Chapter 3, Problem 3.59P

(a)

Interpretation Introduction

Interpretation:

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

$\underset{\_}{}\text{Cu}{\left({\text{NO}}_3\right)}_2\left(aq\right)+\underset{\_}{}\text{KOH}\left(aq\right)\to \underset{\_}{}\text{Cu}{\left(\text{OH}\right)}_2\left(s\right)+\underset{\_}{}{\text{KNO}}_3\left(aq\right)$

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 3.59P

The balanced chemical reaction is as follows:

$\text{Cu}{\left({\text{NO}}_3\right)}_2\left(aq\right)+2\text{KOH}\left(aq\right)\to \text{Cu}{\left(\text{OH}\right)}_2\left(s\right)+2{\text{KNO}}_3\left(aq\right)$

Explanation of Solution

Start with $\text{Cu}{\left({\text{NO}}_3\right)}_2$. The equation implies that $2\text{ N}$ atoms are present on the reactant side while only $1\text{ N}$ atom is present on the product side. Balance nitrogen $\left(\text{N}\right)$ atoms first. To balance $\text{N}$, place the coefficient 2 in front of ${\text{KNO}}_3$. The equation becomes,

$\underset{\_}{}\text{Cu}{\left({\text{NO}}_3\right)}_2\left(aq\right)+\underset{\_}{}\text{KOH}\left(aq\right)\to \underset{\_}{}\text{Cu}{\left(\text{OH}\right)}_2\left(s\right)+\underset{\_}{2}{\text{KNO}}_3\left(aq\right)$

Nitrogen atoms are balanced. Balance potassium $\left(\text{K}\right)$ atoms next. $\text{1 K}$ atom is present on the reactant side while $\text{2 K}$ atoms are present on the product side. To balance $\text{K}$ atoms, place the coefficient 2 in front of $\text{KOH}$. The equation becomes,

$\underset{\_}{}\text{Cu}{\left({\text{NO}}_3\right)}_2\left(aq\right)+\underset{\_}{2}\text{KOH}\left(aq\right)\to \underset{\_}{}\text{Cu}{\left(\text{OH}\right)}_2\left(s\right)+\underset{\_}{2}{\text{KNO}}_3\left(aq\right)$

Balance copper $\left(\text{Cu}\right)$ atoms. $\text{1 Cu}$ atom is present on both sides. Place the coefficient 1 in front of $\text{Cu}{\left({\text{NO}}_3\right)}_2$ and $\text{Cu}{\left(\text{OH}\right)}_2$$\text{Cu}$. The equation becomes,

$\underset{\_}{1}\text{Cu}{\left({\text{NO}}_3\right)}_2\left(aq\right)+\underset{\_}{2}\text{KOH}\left(aq\right)\to \underset{\_}{1}\text{Cu}{\left(\text{OH}\right)}_2\left(s\right)+\underset{\_}{2}{\text{KNO}}_3\left(aq\right)$

Check whether the equation is balanced or not as follows:

$\text{Reactants}\left(1\text{ Cu, 2 K, 2 N, 2 H, 8 O}\right)\to \text{Products}\left(1\text{ Cu, 2 K, 2 N, 2 H, 8 O}\right)$

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

Conclusion

The balanced chemical reaction is as follows:

$\text{Cu}{\left({\text{NO}}_3\right)}_2\left(aq\right)+2\text{KOH}\left(aq\right)\to \text{Cu}{\left(\text{OH}\right)}_2\left(s\right)+2{\text{KNO}}_3\left(aq\right)$

(b)

Interpretation Introduction

Interpretation:

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

$\underset{\_}{}{\text{BCl}}_3\left(g\right)+\underset{\_}{}{\text{H}}_2\text{O}\left(l\right)\to \underset{\_}{}{\text{H}}_3{\text{BO}}_4\left(s\right)+\underset{\_}{}\text{HCl}\left(g\right)$

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 3.59P

The balanced chemical reaction is as follows:

${\text{BCl}}_3\left(g\right)+3{\text{H}}_2\text{O}\left(l\right)\to {\text{H}}_3{\text{BO}}_4\left(s\right)+3\text{HCl}\left(g\right)$

