Chapter 3, Problem 3.59P

(a)

Interpretation Introduction

Interpretation:

The following chemical equation is to be balanced.

$\_{\text{As}}_4{\text{S}}_{\text{6}}(s)+\_{\text{O}}_{\text{2}}\left(g\right)\to \_{\text{As}}_{\text{4}}{\text{O}}_{\text{6}}\left(s\right)+\_{\text{SO}}_{\text{2}}(g)$

Concept introduction:

Balancing is a hit and trial method where a smallest whole number coefficient is used. One element is balanced at a time on both side of the equation.

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: Translate the chemical statement into a skeleton equation. The chemical substances that undergo a change are termed as reactants and the chemical substances that are produced during the chemical change are termed as products. The reactants are specified on the left side of the yield arrow while the products are specified on the right side of the yield arrow. Put a blank before each formula while beginning the balancing process.

Step 2: 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 3: 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 4: 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 5: Check whether the chemical equation is balanced or not by counting the number of atoms of each element on both sides.

Step 6: Specify the states of matter of each chemical substance present in the balanced chemical equation. The table for the abbreviations used for each state is as follows:

$\begin{array}{cc}\text{State}& \text{Abbreviation}\\ \text{Solid}& s\\ \text{Liquid}& l\\ \text{Gas}& g\\ \begin{array}{l}\text{Aqueous}\\ \text{solution}\end{array}& aq\end{array}$

Answer to Problem 3.59P

The balanced chemical equation is:

${\text{As}}_{\text{4}}{\text{S}}_{\text{6}}(s)+9{\text{O}}_{\text{2}}\left(g\right)\to {\text{As}}_{\text{4}}{\text{O}}_{\text{6}}\left(s\right)+{\text{6SO}}_{\text{2}}(g)$

Explanation of Solution

The chemical equation to be balanced is:

$\_{\text{As}}_{\text{4}}{\text{S}}_{\text{6}}(s)+\_{\text{O}}_{\text{2}}\left(g\right)\to \_{\text{As}}_{\text{4}}{\text{O}}_{\text{6}}\left(s\right)+\_{\text{SO}}_{\text{2}}(g)$        (1)

The $6\text{ S}$ atoms in ${\text{As}}_{\text{4}}{\text{S}}_{\text{6}}$ on the left require a coefficient of $\text{6}$ in front of ${\text{SO}}_{\text{2}}$ on the right.

Rewrite the chemical equation (1):

$\_{\text{As}}_4{\text{S}}_{\text{6}}(s)+\_{\text{O}}_{\text{2}}\left(g\right)\to \_{\text{As}}_{\text{4}}{\text{O}}_{\text{6}}\left(s\right)+6{\text{SO}}_{\text{2}}(g)$        (2)

The $\text{18 O}$ atoms on the right ($6$ in ${\text{As}}_{\text{4}}{\text{S}}_{\text{6}}$ and $12$ in ${\text{6 SO}}_{\text{2}}$) require a coefficient of $9$ in front of ${\text{O}}_{\text{2}}$ on the left.

Rewrite the chemical equation (2):

$\_{\text{As}}_{\text{4}}{\text{S}}_{\text{6}}(s)+9{\text{O}}_{\text{2}}\left(g\right)\to \_{\text{As}}_{\text{4}}{\text{O}}_{\text{6}}\left(s\right)+6{\text{SO}}_{\text{2}}(g)$        (3)

There are $\text{4 As}$ atoms on each side. Rewrite the chemical equation (3):

${\text{As}}_{\text{4}}{\text{S}}_{\text{6}}(s)+9{\text{O}}_{\text{2}}\left(g\right)\to {\text{As}}_{\text{4}}{\text{O}}_{\text{6}}\left(s\right)+6{\text{SO}}_{\text{2}}(g)$

Conclusion

The balanced chemical equation is:

${\text{As}}_{\text{4}}{\text{S}}_{\text{6}}(s)+9{\text{O}}_{\text{2}}\left(g\right)\to {\text{As}}_{\text{4}}{\text{O}}_{\text{6}}\left(s\right)+{\text{6SO}}_{\text{2}}(g)$

(b)

Interpretation Introduction

Interpretation:

The following chemical equation is to be balanced.

$\_{\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\left(s\right)+\_{\text{SiO}}_{\text{2}}\left(s\right)+\_\text{C}\left(s\right)\to \_{\text{P}}_{\text{4}}\left(g\right)+\_{\text{CaSiO}}_{\text{3}}\left(l\right)+\_\text{CO}\left(g\right)$

Concept introduction:

Balancing is a hit and trial method where a smallest whole number coefficient is used. One element is balanced at a time on both side of the equation.

