Chapter 12, Problem 12.69P

Interpretation Introduction

Interpretation:

Mass of $\text{Pb}{\left({\text{IO}}_3\right)}_2$ precipitated has to be calculated.

Concept Introduction:

The formula to evaluate moles from molarity is given as follows:

  $n=MV$

Here,

$M$ denotes molarity.

$n$ denotes number of moles.

$V$ denotes volume.

Answer to Problem 12.69P

Mass of $\text{Pb}{\left({\text{IO}}_3\right)}_2$ precipitated is $41.775\text{ g}$ .

Explanation of Solution

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

  $\text{1 L}=\text{1000 mL}$

Hence convert $\text{25 mL}$ to $\text{L}$ as follows:

  $\begin{array}{c}\text{Volume}=\left(\text{25 mL}\right)\left(\frac{1\text{ L}}{1000\text{ mL}}\right)\\ =\text{0}\text{.025 L}\end{array}$

Similarly, convert $\text{30 mL}$ to $\text{L}$ as follows:

  $\begin{array}{c}\text{Volume}=\left(\text{30 mL}\right)\left(\frac{1\text{ L}}{1000\text{ mL}}\right)\\ =\text{0}\text{.030 L}\end{array}$

The formula to evaluate moles from molarity is given as follows:

  $n=MV$        (1)

Substitute $\text{3 mol/L}$ for $M$ and $0.025\text{ L}$ for $V$ in equation (1) to evaluate moles of $\text{Pb}{\left({\text{NO}}_3\right)}_2$.

  $\begin{array}{c}n=\left(\text{3 mol/L}\right)\left(0.025\text{ L}\right)\\ =0.075\text{ mol}\end{array}$

Substitute $\text{5 mol/L}$ for $M$ and $0.030\text{ L}$ for $V$ in equation (1) to calculate moles of ${\text{KIO}}_3$.

  $\begin{array}{c}n=\left(\text{5 mol/L}\right)\left(0.030\text{ L}\right)\\ =0.15\text{ mol}\end{array}$

The balanced chemical reaction is given as follows:

  $\text{Pb}{\left({\text{NO}}_3\right)}_2\left(aq\right)+2{\text{KIO}}_3\left(aq\right)\to \text{Pb}{\left({\text{IO}}_3\right)}_2\left(s\right)+2{\text{KNO}}_3\left(aq\right)$        (2)

Moles of $\text{Pb}{\left({\text{NO}}_3\right)}_2$ is less and thus it acts as limiting reagent. According to the stoichiometry of equation (2), one mole of $\text{Pb}{\left({\text{NO}}_3\right)}_2$ can give $\text{1 mol Pb}{\left({\text{IO}}_3\right)}_2$. Thus moles of $\text{Pb}{\left({\text{IO}}_3\right)}_2$ formed from $0.075\text{ mol Pb}{\left({\text{NO}}_3\right)}_2$ is calculated as follows:

  $\begin{array}{c}\text{Moles of Pb}{\left({\text{IO}}_3\right)}_2=\left(0.075\text{ mol Pb}{\left({\text{NO}}_3\right)}_2\right)\left(\frac{\text{1 mol Pb}{\left({\text{IO}}_3\right)}_2}{\text{1 mol Pb}{\left({\text{NO}}_3\right)}_2}\right)\\ =0.075\text{ mol Pb}{\left({\text{IO}}_3\right)}_2\end{array}$

Similarly, $2{\text{ mol KIO}}_3$ gives $\text{1 mol Pb}{\left({\text{IO}}_3\right)}_2$. Thus moles of $\text{Pb}{\left({\text{IO}}_3\right)}_2$ formed from $0.15{\text{ mol KIO}}_3$ is calculated as follows:

  $\begin{array}{c}\text{Moles of Pb}{\left({\text{IO}}_3\right)}_2=\left(0.15{\text{ mol KIO}}_3\right)\left(\frac{\text{1 mol Pb}{\left({\text{IO}}_3\right)}_2}{{\text{2 mol KIO}}_3}\right)\\ =0.075\text{ mol Pb}{\left({\text{IO}}_3\right)}_2\end{array}$

Thus moles of $\text{Pb}{\left({\text{IO}}_3\right)}_2$ produced either from ${\text{KIO}}_3$ or $\text{Pb}{\left({\text{NO}}_3\right)}_2$ is $0.075\text{ mol}$.

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

  $\text{Mass}=\left(\text{Number of moles}\right)\left(\text{Molar mass}\right)$        (3)

Substitute $0.075\text{ mol}$ for moles and $\text{557 g/mol}$ for molar mass of $\text{Pb}{\left({\text{IO}}_3\right)}_2$ in equation (3).

  $\begin{array}{c}\text{Mass}=\left(0.075\text{ mol}\right)\left(\text{557 g/mol}\right)\\ =41.775\text{ g}\end{array}$