{2}}\text{O}\left(g\right)+\underset{\_}{}{\text{N}}_{\text{2}}\left(g\right)$

Since the left side has 2 $\text{P}$ while the right side has 3 $\text{O}$ atoms so first the stoichiometric coefficient of ${\text{CO}}_{\text{2}}$ and ${\text{H}}_{\text{2}}\text{O}$ is changed to 4 and 10 respectively while the coefficient of ${\text{O}}_2$ on left is made 9 so to make 18 $\text{O}$ atones on each side.

  $\underset{\_}{}{\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}\left(g\right)+\underset{\_}{\text{ 9 }}{\text{O}}_2\left(g\right)\to \underset{\_}{\text{ 4 }}{\text{CO}}_{\text{2}}\left(g\right)+\underset{\_}{\text{ 10 }}{\text{H}}_{\text{2}}\text{O}\left(g\right)+\underset{\_}{}{\text{N}}_{\text{2}}\left(g\right)$

Place 4 as the coefficient of ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}$ so as to make equal 4 $\text{C}$ on each side.

  $\underset{\_}{\text{ 4 }}{\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}\left(g\right)+\underset{\_}{\text{ 9 }}{\text{O}}_2\left(g\right)\to \underset{\_}{\text{ 4 }}{\text{CO}}_{\text{2}}\left(g\right)+\underset{\_}{\text{ 10 }}{\text{H}}_{\text{2}}\text{O}\left(g\right)+\underset{\_}{}{\text{N}}_{\text{2}}\left(g\right)$

Place 2 as the coefficient of ${\text{N}}_{\text{2}}$ so as to make equal 4 $\text{N}$ on each side.

  $\underset{\_}{\text{ 4 }}{\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}\left(g\right)+\underset{\_}{\text{ 9 }}{\text{O}}_2\left(g\right)\to \underset{\_}{\text{ 4 }}{\text{CO}}_{\text{2}}\left(g\right)+\underset{\_}{\text{ 10 }}{\text{H}}_{\text{2}}\text{O}\left(g\right)+\underset{\_}{\text{ 2 }}{\text{N}}_{\text{2}}\left(g\right)$

Since above equation has equal atoms on each side so the equation is now balanced.