Chapter 3, Problem 3.104P

(a)

Interpretation Introduction

Interpretation:

A balanced chemical equation for a reaction between hydrogen sulphide and oxygen to form sulphur dioxide and water vapour is to be written.

Concept introduction:

A balanced chemical equation obeys the law of conservation of mass since the total mass of reactants and products are equal in a balanced chemical equation.

Following are the steps to write a balanced chemical equation.

Step 1: Translate the chemical statement into a skeleton equation. The reactants are the chemical substances that undergo a change, thus, write the reactants on the left side of the yield arrow. The products are the chemical substances that are produced during the chemical change, thus, write the products 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/elements 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/elements 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, adjust 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.

Answer to Problem 3.104P

A balanced chemical equation is as follows:

  ${\text{2H}}_{\text{2}}\text{S}\left(g\right)+3{\text{O}}_2\left(g\right)\stackrel{\Delta }{\to }{\text{2SO}}_{\text{2}}\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(g\right)$

Explanation of Solution

The chemical formulas for hydrogen sulphide and oxygen are ${\text{H}}_{\text{2}}\text{S}$ and ${\text{O}}_{\text{2}}$ respectively and chemical formulas for sulphur dioxide and water vapour are ${\text{SO}}_{\text{2}}$ and ${\text{H}}_{\text{2}}\text{O}$ respectively.

The chemical equation corresponding to the given reaction between hydrogen sulphide and oxygen to form sulphur dioxide and water vapour is :

  ${\text{H}}_{\text{2}}\text{S}\left(g\right)+{\text{O}}_2\left(g\right)\stackrel{\Delta }{\to }{\text{SO}}_{\text{2}}\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)$

Observe there are 2 O atoms in ${\text{O}}_{\text{2}}$ on the left side of the equation and 3 O atoms in ${\text{SO}}_{\text{2}}$ and ${\text{H}}_{\text{2}}\text{O}$ on the right side; so write 2 as the coefficient of ${\text{H}}_{\text{2}}\text{O}$ on the right and also a coefficient of 2 in front of ${\text{O}}_{\text{2}}$ on the left side of the equation. Now this equation has a total of 4 oxygen atoms on each side and the equation becomes:

  ${\text{H}}_{\text{2}}\text{S}\left(g\right)+2{\text{O}}_2\left(g\right)\stackrel{\Delta }{\to }{\text{SO}}_{\text{2}}\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(g\right)$

There are 4H atoms in $2{\text{H}}_{\text{2}}\text{O}$ on the right side so put 2 as a coefficient in front of ${\text{H}}_{\text{2}}\text{S}$ on the left and the equation would become:

  ${\text{2H}}_{\text{2}}\text{S}\left(g\right)+2{\text{O}}_2\left(g\right)\stackrel{\Delta }{\to }{\text{SO}}_{\text{2}}\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(g\right)$

There are 2S atoms as in ${\text{2H}}_{\text{2}}\text{S}$ so a coefficient of 2 is again needed in front of ${\text{SO}}_{\text{2}}$ on the right side of the equation and the equation becomes:

  ${\text{2H}}_{\text{2}}\text{S}\left(g\right)+2{\text{O}}_2\left(g\right)\stackrel{\Delta }{\to }{\text{2SO}}_{\text{2}}\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(g\right)$

Now O atoms on both sides of the equation are no longer equal. There is a total of 6 O atoms on the right ( 4 in ${\text{2SO}}_{\text{2}}$ and 2 in $2{\text{H}}_{\text{2}}\text{O}$) and since there are only 4 ${\text{O}}_2$ on the left side so change the coefficient from 2 to 3 in front of ${\text{O}}_2$.

  ${\text{2H}}_{\text{2}}\text{S}\left(g\right)+3{\text{O}}_2\left(g\right)\stackrel{\Delta }{\to }{\text{2SO}}_{\text{2}}\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(g\right)$

Check whether the equation is balanced or not as follows:

  Reactants $\left(\text{4 H, 2S, 6O}\right)$$\to$ Products $\left(\text{4 H, 2S, 6O}\right)$

Conclusion

A balanced chemical equation always follows the Law of conservation of mass.

