Chapter 12, Problem 12.21P

Interpretation Introduction

Interpretation:

Mass of ${\text{I}}_2$ that can be formed has to be calculated.

Concept Introduction:

The formula to evaluate molarity is given as follows:

  $M=\frac{n}{V}$

Here,

$M$ denotes molarity.

$n$ denotes number of moles.

$V$ denotes volume.

Answer to Problem 12.21P

Mass of ${\text{I}}_2$ that can be formed is $55.52\text{ g}$.

Explanation of Solution

The conversion factor to convert liters to milliliters is as follows:

  $\text{1 L}=\text{1000 mL}$

Hence convert $\text{250 mL}$ to $\text{L}$ as follows:

  $\begin{array}{c}\text{Volume}=\left(\text{250 mL}\right)\left(\frac{1\text{ L}}{\text{1000 mL}}\right)\\ =0.25\text{ L}\end{array}$

The formula to evaluate $n$ from molarity is given as follows:

  $n=MV$        (1)

Substitute $0.25\text{ L}$ for $V$ and $\text{1}\text{.75 mol/L}$ for $M$ in equation (2).

  $\begin{array}{c}n=\left(0.25\text{ L}\right)\left(\text{1}\text{.75 mol}/\text{L}\right)\\ =0.4375\text{ mol}\end{array}$

The formula to convert mass in gram to moles is as follows:

  $\text{Number of moles}=\frac{\text{Given mass}}{\text{molar mass}}$        (3)

Substitute $\text{20 g}$ for mass and $\text{86}\text{.9368 g/mol}$ for molar mass to calculate the moles of ${\text{MnO}}_2$ in equation (3).

  $\begin{array}{c}\text{Number of moles}=\frac{\text{20 g}}{\text{86}\text{.9368 g/mol}}\\ =0.23005\text{ mol}\end{array}$

Substitute $\text{75 g}$ for mass and $\text{166}\text{.0028 g/mol}$ for molar mass of $\text{KI}$ to calculate the moles of $\text{KI}$ in equation (3).

  $\begin{array}{c}\text{Number of moles}=\frac{\text{75 g}}{\text{166}\text{.0028 g/mol}}\\ =0.451799\text{ mol}\end{array}$

The balanced chemical reaction is given as follows:

  ${\text{MnO}}_2\left(s\right)+2\text{KI}(s)+2{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}(aq)\to {\text{MnSO}}_{\text{4}}\left(aq\right)+{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)+{\text{I}}_2\left(s\right)$

According to the stoichiometry above balanced equation, one mole of ${\text{I}}_2$ is produced from ${\text{1 mol MnO}}_2$. Hence moles of ${\text{I}}_2$ produced from $0.0375\text{ mol HCl}$ is calculated as follows:

  $\begin{array}{c}{\text{Moles of I}}_2=\left(0.23005{\text{ mol I}}_2\right)\left(\frac{1{\text{ mol I}}_2}{{\text{1 mol MnO}}_2}\right)\\ =0.23005{\text{ mol I}}_2\end{array}$

Similarly moles of ${\text{I}}_2$ produced from $0.451799\text{ mol KI}$ is calculated as follows:

  $\begin{array}{c}{\text{Moles of I}}_2=\left(0.451799\text{ mol KI}\right)\left(\frac{{\text{1 mol I}}_2}{2\text{ mol KI }}\right)\\ =0.22589{\text{ mol I}}_2\end{array}$

Likewise, moles of ${\text{I}}_2$ produced from $0.4375{\text{ mol H}}_2{\text{SO}}_4$ is calculated as follows:

  $\begin{array}{c}{\text{Moles of I}}_2=\left(0.4375{\text{ mol H}}_2{\text{SO}}_4\right)\left(\frac{1{\text{ mol I}}_2}{{\text{2 mol H}}_2{\text{SO}}_4}\right)\\ =0.21875{\text{ mol I}}_2\end{array}$

As ${\text{H}}_2{\text{SO}}_4$ produces less moles hence it must be limiting reagent and moles of ${\text{I}}_2$ produced from ${\text{H}}_2{\text{SO}}_4$ is $0.21875\text{ mol}$.

The formula to convert mass in gram to moles is as follows:

  $\text{Mass}=\left(\text{Number of moles}\right)\left(\text{Molar mass}\right)$        (5)

Substitute $0.21875\text{ mol}$ for moles and $\text{253}\text{.8089 g/mol}$ for molar mass of ${\text{I}}_2$ in equation (5).

  $\begin{array}{c}\text{Mass}=\left(0.21875\text{ mol}\right)\left(\text{253}\text{.8089 g/mol}\right)\\ =55.52\text{ g}\end{array}$