Chapter 3, Problem 3.58P

(a)

Interpretation Introduction

Interpretation:

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

$\underset{\_}{}{\text{SO}}_2\left(g\right)+\underset{\_}{}{\text{O}}_2\left(g\right)\to \underset{\_}{}{\text{SO}}_3\left(g\right)$

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.

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

A balanced equation for the following reaction is as follows:

$2{\text{SO}}_2\left(g\right)+{\text{O}}_2\left(g\right)\to 2{\text{SO}}_3\left(g\right)$

Explanation of Solution

The equation implies that four oxygen $\left(\text{O}\right)$ atoms are present on the reactant side while three $\text{O}$ atoms are present on the product side. Balance $\text{O}$ atoms first. To balance $\text{O}$, place the coefficient 2 in front of ${\text{SO}}_3$. The equation becomes,

$\underset{\_}{}{\text{SO}}_2\left(g\right)+\underset{\_}{}{\text{O}}_2\left(g\right)\to \underset{\_}{2}{\text{SO}}_3\left(g\right)$

Since one sulfur $\left(\text{S}\right)$ atom is present on the reactant side while two $\text{S}$ atoms are present on the product side. To balance $\text{S}$ atoms, place the coefficient 2 in front of ${\text{SO}}_2$. The equation becomes,

$\underset{\_}{2}{\text{SO}}_2\left(g\right)+\underset{\_}{}{\text{O}}_2\left(g\right)\to \underset{\_}{2}{\text{SO}}_3\left(g\right)$

Place the coefficient 1 in front of ${\text{O}}_2$ as follows:

$\underset{\_}{2}{\text{SO}}_2\left(g\right)+\underset{\_}{1}{\text{O}}_2\left(g\right)\to \underset{\_}{2}{\text{SO}}_3\left(g\right)$

Check whether the equation is balanced or not as follows:

$\text{Reactants}\left(2\text{ S, 6 O}\right)\to \text{Products}\left(2\text{ S, 6 O}\right)$

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

Conclusion

A balanced equation for the following reaction is as follows:

$2{\text{SO}}_2\left(g\right)+{\text{O}}_2\left(g\right)\to 2{\text{SO}}_3\left(g\right)$

(b)

Interpretation Introduction

Interpretation:

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

$\underset{\_}{}{\text{Sc}}_2{\text{O}}_3\left(s\right)+\underset{\_}{}{\text{H}}_2\text{O}\left(l\right)\to \underset{\_}{}\text{Sc}{\left(\text{OH}\right)}_3\left(s\right)$

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.

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:

${\text{Sc}}_2{\text{O}}_3\left(s\right)+3{\text{H}}_2\text{O}\left(l\right)\to 2\text{Sc}{\left(\text{OH}\right)}_3\left(s\right)$

Explanation of Solution

The equation implies that two scandium $\left(\text{Sc}\right)$ atoms are present on the reactant side while only one $\text{Sc}$ atom is present on the product side. Balance $\text{Sc}$ atoms first. To balance $\text{Sc}$, place the coefficient 2 in front of $\text{Sc}{\left(\text{OH}\right)}_3$. The equation becomes,

$\underset{\_}{}{\text{Sc}}_2{\text{O}}_3\left(s\right)+\underset{\_}{}{\text{H}}_2\text{O}\left(l\right)\to \underset{\_}{2}\text{Sc}{\left(\text{OH}\right)}_3\left(s\right)$

Next, balance hydrogen $\left(\text{H}\right)$ atoms. Two $\text{H}$ atoms are present on the reactant side while six $\text{H}$ 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,

$\underset{\_}{}{\text{Sc}}_2{\text{O}}_3\left(s\right)+\underset{\_}{3}{\text{H}}_2\text{O}\left(l\right)\to \underset{\_}{2}\text{Sc}{\left(\text{OH}\right)}_3\left(s\right)$

Hydrogen atoms are balanced. Six oxygen $\left(\text{O}\right)$ atoms are present on each side of the reaction. Place coefficient 1 in front of ${\text{Sc}}_2{\text{O}}_3$.

$\underset{\_}{1}{\text{Sc}}_2{\text{O}}_3\left(s\right)+\underset{\_}{3}{\text{H}}_2\text{O}\left(l\right)\to \underset{\_}{2}\text{Sc}{\left(\text{OH}\right)}_3\left(s\right)$

Check whether the equation is balanced or not as follows:

$\text{Reactants}\left(2\text{ Sc, 6 H, 6 O}\right)\to \text{Products}\left(1\text{ Sc, 6 H, 6 O}\right)$

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

Conclusion

The balanced chemical reaction is as follows:

${\text{Sc}}_2{\text{O}}_3\left(s\right)+3{\text{H}}_2\text{O}\left(l\right)\to 2\text{Sc}{\left(\text{OH}\right)}_3\left(s\right)$

(c)

Interpretation Introduction

Interpretation:

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

$\underset{\_}{}{\text{H}}_3{\text{PO}}_4\left(aq\right)+\underset{\_}{}\text{NaOH}\left(aq\right)\to \underset{\_}{}{\text{Na}}_2{\text{HPO}}_4\left(aq\right)+\underset{\_}{}{\text{H}}_2\text{O}\left(l\right)$

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:

${\text{H}}_3{\text{PO}}_4\left(aq\right)+2\text{NaOH}\left(aq\right)\to {\text{Na}}_2{\text{HPO}}_4\left(aq\right)+2{\text{H}}_2\text{O}\left(l\right)$

Explanation of Solution

Start with $\text{Na}$ atoms. The equation implies that one sodium $\left(\text{Na}\right)$ atom is present on the reactant side while two $\text{Na}$ atoms are presents on the product side. To balance $\text{Na}$, place the coefficient 2 in front of $\text{NaOH}$. The equation becomes,