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: Translate the chemical statement into a skeleton equation. The chemical substances that undergo a change are termed as reactants and the chemical substances that are produced during the chemical change are termed as products. The reactants are specified on the left side of the yield arrow while the products are specified on the right side of the yield arrow. Put a blank before each formula while beginning the balancing process.

Step 2: 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 3: 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 4: 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 5: Check whether the chemical equation is balanced or not by counting the number of atoms of each element on both sides.

Step 6: Specify the states of matter of each chemical substance present in the balanced chemical equation. The table for the abbreviations used for each state is as follows:

$\begin{array}{cc}\text{State}& \text{Abbreviation}\\ \text{Solid}& s\\ \text{Liquid}& l\\ \text{Gas}& g\\ \begin{array}{l}\text{Aqueous}\\ \text{solution}\end{array}& aq\end{array}$

Answer to Problem 3.59P

The balanced chemical equation is:

${\text{2Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\left(s\right)+{\text{6SiO}}_{\text{2}}\left(s\right)+10\text{C}\left(s\right)\to {\text{P}}_{\text{4}}\left(g\right)+{\text{6CaSiO}}_{\text{3}}\left(l\right)+10\text{CO}\left(g\right)$

Explanation of Solution

The chemical equation to be balanced is:

$\_{\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\left(s\right)+\_{\text{SiO}}_{\text{2}}\left(s\right)+\_\text{C}\left(s\right)\to \_{\text{P}}_{\text{4}}\left(g\right)+\_{\text{CaSiO}}_{\text{3}}\left(l\right)+\_\text{CO}\left(g\right)$        (4)

The $4\text{ P}$ atoms in ${\text{P}}_{\text{4}}$ on the left require a coefficient of $2$ in front of ${\text{Ca}}_{\text{3}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{2}}$ on the left.

Rewrite the chemical equation (4):

${\text{2Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\left(s\right)+\_{\text{SiO}}_{\text{2}}\left(s\right)+\_\text{C}\left(s\right)\to \_{\text{P}}_{\text{4}}\left(g\right)+\_{\text{CaSiO}}_{\text{3}}\left(l\right)+\_\text{CO}\left(g\right)$        (5)

The $\text{6 Ca}$ atoms in ${\text{2 Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ on the left require a coefficient of $6$ in front of ${\text{CaSiO}}_{\text{3}}$ on the right.

Rewrite the chemical equation (5):

${\text{2Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\left(s\right)+\_{\text{SiO}}_{\text{2}}\left(s\right)+\_\text{C}\left(s\right)\to \_{\text{P}}_{\text{4}}\left(g\right)+6{\text{CaSiO}}_{\text{3}}\left(l\right)+\_\text{CO}\left(g\right)$        (6)

The $\text{6 Si}$ atoms in ${\text{6 CaSiO}}_{\text{3}}$ on the right require a coefficient of $6$ in front of ${\text{SiO}}_{\text{2}}$ on the left.

Rewrite the chemical equation (6):

${\text{2Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\left(s\right)+6{\text{SiO}}_{\text{2}}\left(s\right)+\_\text{C}\left(s\right)\to \_{\text{P}}_{\text{4}}\left(g\right)+6{\text{CaSiO}}_{\text{3}}\left(l\right)+\_\text{CO}\left(g\right)$        (7)

There are $\text{28 O}$ atoms on the left ($16$ in ${\text{2 Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ and $12$ in ${\text{6 SiO}}_{\text{2}}$), there are $18\text{ O}$ atoms on the right in ${\text{6 CaSiO}}_{\text{3}}$. A coefficient of $10$ in front of $\text{CO}$ on the right requires a coefficient of $10$ in front of $\text{C}$ on the left.

Rewrite the chemical equation (7).

${\text{2Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\left(s\right)+{\text{6SiO}}_{\text{2}}\left(s\right)+10\text{C}\left(s\right)\to {\text{P}}_{\text{4}}\left(g\right)+{\text{6CaSiO}}_{\text{3}}\left(l\right)+10\text{CO}\left(g\right)$

Conclusion

The balanced chemical equation is:

${\text{2Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}\left(s\right)+{\text{6SiO}}_{\text{2}}\left(s\right)+10\text{C}\left(s\right)\to {\text{P}}_{\text{4}}\left(g\right)+{\text{6CaSiO}}_{\text{3}}\left(l\right)+10\text{CO}\left(g\right)$

(c)

Interpretation Introduction

Interpretation:

The following chemical equation is to be balanced.

