Chapter 3, Problem 3.119QP

(a)

Interpretation Introduction

Interpretation:

The produced mass of ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ should be calculated.

Concept introduction:

Mole:

Number of atoms present in gram atomic mass of element is known as Avogadro number.

Avogadro number is $6.022136\times {10}^{23}$

One mole equal of atom equal to Avogadro number $(6.022136\times {10}^{23})$ hence, 1 mole of Iron $\text{(55}\text{.9 g)}$ contain atom $6.022136\times {10}^{23}$ $\text{Fe}$ atoms.

The mole of taken gram mass of compound is given by ratio between taken mass of compound to molar mass of compound.

$\text{Mole}\text{=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Answer to Problem 3.119QP

The produced mass of ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\text{12}\text{g}$

Explanation of Solution

To determine the produced mass of ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$

The given reaction is,

$\text{2}\text{KOH+}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\stackrel{}{\to }{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{ + 2H}}_{\text{2}}\text{O}$

Given,

Mass of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\text{25 g}$

Mass of $\text{KOH}$ is $\text{7}\text{.7}\text{g}$

To calculate the mass of ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ that can be produced respect with $\text{25 g}$ of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is,

Molar mass of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\text{44}\text{.02 mg}$

Molar mass of ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\text{80}\text{.05 mg}$

$\begin{array}{l}\text{=}\text{25}\text{g}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{×}\frac{\text{1}\text{mole}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{98}\text{.072}\text{g}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}\text{×}\frac{\text{1}\text{mole}{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{1}\text{mole}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}\text{×}\frac{\text{174}\text{.26}\text{g}{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{1}\text{mmole}{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}}\\ \text{=}\text{44}\text{.0}\text{g}{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}\end{array}$

The molar masses of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ and ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$, mole ratio of reactant require to produce ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ plugged in above equation to give the amount of ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ produced in the reaction.

The produced mass of ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ $\text{44}\text{.0}\text{g}$

To calculate the mass of ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ that can be produced respect with $\text{7}\text{.7 g}$ of $\text{KOH}$ is,

Molar mass of $\text{KOH}$ is $\text{56}\text{.108 g}$

$\begin{array}{l}\text{=}\text{7}\text{.7}\text{g}\text{KOH}\text{×}\frac{\text{1}\text{mole}\text{KOH}}{\text{56}\text{.108}\text{g}\text{KOH}}\text{×}\frac{\text{1}\text{mole}{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{2}\text{mole}\text{KOH}}\text{×}\frac{\text{174}\text{.26}\text{g}{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{1}\text{mmole}{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}}\\ \text{=}\text{11}\text{.96}\text{g}{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}\end{array}$

The molar masses of $\text{KOH}$ and ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$, mole ratio of reactant require to produce ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ are plugged in above equation to give the amount of ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ produced in the reaction.

The produced mass of ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ $\text{12}\text{.0}\text{g}$

Form the above calculations, the mass of ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ produced respect with $\text{KOH}$ is lower than ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ so $\text{KOH}$ is the limiting reagent.

Hence, ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ produced $\text{12}\text{.0}\text{g}$

Conclusion

The produced mass of ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ was calculated.

(b)

Interpretation Introduction

Interpretation:

After completion of reaction, the excess mass of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ should be calculated.

Concept introduction:

Mole:

Number of atoms present in gram atomic mass of element is known as Avogadro number.

Avogadro number is $6.022136\times {10}^{23}$

One mole equal of atom equal to Avogadro number $(6.022136\times {10}^{23})$ hence, 1 mole of Iron $\text{(55}\text{.9 g)}$ contain atom $6.022136\times {10}^{23}$ $\text{Fe}$ atoms.

The mole of taken gram mass of compound is given by ratio between taken mass of compound to molar mass of compound.

$\text{Mole}\text{=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Answer to Problem 3.119QP

Excess mass of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$, after completion of reaction is $\text{18}\text{g}$

Explanation of Solution

To determine the excess mass of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$, after completion of reaction.