(b)

Interpretation Introduction

Interpretation:

A balanced chemical equation for a reaction when potassium chlorate is heated forming potassium chloride and potassium perchlorate is to be written.

Concept introduction:

A balanced chemical equation obeys the law of conservation of mass since the total mass of reactants and products are equal in a balanced chemical equation.

Following are the steps to write a balanced chemical equation.

Step 1: Translate the chemical statement into a skeleton equation. The reactants are the chemical substances that undergo a change, thus, write the reactants on the left side of the yield arrow. The products are the chemical substances that are produced during the chemical change, thus, write the products 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/elements 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/elements 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, adjust 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.

Answer to Problem 3.104P

The balanced chemical equation is as follows:

  ${\text{4KClO}}_{\text{3}}\left(s\right)\stackrel{\Delta }{\to }\text{KCl}\left(s\right)+3{\text{KClO}}_{\text{4}}\left(s\right)$

Explanation of Solution

The chemical formulas for potassium chlorate, potassium chloride, and potassium perchlorate are ${\text{KClO}}_3$, $\text{KCl}$ and  ${\text{KClO}}_{\text{4}}$ respectively.

The chemical equation corresponding to the given reaction when potassium chlorate  is heated forming potassium chloride and potassium perchlorate is:

  ${\text{KClO}}_{\text{3}}\left(s\right)\stackrel{\Delta }{\to }\text{KCl}\left(s\right)+{\text{KClO}}_{\text{4}}\left(s\right)$

Observe that there are 3 O atoms in ${\text{KClO}}_{\text{3}}$ on the left side of the equation and 4 O atoms in ${\text{KClO}}_{\text{3}}$ on the right side of it, so place 4 as a coefficient in front of ${\text{KClO}}_{\text{3}}$ and 3 as a coefficient in front of ${\text{KClO}}_{\text{4}}$ and the equation becomes:

  ${\text{4KClO}}_{\text{3}}\left(s\right)\stackrel{\Delta }{\to }\text{KCl}\left(s\right)+3{\text{KClO}}_{\text{4}}\left(s\right)$

Now observe that there are 12 O atoms on both sides of the equation now. Also, there are 4 $\text{Cl}$ and 4 $\text{K}$ atoms on both sides of the equation hence it is balanced now.

Conclusion

A balanced chemical equation always follows the Law of conservation of mass.

(c)

Interpretation Introduction

Interpretation:

A balanced chemical equation for a reaction between hydrogen gas and $\text{iron}\left(\text{III}\right)\text{oxide}$ to form iron metal and water vapor is to be written.

Concept introduction:

A balanced chemical equation obeys the law of conservation of mass since the total mass of reactants and products are equal in a balanced chemical equation.

Following are the steps to write a balanced chemical equation.

Step 1: Translate the chemical statement into a skeleton equation. The reactants are the chemical substances that undergo a change, thus, write the reactants on the left side of the yield arrow. The products are the chemical substances that are produced during the chemical change, thus, write the products 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/elements 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/elements 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, adjust 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.

Answer to Problem 3.104P

The balanced chemical equation is as follows:

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

Explanation of Solution

The chemical formulas for hydrogen gas, $\text{iron}\left(\text{III}\right)\text{oxide}$, iron metal and water vapor are ${\text{H}}_{\text{2}}$, ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$, $\text{Fe}$, ${\text{H}}_{\text{2}}\text{O}$ respectively.

The chemical equation corresponding to the given reaction is:

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

Put a coefficient of 2 in front of iron metal on the left side of the equation as there are 2 $\text{Fe}$ atoms in ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ and the equation would become:

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

Next, observe that there are 3 oxygen atoms on the left so place a coefficient of 3 in front of ${\text{H}}_{\text{2}}\text{O}$ on the right side and the equation would become:

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

Now there are 6 hydrogen atoms on the right so place a coefficient of 3 in front of ${\text{H}}_{\text{2}}$ on the left side in order to make equal hydrogen atoms on both sides and the equation would appear as:

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

Now check whether there is an equal number of each of the three atoms on both sides:

  Reactants $\left(\text{6H, 2Fe, 3O}\right)$$\to$ Products $\left(\text{6H, 2Fe, 3O}\right)$

Conclusion

A balanced chemical equation always follows the Law of conservation of mass.