$\_\text{Fe}\left(s\right)\text{ + }\_{\text{H}}_{\text{2}}\text{O}\left(g\right)\to \_{\text{Fe}}_{\text{3}}{\text{O}}_{\text{4}}\left(s\right)+\_{\text{H}}_{\text{2}}\left(g\right)$

Concept introduction:

Balancing is a hit and trial method where a smallest whole number coefficient is used. One element is balanced at a time on both side of the equation.

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: Translate the chemical statement into a skeleton equation. The chemical substances that undergo a change are termed as reactants and the chemical substances that are produced during the chemical change are termed as products. The reactants are specified on the left side of the yield arrow while the products are specified on the right side of the yield arrow. Put a blank before each formula while beginning the balancing process.

Step 2: 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 3: 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 4: 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 5: Check whether the chemical equation is balanced or not by counting the number of atoms of each element on both sides.

Step 6: Specify the states of matter of each chemical substance present in the balanced chemical equation. The table for the abbreviations used for each state is as follows:

$\begin{array}{cc}\text{State}& \text{Abbreviation}\\ \text{Solid}& s\\ \text{Liquid}& l\\ \text{Gas}& g\\ \begin{array}{l}\text{Aqueous}\\ \text{solution}\end{array}& aq\end{array}$

Answer to Problem 3.59P

The balanced chemical equation is:

$3\text{Fe}\left(s\right){\text{ + 4H}}_{\text{2}}\text{O}\left(g\right)\to {\text{Fe}}_{\text{3}}{\text{O}}_{\text{4}}\left(s\right)+4{\text{H}}_{\text{2}}\left(g\right)$

Explanation of Solution

The chemical equation to be balanced is:

$\_\text{Fe}\left(s\right)\text{ + }\_{\text{H}}_{\text{2}}\text{O}\left(g\right)\to \_{\text{Fe}}_{\text{3}}{\text{O}}_{\text{4}}\left(s\right)+\_{\text{H}}_{\text{2}}\left(g\right)$        (8)

The $\text{3 Fe}$ atoms in ${\text{Fe}}_{\text{3}}{\text{O}}_{\text{4}}$ on the right require a coefficient of $3$ in front of $\text{Fe}$ on the left.

Rewrite the chemical equation (8):

$\text{3Fe}\left(s\right)\text{ + }\_{\text{H}}_{\text{2}}\text{O}\left(g\right)\to \_{\text{Fe}}_{\text{3}}{\text{O}}_{\text{4}}\left(s\right)+\_{\text{H}}_{\text{2}}\left(g\right)$        (9)

The $\text{4 O}$ atoms in ${\text{Fe}}_{\text{3}}{\text{O}}_{\text{4}}$ on the right require a coefficient of $4$ in front of ${\text{H}}_{\text{2}}\text{O}$ on the left.

Rewrite the chemical equation (9):

$\text{3Fe}\left(s\right){\text{ + 4H}}_{\text{2}}\text{O}\left(g\right)\to \_{\text{Fe}}_{\text{3}}{\text{O}}_{\text{4}}\left(s\right)+\_{\text{H}}_{\text{2}}\left(g\right)$        (10)

The $\text{8 H}$ atoms on the left in ${\text{4 H}}_{\text{2}}\text{O}$ require a coefficient of $4$ in front of ${\text{H}}_{\text{2}}$ on the right.

Rewrite the chemical equation (10):

$3\text{Fe}\left(s\right){\text{ + 4H}}_{\text{2}}\text{O}\left(g\right)\to {\text{Fe}}_{\text{3}}{\text{O}}_{\text{4}}\left(s\right)+4{\text{H}}_{\text{2}}\left(g\right)$

Conclusion

The balanced chemical equation is:

$3\text{Fe}\left(s\right){\text{ + 4H}}_{\text{2}}\text{O}\left(g\right)\to {\text{Fe}}_{\text{3}}{\text{O}}_{\text{4}}\left(s\right)+4{\text{H}}_{\text{2}}\left(g\right)$

(d)

Interpretation Introduction

Interpretation:

The following chemical equation is to be balanced.

$\_{\text{S}}_{\text{2}}{\text{Cl}}_{\text{2}}\left(l\right)+\_{\text{NH}}_{\text{3}}\left(g\right)\to \_{\text{S}}_{\text{4}}{\text{N}}_{\text{4}}\left(s\right)+\_{\text{S}}_{\text{8}}\left(s\right)+\_{\text{NH}}_{\text{4}}\text{Cl}\left(s\right)$

Concept introduction:

Balancing is a hit and trial method where a smallest whole number coefficient is used. One element is balanced at a time on both side of the equation.