$\underset{\_}{}{\text{H}}_3{\text{PO}}_4\left(aq\right)+\underset{\_}{2}\text{NaOH}\left(aq\right)\to \underset{\_}{}{\text{Na}}_2{\text{HPO}}_4\left(aq\right)+\underset{\_}{}{\text{H}}_2\text{O}\left(l\right)$

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

$\underset{\_}{}{\text{H}}_3{\text{PO}}_4\left(aq\right)+\underset{\_}{2}\text{NaOH}\left(aq\right)\to \underset{\_}{}{\text{Na}}_2{\text{HPO}}_4\left(aq\right)+\underset{\_}{2}{\text{H}}_2\text{O}\left(l\right)$

Five hydrogen $\left(\text{H}\right)$ atoms are present on both sides. Place coefficient 1 in front of ${\text{H}}_3{\text{PO}}_4$ and ${\text{Na}}_2{\text{HPO}}_4$ as follows:

$\underset{\_}{1}{\text{H}}_3{\text{PO}}_4\left(aq\right)+\underset{\_}{2}\text{NaOH}\left(aq\right)\to \underset{\_}{1}{\text{Na}}_2{\text{HPO}}_4\left(aq\right)+\underset{\_}{2}{\text{H}}_2\text{O}\left(l\right)$

Check whether the equation is balanced or not as follows:

$\text{Reactants}\left(2\text{ Na, }1\text{ P, 5 H, 6 O}\right)\to \text{Products}\left(2\text{ Na, }1\text{ P, 5 H, 6 O}\right)$

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

Conclusion

The balanced chemical reaction is as follows:

${\text{H}}_3{\text{PO}}_4\left(aq\right)+2\text{NaOH}\left(aq\right)\to {\text{Na}}_2{\text{HPO}}_4\left(aq\right)+2{\text{H}}_2\text{O}\left(l\right)$

(d)

Interpretation Introduction

Interpretation:

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

$\underset{\_}{}{\text{C}}_6{\text{H}}_{10}{\text{O}}_5\left(s\right)+\underset{\_}{}{\text{O}}_2\left(g\right)\to \underset{\_}{}{\text{CO}}_2\left(g\right)+\underset{\_}{}{\text{H}}_2\text{O}\left(g\right)$

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:

${\text{C}}_6{\text{H}}_{10}{\text{O}}_5\left(s\right)+6{\text{O}}_2\left(g\right)\to 6{\text{CO}}_2\left(g\right)+5{\text{H}}_2\text{O}\left(g\right)$

Explanation of Solution

Start with carbon $\left(\text{C}\right)$ atoms. The equation implies that six $\text{C}$ atoms are present on the reactant side while only one $\text{C}$ atom is present on the product side. Balance $\text{C}$ atoms first and oxygen $\left(\text{O}\right)$ atoms in last. To balance $\text{C}$, place the coefficient 6 in front of ${\text{CO}}_2$. The equation becomes,

$\underset{\_}{}{\text{C}}_6{\text{H}}_{10}{\text{O}}_5\left(s\right)+\underset{\_}{}{\text{O}}_2\left(g\right)\to \underset{\_}{6}{\text{CO}}_2\left(g\right)+\underset{\_}{}{\text{H}}_2\text{O}\left(g\right)$

Carbon atoms are balanced. Next, balance hydrogen $\left(\text{H}\right)$ atoms. The equation implies that ten $\text{H}$ atoms are present on the reactant side while only two $\text{H}$ atoms are presents on the product side. To balance $\text{H}$, place the coefficient 5 in front of ${\text{H}}_2\text{O}$. The equation becomes,

$\underset{\_}{}{\text{C}}_6{\text{H}}_{10}{\text{O}}_5\left(s\right)+\underset{\_}{}{\text{O}}_2\left(g\right)\to \underset{\_}{6}{\text{CO}}_2\left(g\right)+\underset{\_}{5}{\text{H}}_2\text{O}\left(g\right)$

Next, balance oxygen $\left(\text{O}\right)$ atoms. Seven $\text{O}$ atoms are present on the reactant side while seventeen $\text{O}$ atoms are present on the product side. Hence, place the coefficient 6 in front of ${\text{O}}_2$. The equation becomes,

$\underset{\_}{}{\text{C}}_6{\text{H}}_{10}{\text{O}}_5\left(s\right)+\underset{\_}{6}{\text{O}}_2\left(g\right)\to \underset{\_}{6}{\text{CO}}_2\left(g\right)+\underset{\_}{5}{\text{H}}_2\text{O}\left(g\right)$

Oxygen atoms are balanced. Place coefficient 1 in front of ${\text{C}}_6{\text{H}}_{10}{\text{O}}_5$.

$\underset{\_}{1}{\text{C}}_6{\text{H}}_{10}{\text{O}}_5\left(s\right)+\underset{\_}{6}{\text{O}}_2\left(g\right)\to \underset{\_}{6}{\text{CO}}_2\left(g\right)+\underset{\_}{5}{\text{H}}_2\text{O}\left(g\right)$

Check whether the equation is balanced or not as follows:

$\text{Reactants}\left(6\text{ C, 10 H, 17 O}\right)\to \text{Products}\left(6\text{ C, 10 H, 17 O}\right)$

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

Conclusion

The balanced chemical reaction is as follows:

${\text{C}}_6{\text{H}}_{10}{\text{O}}_5\left(s\right)+6{\text{O}}_2\left(g\right)\to 6{\text{CO}}_2\left(g\right)+5{\text{H}}_2\text{O}\left(g\right)$