(d)

Interpretation Introduction

Interpretation:

The chemical equation for the combustion reaction of gaseous ethane to form carbon dioxide and water vapor is to be balanced.

Concept introduction:

A balanced chemical equation obeys the law of conservation of mass since the total

mass of reactants and products are equal in a balanced chemical equation.

Following are the steps to write a balanced chemical equation.

Step 1: Translate the chemical statement into a skeleton equation. The reactants are the chemical substances that undergo a change, thus, write the reactants on the left side of the yield arrow. The products are the chemical substances that are produced during the chemical change, thus, write the products 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/elements 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/elements 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, adjust 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.

Answer to Problem 3.104P

The balanced chemical equation is as follows:

  ${\text{2C}}_{\text{2}}{\text{H}}_{\text{6}}\left(g\right)+7{\text{O}}_{\text{2}}\left(g\right)\stackrel{\Delta }{\to }{\text{4CO}}_{\text{2}}\left(g\right)+6{\text{H}}_{\text{2}}\text{O}\left(g\right)$

Explanation of Solution

The chemical formulas for ethane, oxygen, carbon dioxide, and water vapor would be ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}$, ${\text{O}}_{\text{2}}$, ${\text{CO}}_{\text{2}}$, ${\text{H}}_{\text{2}}\text{O}$ respectively.

First write the reaction corresponding to the given combustion reaction as:

  ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)\stackrel{\Delta }{\to }{\text{CO}}_{\text{2}}\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)$

There are two carbon atoms on the left side so place 2 as the coefficient for ${\text{CO}}_{\text{2}}$ in the left side of this equation:

  ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)\stackrel{\Delta }{\to }{\text{2CO}}_{\text{2}}\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(g\right)$

Next, observe that there are 6 hydrogen atoms so place 3 as a coefficient in front of ${\text{H}}_{\text{2}}\text{O}$ on the left side and equation becomes:

  ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\left(g\right)+{\text{O}}_{\text{2}}\left(g\right)\stackrel{\Delta }{\to }{\text{2CO}}_{\text{2}}\left(g\right)+3{\text{H}}_{\text{2}}\text{O}\left(g\right)$

Now in the right side, there are a total of 7 oxygen atoms (4 in ${\text{2CO}}_{\text{2}}$ and 3 in $3{\text{H}}_{\text{2}}\text{O}$) so since there are already 2 oxygen atoms on left side so left side requires a coefficient of $\text{7/2}$ and the equation will then have seven oxygen atoms on each side as:

  ${\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\left(g\right)+\frac{7}{2}{\text{O}}_{\text{2}}\left(g\right)\stackrel{\Delta }{\to }{\text{4CO}}_{\text{2}}\left(g\right)+6{\text{H}}_{\text{2}}\text{O}\left(g\right)$        (1)

Finally in order to have only whole numbers as coefficients multiply both sides of equation (1) by 2 and equation would be:

  ${\text{2C}}_{\text{2}}{\text{H}}_{\text{6}}\left(g\right)+7{\text{O}}_{\text{2}}\left(g\right)\stackrel{\Delta }{\to }{\text{4CO}}_{\text{2}}\left(g\right)+6{\text{H}}_{\text{2}}\text{O}\left(g\right)$

Conclusion

A balanced chemical equation always follows the Law of conservation of mass.

(e)

Interpretation Introduction

Interpretation:

The chemical equation for the reaction when $\text{iron}\left(\text{II}\right)\text{chloride}$ is treated with chlorine trifluoride gas to form $\text{iron}\left(\text{III}\right)\text{fluoride}$ is to be balanced.

Concept introduction:

A balanced chemical equation obeys the law of conservation of mass since the total

mass of reactants and products are equal in a balanced chemical equation.

Following are the steps to write a balanced chemical equation.