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: Translate the chemical statement into a skeleton equation. The chemical substances that undergo a change are termed as reactants and the chemical substances that are produced during the chemical change are termed as products. The reactants are specified on the left side of the yield arrow while the products are specified on the right side of the yield arrow. Put a blank before each formula while beginning the balancing process.

Step 2: 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 3: 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 4: 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 5: Check whether the chemical equation is balanced or not by counting the number of atoms of each element on both sides.

Step 6: Specify the states of matter of each chemical substance present in the balanced chemical equation. The table for the abbreviations used for each state is as follows:

$\begin{array}{cc}\text{State}& \text{Abbreviation}\\ \text{Solid}& s\\ \text{Liquid}& l\\ \text{Gas}& g\\ \begin{array}{l}\text{Aqueous}\\ \text{solution}\end{array}& aq\end{array}$

Answer to Problem 3.59P

The balanced chemical equation is:

${\text{6S}}_{\text{2}}{\text{Cl}}_{\text{2}}\left(l\right)+16{\text{NH}}_{\text{3}}\left(g\right)\to {\text{S}}_{\text{4}}{\text{N}}_{\text{4}}\left(s\right)+{\text{S}}_{\text{8}}\left(s\right)+12{\text{NH}}_{\text{4}}\text{Cl}\left(s\right)$

Explanation of Solution

The chemical equation to be balanced is:

$\_{\text{S}}_{\text{2}}{\text{Cl}}_{\text{2}}\left(l\right)+\_{\text{NH}}_{\text{3}}\left(g\right)\to \_{\text{S}}_{\text{4}}{\text{N}}_{\text{4}}\left(s\right)+\_{\text{S}}_{\text{8}}\left(s\right)+\_{\text{NH}}_{\text{4}}\text{Cl}\left(s\right)$        (11)

The $\text{12 S}$ atoms on the right ($4$ in ${\text{S}}_{\text{4}}{\text{N}}_{\text{4}}$ and $8$ in ${\text{S}}_{\text{8}}$) require a coefficient of $6$ in front of ${\text{S}}_{\text{2}}{\text{Cl}}_{\text{2}}$ on the left.

Rewrite the chemical equation (11):

${\text{6S}}_{\text{2}}{\text{Cl}}_{\text{2}}\left(l\right)+\_{\text{NH}}_{\text{3}}\left(g\right)\to \_{\text{S}}_{\text{4}}{\text{N}}_{\text{4}}\left(s\right)+\_{\text{S}}_{\text{8}}\left(s\right)+\_{\text{NH}}_{\text{4}}\text{Cl}\left(s\right)$        (12)

The $\text{12 C}$ atoms in ${\text{6 S}}_{\text{2}}{\text{Cl}}_{\text{2}}$ on the left require a coefficient of $12$ in front of ${\text{NH}}_{\text{4}}\text{Cl}$ on the right.

Rewrite the chemical equation (12):

${\text{6S}}_{\text{2}}{\text{Cl}}_{\text{2}}\left(l\right)+\_{\text{NH}}_{\text{3}}\left(g\right)\to \_{\text{S}}_{\text{4}}{\text{N}}_{\text{4}}\left(s\right)+\_{\text{S}}_{\text{8}}\left(s\right)+12{\text{NH}}_{\text{4}}\text{Cl}\left(s\right)$        (13)

The $\text{16 N}$ atoms on the right ($4$ in ${\text{S}}_{\text{4}}{\text{N}}_{\text{4}}$ and $12$ in ${\text{NH}}_{\text{4}}\text{Cl}$) require a coefficient of $16$ in front of ${\text{NH}}_{\text{3}}$ on the left.

Rewrite the chemical equation (13):

${\text{6S}}_{\text{2}}{\text{Cl}}_{\text{2}}\left(l\right)+16{\text{NH}}_{\text{3}}\left(g\right)\to {\text{S}}_{\text{4}}{\text{N}}_{\text{4}}\left(s\right)+{\text{S}}_{\text{8}}\left(s\right)+12{\text{NH}}_{\text{4}}\text{Cl}\left(s\right)$

Conclusion

The balanced chemical equation is:

${\text{6S}}_{\text{2}}{\text{Cl}}_{\text{2}}\left(l\right)+16{\text{NH}}_{\text{3}}\left(g\right)\to {\text{S}}_{\text{4}}{\text{N}}_{\text{4}}\left(s\right)+{\text{S}}_{\text{8}}\left(s\right)+12{\text{NH}}_{\text{4}}\text{Cl}\left(s\right)$