Chapter 15, Problem 54A

(a)

Interpretation Introduction

Interpretation: The number of moles of chloride ions and number of moles of iron ions has to be calculated

Concept Introduction: Molarity: Molarity is defined as the number of moles of solute in one liter of solution. Molarity is the preferred concentration unit for stoichiometry calculations. The formula is,

$\text{Molarity=}\frac{\text{Mass}\text{of}\text{solute(in}\text{moles)}}{\text{Volume}\text{of}\text{solution}\text{(in}\text{litres)}}$

Answer to Problem 54A

The number of moles of iron ions is $\text{3}\text{.00}\text{mol}{\text{Fe}}^{\text{3+}}\text{.}$

The number of moles of chloride ions is $\text{9}\text{.00}\text{mol}{\text{Cl}}^{\text{-}}\text{.}$

Explanation of Solution

The molarity of iron (III) chloride is $\text{2}\text{.00}\text{M}\text{.}$

The volume of iron (III) chloride is $\text{1}\text{.50}\text{L}\text{.}$

The molarity of ${\text{FeCl}}_{\text{3}}$ is calculated by writing a proportion, then cross-multiplying and dividing.

$\begin{array}{l}\frac{\text{n}\text{mol}}{\text{1}\text{.50}\text{L}}\text{=}\frac{\text{2}\text{.00}\text{mol}}{\text{1}\text{L}}\\ \text{n}\left(\text{mol×L}\right)\text{=3}\text{.00}\left(\text{mol×L}\right)\\ \text{n=3}\text{.00}\end{array}$

The solution contains 3.00 moles of ${\text{FeCl}}_{\text{3}}.$ Each molecule contains one iron ion and three chloride ions, so the solution contains $\text{1×3}\text{.00}\text{mol}{\text{Fe}}^{\text{3+}}\text{=3}\text{.00}\text{mol}{\text{Fe}}^{\text{3+}}$ and $\text{3×3}\text{.00}\text{mol}{\text{Cl}}^{\text{-}}\text{=9}\text{.00}\text{mol}{\text{Cl}}^{\text{-}}\text{.}$

The number of moles of iron ions is $\text{3}\text{.00}\text{mol}{\text{Fe}}^{\text{3+}}\text{.}$

The number of moles of chloride ions is $\text{9}\text{.00}\text{mol}{\text{Cl}}^{\text{-}}\text{.}$

(b)

Interpretation Introduction

Interpretation: The mass of solid product formed (in grams) has to be determined.

Concept Introduction : Molarity: Molarity is defined as the number of moles of solute in one liter of solution. Molarity is the preferred concentration unit for stoichiometry calculations. The formula is,

$\text{Molarity=}\frac{\text{Mass}\text{of}\text{solute(in}\text{moles)}}{\text{Volume}\text{of}\text{solution}\text{(in}\text{litres)}}$

Answer to Problem 54A

The mass of solid product formed (in grams) is $\text{556}\text{.2}\text{g}\text{.}$

Explanation of Solution

Iron (III) chloride contains ${\text{Fe}}^{3+}$ and ${\text{Cl}}^{-}.$ When $\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is dissolved in this solution, ${\text{Pb}}^{2+}$ and ${\text{NO}}_3{}^{-}$ ions are added. The ${\text{Pb}}^{2+}$ and ${\text{Cl}}^{-}$ ions react to form solid ${\text{PbCl}}_{\text{2}}\text{.}$ The net ionic equation is,

${\text{Pb}}^{2+}\left(aq\right)+{\text{Cl}}^{-}\to {\text{PbCl}}_{\text{2}}\left(s\right)$

The limiting reagent is identified by calculating the moles of ${\text{Pb}}^{2+}$ and ${\text{Cl}}^{-}.$

$\text{4}\text{.00}\text{M}$ of lead (II) nitrate contains $\text{4}\text{.00}\text{M}{\text{Pb}}^{2+}$ ions in $\text{0}\text{.500}\text{L}$ of solution. The moles of ${\text{Pb}}^{2+}$ ions is,

Moles= $\text{0}\text{.500}\text{L×}\frac{\text{4}\text{.00}\text{mol}{\text{Pb}}^{\text{2+}}}{\text{1}\text{L}}$

Moles= $\text{2}\text{mol}{\text{Pb}}^{\text{2+}}$

The moles of lead (II) ions is $\text{2}\text{mol}\text{.}$

$\text{2}\text{.00}\text{M}$ of iron (III) chloride contains $\text{2}\text{.00}\text{M}{\text{Cl}}^{-}$ ions in $\text{1}\text{.50}\text{L}$ of solution. The moles of ${\text{Cl}}^{-}$ ions is,

$\begin{array}{l}\text{Moles=1}\text{.50}\text{L×}\frac{\text{2}\text{.00}\text{mol}{\text{Cl}}^{\text{-}}}{\text{L}}\\ \text{Moles=3}\text{mol}{\text{Cl}}^{\text{-}}\end{array}$

The solution contains $\text{2×3}\text{.00}\text{mol=6}\text{mol}{\text{Cl}}^{\text{-}}.$

The moles of chloride ions is $\text{6}\text{mol}\text{.}$

The chloride ions are present in excess, so only 2.00 moles of solid ${\text{PbCl}}_{\text{2}}$ will be formed.

The molar mass of lead (II) chloride is $\text{278}\text{.1}\text{g/mol}\text{.}$

Mass of solid lead (II) chloride is calculated as,

Mass of solid lead (II) chloride= $\text{2}\text{.00}\text{mol}{\text{PbCl}}_{\text{2}}\text{×}\frac{\text{278}\text{.1}\text{g}{\text{PbCl}}_{\text{2}}}{\text{1}\text{mol}{\text{PbCl}}_{\text{2}}}$

Mass of solid lead (II) chloride= $\text{556}\text{.2}\text{g}$

The mass of solid product formed (in grams) is $\text{556}\text{.2}\text{g}\text{.}$