Step 1: Translate the chemical statement into a skeleton equation. The reactants are the chemical substances that undergo a change, thus, write the reactants on the left side of the yield arrow. The products are the chemical substances that are produced during the chemical change, thus, write the products 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/elements 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/elements 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, adjust 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.

Answer to Problem 3.104P

The balanced chemical equation is as follows:

  ${\text{2FeCl}}_{\text{2}}\left(s\right)+2{\text{ClF}}_{\text{3}}\left(g\right)\to 2{\text{FeF}}_{\text{3}}\left(s\right)+3{\text{Cl}}_{\text{2}}\left(g\right)$

Explanation of Solution

The chemical formulas for $\text{iron}\left(\text{II}\right)\text{chloride}$, chlorine trifluoride $\text{iron}\left(\text{III}\right)\text{fluoride}$ and chlorine gas are ${\text{FeCl}}_{\text{2}}$, ${\text{ClF}}_{\text{3}}$, ${\text{FeF}}_{\text{3}}$, ${\text{Cl}}_{\text{2}}$ respectively.

First, write the reaction corresponding to the given reaction when $\text{iron}\left(\text{II}\right)\text{chloride}$ is treated with chlorine trifluoride gas to form $\text{iron}\left(\text{III}\right)\text{fluoride}$ as follows:

  ${\text{FeCl}}_{\text{2}}\left(s\right)+{\text{ClF}}_{\text{3}}\left(g\right)\to {\text{FeF}}_{\text{3}}\left(s\right)+{\text{Cl}}_{\text{2}}\left(g\right)$

Observe that there are 3 $\text{Cl}$ atoms on the left side and 2 $\text{Cl}$ atoms on the right side in ${\text{Cl}}_{\text{2}}$ molecule thus a coefficient of 2 is required before ${\text{Cl}}_{\text{2}}$ and also the coefficient of 2 before ${\text{ClF}}_{\text{3}}$ on the left side so that the equation becomes:

  ${\text{FeCl}}_{\text{2}}\left(s\right)+2{\text{ClF}}_{\text{3}}\left(g\right)\to {\text{FeF}}_{\text{3}}\left(s\right)+2{\text{Cl}}_{\text{2}}\left(g\right)$

Next, observe that there are a total of 6 fluorine atoms (1 in ${\text{FeCl}}_{\text{2}}$ and 3 in ${\text{ClF}}_{\text{3}}$) on the right side of the equation so place a coefficient of 2 in front of ${\text{FeF}}_{\text{3}}$ on the right so that the equation now has 6 Fluorine atoms on the right side too.

  ${\text{FeCl}}_{\text{2}}\left(s\right)+2{\text{ClF}}_{\text{3}}\left(g\right)\to 2{\text{FeF}}_{\text{3}}\left(s\right)+2{\text{Cl}}_{\text{2}}\left(g\right)$

Now on the left side, 2 $\text{Fe}$ atoms are present and just one in the right side so place 2 as the coefficient of ${\text{FeCl}}_{\text{2}}$ on the left side and the equation would now become:

  ${\text{2FeCl}}_{\text{2}}\left(s\right)+2{\text{ClF}}_{\text{3}}\left(g\right)\to 2{\text{FeF}}_{\text{3}}\left(s\right)+2{\text{Cl}}_{\text{2}}\left(g\right)$

Now in order to have 6 chlorine atoms on the right side change the coefficient of ${\text{Cl}}_{\text{2}}$ on the right to 3 so that there are a total of 6 chlorine atoms on each side of the equation.

  ${\text{2FeCl}}_{\text{2}}\left(s\right)+2{\text{ClF}}_{\text{3}}\left(g\right)\to 2{\text{FeF}}_{\text{3}}\left(s\right)+3{\text{Cl}}_{\text{2}}\left(g\right)$

Finally check that there are 2$\text{Fe}$, 6 $\text{Cl}$ and 6 $\text{F}$ atoms on both sides of the above equation so now this equation has been balanced.

Conclusion

A balanced chemical equation always follows the Law of conservation of mass.