Explanation of Solution

The equation implies that $3\text{ Cl}$ atoms are present on the reactant side while only $1\text{ Cl}$ atom is present on the product side. Balance chlorine $\left(\text{Cl}\right)$ atoms first. To balance $\text{Cl}$, place the coefficient 3 in front of $\text{HCl}$. The equation becomes,

$\underset{\_}{}{\text{BCl}}_3\left(g\right)+\underset{\_}{}{\text{H}}_2\text{O}\left(l\right)\to \underset{\_}{}{\text{H}}_3{\text{BO}}_4\left(s\right)+\underset{\_}{3}\text{HCl}\left(g\right)$

Chlorine atoms are balanced. Next, balance hydrogen $\left(\text{H}\right)$ atoms. $\text{2 H}$ atoms are present on the reactant side while $\text{6 H}$ 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,

$\underset{\_}{}{\text{BCl}}_3\left(g\right)+\underset{\_}{3}{\text{H}}_2\text{O}\left(l\right)\to \underset{\_}{}{\text{H}}_3{\text{BO}}_4\left(s\right)+\underset{\_}{3}\text{HCl}\left(g\right)$

Hydrogen atoms are balanced. Place coefficient 1 in front of ${\text{BCl}}_3$ and ${\text{H}}_3{\text{BO}}_4$.

$\underset{\_}{1}{\text{BCl}}_3\left(g\right)+\underset{\_}{3}{\text{H}}_2\text{O}\left(l\right)\to \underset{\_}{1}{\text{H}}_3{\text{BO}}_4\left(s\right)+\underset{\_}{3}\text{HCl}\left(g\right)$

Check whether the equation is balanced or not as follows:

$\text{Reactants}\left(1\text{ B, 3 Cl, 6 H, 3 O}\right)\to \text{Products}\left(1\text{ B, 3 Cl, 6 H, 3 O}\right)$

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

Conclusion

The balanced chemical reaction is as follows:

${\text{BCl}}_3\left(g\right)+3{\text{H}}_2\text{O}\left(l\right)\to {\text{H}}_3{\text{BO}}_4\left(s\right)+3\text{HCl}\left(g\right)$

(c)

Interpretation Introduction

Interpretation:

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

$\underset{\_}{}{\text{CaSiO}}_3\left(s\right)+\underset{\_}{}\text{HF}\left(g\right)\to \underset{\_}{}{\text{SiF}}_4\left(g\right)+\underset{\_}{}{\text{CaF}}_2\left(s\right)+\underset{\_}{}{\text{H}}_2\text{O}\left(l\right)$

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 3.59P

The balanced chemical reaction is as follows:

${\text{CaSiO}}_3\left(s\right)+6\text{HF}\left(g\right)\to {\text{SiF}}_4\left(g\right)+{\text{CaF}}_2\left(s\right)+3{\text{H}}_2\text{O}\left(l\right)$

Explanation of Solution

The equation implies that $1\text{ F}$ atom is present on the reactant side while only $6\text{ F}$ atom is present on the product side. Balance fluorine $\left(\text{F}\right)$ atoms first and oxygen $\left(\text{O}\right)$ atoms in last. To balance $\text{F}$, place the coefficient 6 in front of $\text{HF}$. The equation becomes,

$\underset{\_}{}{\text{CaSiO}}_3\left(s\right)+\underset{\_}{6}\text{HF}\left(g\right)\to \underset{\_}{}{\text{SiF}}_4\left(g\right)+\underset{\_}{}{\text{CaF}}_2\left(s\right)+\underset{\_}{}{\text{H}}_2\text{O}\left(l\right)$

Fluorine 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,

$\underset{\_}{}{\text{CaSiO}}_3\left(s\right)+\underset{\_}{6}\text{HF}\left(g\right)\to \underset{\_}{}{\text{SiF}}_4\left(g\right)+\underset{\_}{}{\text{CaF}}_2\left(s\right)+\underset{\_}{3}{\text{H}}_2\text{O}\left(l\right)$

Hydrogen atoms are balanced. One calcium $\left(\text{Ca}\right)$ atom, one silicon $\left(\text{Si}\right)$ atom, and three oxygen $\left(\text{O}\right)$ atoms. Place coefficient 1 in front of ${\text{CaSiO}}_3$, ${\text{SiF}}_4$, and ${\text{CaF}}_2$ as,