The given reaction is,

$\text{2}\text{KOH+}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\stackrel{}{\to }{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{ + 2H}}_{\text{2}}\text{O}$

Given,

Mass of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\text{25 g}$

Mass of $\text{KOH}$ is $\text{7}\text{.7 g}$

To calculate the mass of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ that react with $\text{7}\text{.7 g}$ of $\text{KOH}$ is,

Molar mass of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\text{44}\text{.02 mg}$

Molar mass of $\text{KOH}$ is $\text{56}\text{.108 g}$

$\begin{array}{l}\text{=}\text{7}\text{.7}\text{g}\text{KOH}\text{×}\frac{\text{1}\text{mole}\text{KOH}}{\text{56}\text{.108}\text{KOH}}\text{×}\frac{\text{1}\text{mole}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{2}\text{mole}\text{KOH}}\text{×}\frac{\text{98}\text{.072}\text{g}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{1}\text{mole}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}\\ \text{=}\text{6}\text{.73}\text{g}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\end{array}$

The molar masses of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ and $\text{KOH}$, mole ratio of reactant require to produce $\text{KOH}$ plugged in above equation to give the mass of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ that react with $\text{7}\text{.7 g}$ of $\text{KOH}$.

The calculated reacted mass of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is subtract from taken mass to give the excess mass of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$, after completion of reaction.

$\begin{array}{l}\text{=}\text{25}\text{g-}\text{6}\text{.73}\text{g}\\ \text{=}\text{18}\text{g}\end{array}$

Excess mass of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$, after completion of reaction is $\text{18}\text{g}$

Conclusion

After completion of reaction, the excess mass of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ was calculated.

(c)

Interpretation Introduction

Interpretation:

The theoretical yield of give reaction should be calculated.

Concept introduction:

Percentage yield:

The percentage ratio between actual yields of the reaction to calculated theoretical yield of the reaction is known as percentage yield.

Percentage yield gives the information about the efficiency of the reaction.

$\text{percentage yield }\text{=}\frac{\text{Actual}\text{yield}}{\text{Theoretical}\text{yield}}\text{×100}$

Answer to Problem 3.119QP

The theoretical yield of the given reaction is $\text{8}\text{.0}\text{g}$

Explanation of Solution

To determine the produced mass of ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$

The given reaction is,

$\text{2}\text{KOH+}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\stackrel{}{\to }{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{ + 2H}}_{\text{2}}\text{O}$

Given,

Mass of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\text{25 g}$

Mass of $\text{KOH}$ is $\text{7}\text{.7}\text{g}$

To calculate the mass of ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ that can be produced respect with $\text{25 g}$ of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is,

Molar mass of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\text{44}\text{.02 mg}$

Molar mass of ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\text{80}\text{.05 mg}$

$\begin{array}{l}\text{=}\text{25}\text{g}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{×}\frac{\text{1}\text{mole}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{98}\text{.072}\text{g}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}\text{×}\frac{\text{1}\text{mole}{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{1}\text{mole}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}\text{×}\frac{\text{174}\text{.26}\text{g}{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{1}\text{mmole}{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}}\\ \text{=}\text{44}\text{.0}\text{g}{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}\end{array}$

The molar masses of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ and ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$, mole ratio of reactant require to produce ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ plugged in above equation to give the amount of ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ produced in the reaction.

The produced mass of ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ $\text{44}\text{.0}\text{g}$

To calculate the mass of ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ that can be produced respect with $\text{7}\text{.7 g}$ of $\text{KOH}$ is,

Molar mass of $\text{KOH}$ is $\text{56}\text{.108 g}$

$\begin{array}{l}\text{=}\text{7}\text{.7}\text{g}\text{KOH}\text{×}\frac{\text{1}\text{mole}\text{KOH}}{\text{56}\text{.108}\text{g}\text{KOH}}\text{×}\frac{\text{1}\text{mole}{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{2}\text{mole}\text{KOH}}\text{×}\frac{\text{174}\text{.26}\text{g}{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{1}\text{mmole}{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}}\\ \text{=}\text{11}\text{.96}\text{g}{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}\end{array}$

The molar masses of $\text{KOH}$ and ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$, mole ratio of reactant require to produce ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ are plugged in above equation to give the amount of ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ produced in the reaction.

The produced mass of ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ $\text{12}\text{.0}\text{g}$

Form the above calculations, the mass of ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ produced respect with $\text{KOH}$ is lower than ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ so $\text{KOH}$ is the limiting reagent.

Hence, ${\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ produced $\text{12}\text{.0}\text{g}$

To calculate the theoretical yield of given reaction.

Given,

Percentage yield is $\text{67}\text{.1}\text{g}$

$\begin{array}{l}\text{=}\text{12}\text{g}\text{×0}\text{.671}\\ \text{=}\text{8}\text{.0}\text{g}\end{array}$

The theoretical yield of the given reaction is $\text{8}\text{.0}\text{g}$

Conclusion

The theoretical yield of the given reaction was calculated.