Chapter 3, Problem 3.38P

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

1. $\text{\_\_Cu}{\left({\text{NO}}_3\right)}_2\left(aq\right)+\_\_\text{KOH}\left(aq\right)\to \_\_\text{Cu}{\left(\text{OH}\right)}_2\left(s\right)+\_\_{\text{KNO}}_3\left(aq\right)$ ${\text{\_\_BCl}}_3\left(g\right)+\_\_{\text{H}}_2\text{O}\left(\text{l}\right)\to \_\_{\text{H}}_3{\text{BO}}_3\left(s\right)+\_\_\text{HCl}\left(g\right)$ ${\text{\_\_CaSiO}}_3\left(g\right)+\_\_\text{HF}\left(g\right)\to \_\_{\text{SiF}}_{\text{4}}\left(g\right)+\_\_{\text{CaF}}_2\left(s\right)+\_\_{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)$ $\text{\_\_}{\left(\text{CN}\right)}_2\left(g\right)+\_\_{\text{H}}_2\text{O}\left(\text{l}\right)\to \_\_{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\left(aq\right)+\_\_{\text{NH}}_3\left(g\right)$

Interpretation Introduction

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

  $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_2\left(aq\right)+\underset{\_}{}\text{KOH}\left(aq\right)\to \text{Cu}{\left(\text{OH}\right)}_2\left(s\right)+\underset{\_}{}{\text{KNO}}_{\text{3}}\left(aq\right)$

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 3.38P

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

  $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_2\left(aq\right)+\underset{\_}{\text{ 2 }}\text{KOH}\left(aq\right)\to \text{Cu}{\left(\text{OH}\right)}_2\left(s\right)+\underset{\_}{\text{ 2 }}{\text{KNO}}_{\text{3}}\left(aq\right)$

Explanation of Solution

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

  $\underset{\_}{}\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_2\left(aq\right)+\underset{\_}{}\text{KOH}\left(aq\right)\to \underset{\_}{}\text{Cu}{\left(\text{OH}\right)}_2\left(s\right)+\underset{\_}{}{\text{KNO}}_{\text{3}}\left(aq\right)$

Since the left side has 2 ${\text{NO}}_{\text{3}}$ while the right side has 1 ${\text{NO}}_{\text{3}}$ atom so first the stoichiometric coefficient of ${\text{NO}}_{\text{3}}$ on the right is changed to 2. This is done when 2 is placed as a coefficient of ${\text{KNO}}_{\text{3}}$ .

  $\underset{\_}{}\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_2\left(aq\right)+\underset{\_}{}\text{KOH}\left(aq\right)\to \underset{\_}{}\text{Cu}{\left(\text{OH}\right)}_2\left(s\right)+\underset{\_}{\text{ 2 }}{\text{KNO}}_{\text{3}}\left(aq\right)$

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

  $\underset{\_}{}\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_2\left(aq\right)+\underset{\_}{\text{ 2 }}\text{KOH}\left(aq\right)\to \underset{\_}{}\text{Cu}{\left(\text{OH}\right)}_2\left(s\right)+\underset{\_}{\text{ 2 }}{\text{KNO}}_{\text{3}}\left(aq\right)$

Coefficient of $\text{Cu}{\left(\text{OH}\right)}_2$ will remain 1 on each side and thus final balanced equation is written as follows:

  $\text{Cu}{\left({\text{NO}}_{\text{3}}\right)}_2\left(aq\right)+\underset{\_}{\text{ 2 }}\text{KOH}\left(aq\right)\to \text{Cu}{\left(\text{OH}\right)}_2\left(s\right)+\underset{\_}{\text{ 2 }}{\text{KNO}}_{\text{3}}\left(aq\right)$

Since above equation has equal 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.

  $\underset{\_}{}{\text{BCl}}_{\text{3}}\left(g\right)+\underset{\_}{}{\text{H}}_{\text{2}}\text{O}\left(l\right)\to \underset{\_}{}{\text{H}}_{\text{3}}{\text{BO}}_3\left(s\right)+\underset{\_}{}\text{HCl}\left(g\right)$

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 3.38P

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

  ${\text{BCl}}_{\text{3}}\left(g\right)+\underset{\_}{\text{ 3 }}{\text{H}}_{\text{2}}\text{O}\left(l\right)\to {\text{H}}_{\text{3}}{\text{BO}}_3\left(s\right)+\underset{\_}{\text{ 3 }}\text{HCl}\left(g\right)$

Explanation of Solution

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

  $\underset{\_}{}{\text{BCl}}_{\text{3}}\left(g\right)+\underset{\_}{}{\text{H}}_{\text{2}}\text{O}\left(l\right)\to \underset{\_}{}{\text{H}}_{\text{3}}{\text{BO}}_3\left(s\right)+\underset{\_}{}\text{HCl}\left(g\right)$

Since the left side has 3 $\text{Cl}$ while the right side has 3 $\text{Cl}$ atoms so first the 4 is placed as the stoichiometric coefficient of $\text{HCl}$ .