$\underset{\_}{1}{\text{CaSiO}}_3\left(s\right)+\underset{\_}{6}\text{HF}\left(g\right)\to \underset{\_}{1}{\text{SiF}}_4\left(g\right)+\underset{\_}{1}{\text{CaF}}_2\left(s\right)+\underset{\_}{3}{\text{H}}_2\text{O}\left(l\right)$

Check whether the equation is balanced or not as follows:

$\text{Reactants}\left(1\text{ Ca, }1\text{ Si, 6 H, 6 F, 3 O}\right)\to \text{Products}\left(1\text{ Ca, }1\text{ Si, 6 H, 6 F, 3 O}\right)$

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

Conclusion

The balanced chemical reaction is as follows:

${\text{CaSiO}}_3\left(s\right)+6\text{HF}\left(g\right)\to {\text{SiF}}_4\left(g\right)+{\text{CaF}}_2\left(s\right)+3{\text{H}}_2\text{O}\left(l\right)$

(d)

Interpretation Introduction

Interpretation:

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

$\underset{\_}{}{\left(\text{CN}\right)}_2\left(g\right)+\underset{\_}{}{\text{H}}_2\text{O}\left(l\right)\to \underset{\_}{}{\text{H}}_2{\text{C}}_2{\text{O}}_4\left(aq\right)+\underset{\_}{}{\text{H}}_2\text{O}\left(g\right)+\underset{\_}{}{\text{NH}}_3\left(g\right)$

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 3.59P

The balanced chemical reaction is as follows:

${\left(\text{CN}\right)}_2\left(g\right)+4{\text{H}}_2\text{O}\left(l\right)\to {\text{H}}_2{\text{C}}_2{\text{O}}_4\left(aq\right)+2{\text{NH}}_3\left(g\right)$

Explanation of Solution

The equation implies that $2\text{ N}$ atoms is present on the reactant side while $1\text{ N}$ atom is 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 ${\text{NH}}_3$. The equation becomes,

$\underset{\_}{}{\left(\text{CN}\right)}_2\left(g\right)+\underset{\_}{}{\text{H}}_2\text{O}\left(l\right)\to \underset{\_}{}{\text{H}}_2{\text{C}}_2{\text{O}}_4\left(aq\right)+\underset{\_}{2}{\text{NH}}_3\left(g\right)$

Nitrogen atoms are balanced, hence, place the coefficient 1 in front of ${\left(\text{CN}\right)}_2$. Next, balance oxygen $\left(\text{O}\right)$ atoms. $\text{1 O}$ atom is present on the reactant side while $\text{4 O}$ atoms are present on the product side. Hence, place the coefficient 4 in front of ${\text{H}}_2\text{O}$. The equation becomes,

$\underset{\_}{}{\left(\text{CN}\right)}_2\left(g\right)+\underset{\_}{4}{\text{H}}_2\text{O}\left(l\right)\to \underset{\_}{}{\text{H}}_2{\text{C}}_2{\text{O}}_4\left(aq\right)+\underset{\_}{2}{\text{NH}}_3\left(g\right)$

Oxygen atoms are balanced. Two carbon $\left(\text{C}\right)$ atoms and 8 hydrogens $\left(\text{H}\right)$ are present on both sides. Place coefficient 1 in front of ${\left(\text{CN}\right)}_2$ and ${\text{H}}_2{\text{C}}_2{\text{O}}_4$ as:

$\underset{\_}{1}{\left(\text{CN}\right)}_2\left(g\right)+\underset{\_}{4}{\text{H}}_2\text{O}\left(l\right)\to \underset{\_}{1}{\text{H}}_2{\text{C}}_2{\text{O}}_4\left(aq\right)+\underset{\_}{2}{\text{NH}}_3\left(g\right)$

Check whether the equation is balanced or not as follows:

$\text{Reactants}\left(2\text{ C, }2\text{ N, 8 H, 4 O}\right)\to \text{Products}\left(2\text{ C, }2\text{ N, 8 H, 4 O}\right)$

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

Conclusion

The balanced chemical reaction is as follows:

${\left(\text{CN}\right)}_2\left(g\right)+4{\text{H}}_2\text{O}\left(l\right)\to {\text{H}}_2{\text{C}}_2{\text{O}}_4\left(aq\right)+2{\text{NH}}_3\left(g\right)$