  $\underset{\_}{}{\text{BCl}}_{\text{3}}\left(g\right)+\underset{\_}{}{\text{H}}_{\text{2}}\text{O}\left(l\right)\to \underset{\_}{}{\text{H}}_{\text{3}}{\text{BO}}_3\left(s\right)+\underset{\_}{\text{ 3 }}\text{HCl}\left(g\right)$

Place 3 as coefficient of ${\text{H}}_{\text{2}}\text{O}$ so as to make equal 6 $\text{H}$ on each side and coefficient of other species will remain 1.

  ${\text{BCl}}_{\text{3}}\left(g\right)+\underset{\_}{\text{ 3 }}{\text{H}}_{\text{2}}\text{O}\left(l\right)\to {\text{H}}_{\text{3}}{\text{BO}}_3\left(s\right)+\underset{\_}{\text{ 3 }}\text{HCl}\left(g\right)$

Since above equation has equal 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.

  $\underset{\_}{}{\text{CaSiO}}_3\left(s\right)+\underset{\_}{}\text{HF}\left(aq\right)\to \underset{\_}{}{\text{SiF}}_{\text{4}}\left(aq\right)+\underset{\_}{}{\text{CaF}}_{\text{2}}+\underset{\_}{}{\text{H}}_{\text{2}}\text{O}\left(l\right)$

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 3.38P

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

  ${\text{CaSiO}}_3\left(s\right)+\underset{\_}{\text{ 6 }}\text{HF}\left(aq\right)\to {\text{SiF}}_{\text{4}}\left(aq\right)+{\text{CaF}}_{\text{2}}+\underset{\_}{\text{ 3 }}{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Explanation of Solution

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

  $\underset{\_}{}{\text{CaSiO}}_3\left(s\right)+\underset{\_}{}\text{HF}\left(aq\right)\to \underset{\_}{}{\text{SiF}}_{\text{4}}\left(aq\right)+\underset{\_}{}{\text{CaF}}_{\text{2}}+\underset{\_}{}{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Since the left side has 3 $\text{O}$ while the right side has 1 $\text{O}$ atom so first 2 is placed as the stoichiometric coefficient of ${\text{H}}_{\text{2}}\text{O}$ .

  $\underset{\_}{}{\text{CaSiO}}_3\left(s\right)+\underset{\_}{}\text{HF}\left(aq\right)\to \underset{\_}{}{\text{SiF}}_{\text{4}}\left(aq\right)+\underset{\_}{}{\text{CaF}}_{\text{2}}+\underset{\_}{\text{ 3 }}{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Place 6 as the coefficient of $\text{HF}$ so as to make 6 $\text{H}$ on each side.

  ${\text{CaSiO}}_3\left(s\right)+\underset{\_}{\text{ 6 }}\text{HF}\left(aq\right)\to {\text{SiF}}_{\text{4}}\left(aq\right)+{\text{CaF}}_{\text{2}}+\underset{\_}{\text{ 3 }}{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Since above equation has equal on each side so the equation is now balanced and the coefficient of each of the species left will be 1.

Interpretation Introduction

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

  $\underset{\_}{}{\left(\text{CN}\right)}_2\left(g\right)+\underset{\_}{}{\text{H}}_2\text{O}\left(l\right)\to \underset{\_}{}{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\left(aq\right)+\underset{\_}{}{\text{NH}}_3\left(g\right)$

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 3.38P

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

  ${\left(\text{CN}\right)}_2\left(g\right)+\underset{\_}{\text{ 4 }}{\text{H}}_2\text{O}\left(l\right)\to {\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\left(aq\right)+\underset{\_}{\text{ 2 }}{\text{NH}}_3\left(g\right)$

Explanation of Solution

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

  $\underset{\_}{}{\left(\text{CN}\right)}_2\left(g\right)+\underset{\_}{}{\text{H}}_2\text{O}\left(l\right)\to \underset{\_}{}{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\left(aq\right)+\underset{\_}{}{\text{NH}}_3\left(g\right)$

Since the left side has 2 $\text{N}$ while the right side has 3 $\text{N}$ atoms so first the 2 is placed as the stoichiometric coefficient of ${\text{NH}}_3$ .

  $\underset{\_}{}{\left(\text{CN}\right)}_2\left(g\right)+\underset{\_}{}{\text{H}}_2\text{O}\left(l\right)\to \underset{\_}{}{\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\left(aq\right)+\underset{\_}{\text{ 2 }}{\text{NH}}_3\left(g\right)$

Place 4 as the coefficient of ${\text{H}}_{\text{2}}\text{O}$ so as to make equal 8 $\text{H}$ on each side and coefficient of other species will remain 1.

  ${\left(\text{CN}\right)}_2\left(g\right)+\underset{\_}{\text{ 4 }}{\text{H}}_2\text{O}\left(l\right)\to {\text{H}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\left(aq\right)+\underset{\_}{\text{ 2 }}{\text{NH}}_3\left(g\right)$

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