Chapter 3, Problem 92QRT

You mix 25.0 mL of 0.234-M FeCl₃ solution with 42.5 mL of 0.453-M NaOH.

1. (a) Calculate the maximum mass, in grams, of Fe(OH)₃ that will precipitate.
2. (b) Determine which reactant is in excess.
3. (c) Calculate the concentration of the excess reactant remaining in solution after the maximum mass of Fe(OH)₃ has precipitated.

Interpretation Introduction

Interpretation:

The maximum mass in grams of $\text{Fe}{\left(\text{OH}\right)}_{\text{3}}$ that will precipitate has to be calculated.

Concept introduction:

Limiting reagent:

A limiting reactant is a reactant that is completely converted to products. Once all the limiting reactant is converted to products there is no other reactant to react.

Number of moles can be calculated by using following formula,

  $\text{No}\text{.of Moles = }\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Explanation of Solution

The reaction is shown below,

  ${\text{FeCl}}_{\text{3}}\text{(aq) + 3 NaOH(aq) }\to {\text{Fe(OH)}}_{\text{3}}\text{(s) + 3 NaCl(aq)}$

  Mixing of $\text{25}\text{.0 mL of 0}{\text{.234-M FeCl}}_{\text{3}}$ solution with $\text{42}\text{.5 mL of 0}\text{.453-M NaOH}$.

According to the balanced equation, 1 moles of ${\text{FeCl}}_{\text{3}}$ reaction with 3 moles of $\text{NaOH}$ gives one mole of ${\text{Fe(OH)}}_{\text{3}}$ and 3 moles of sodium chloride. Therefore,

Moles of ${\text{Fe(OH)}}_{\text{3}}$ is calculated as follows,

    $\begin{array}{l}\text{Moles}\text{of}{\text{Fe(OH)}}_{\text{3}}\text{=}\text{0}\text{.025}\text{L}\text{of}{\text{FeCl}}_{\text{3}}\text{×}\frac{0.234\text{mol}{\text{FeCl}}_{\text{3}}}{1L{\text{FeCl}}_{\text{3}}}\text{×}\frac{\text{1}\text{mol}{\text{Fe(OH)}}_{\text{3}}}{1\text{mol}{\text{FeCl}}_{\text{3}}}\\ \text{Moles}\text{of}{\text{Fe(OH)}}_{\text{3}}\text{=}\text{5}\text{.85}\times {\text{10}}^{-3}\text{mol}\end{array}$

  $\begin{array}{l}\text{Moles}\text{of}{\text{Fe(OH)}}_{\text{3}}\text{=}\text{0}\text{.0425}\text{L}\text{of}\text{NaOH×}\frac{\text{0}\text{.453}\text{mol}\text{NaOH}}{\text{1L}\text{NaOH}}\text{×}\frac{\text{1}\text{mol}{\text{Fe(OH)}}_{\text{3}}}{\text{3}\text{mol}\text{NaOH}}\\ \text{Moles}\text{of}{\text{Fe(OH)}}_{\text{3}}\text{=}\text{6}{\text{.42×10}}^{\text{-3}}\text{mol}\end{array}$

According to the mole calculation, ${\text{FeCl}}_{\text{3}}$ is smaller than the Sodium hydroxide, therefore, ${\text{FeCl}}_{\text{3}}$ is the limiting reagent.

Molar Mass ${\text{Fe(OH)}}_{\text{3}}=\text{106}\text{.867 g/mol}$

The mass in grams of $\text{Fe}{\left(\text{OH}\right)}_{\text{3}}$ is calculated as follows,

  $\begin{array}{l}\text{Mass = No}\text{.of Moles}\times \text{Molar}\text{mass}\\ \text{Mass = 5}\text{.85}\times {\text{10}}^{-3}\text{mol}\times \text{106}\text{.867 g/mol }\\ \text{Mass = 0}\text{.625 g}\text{of}{\text{Fe(OH)}}_{\text{3}}\end{array}$

The mass in grams of $\text{Fe}{\left(\text{OH}\right)}_{\text{3}}$ is $\text{0}\text{.625 g}$.

Interpretation Introduction

Interpretation:

The excess reactant has to be identified.

Concept introduction:

Refer to part (a)

Explanation of Solution

The reaction is shown below,

  ${\text{FeCl}}_{\text{3}}\text{(aq) + 3 NaOH(aq) }\to {\text{Fe(OH)}}_{\text{3}}\text{(s) + 3 NaCl(aq)}$

  Mixing of $\text{25}\text{.0 mL of 0}{\text{.234-M FeCl}}_{\text{3}}$ solution with $\text{42}\text{.5 mL of 0}\text{.453-M NaOH}$.

According to the balanced equation, 1 moles of ${\text{FeCl}}_{\text{3}}$ reaction with 3 moles of $\text{NaOH}$ gives one mole of ${\text{Fe(OH)}}_{\text{3}}$ and 3 moles of sodium chloride. Therefore,

Moles of ${\text{Fe(OH)}}_{\text{3}}$ is calculated as follows,

    $\begin{array}{l}\text{Moles}\text{of}{\text{Fe(OH)}}_{\text{3}}\text{=}\text{0}\text{.025}\text{L}\text{of}{\text{FeCl}}_{\text{3}}\text{×}\frac{0.234\text{mol}{\text{FeCl}}_{\text{3}}}{1L{\text{FeCl}}_{\text{3}}}\text{×}\frac{\text{1}\text{mol}{\text{Fe(OH)}}_{\text{3}}}{1\text{mol}{\text{FeCl}}_{\text{3}}}\\ \text{Moles}\text{of}{\text{Fe(OH)}}_{\text{3}}\text{=}\text{5}\text{.85}\times {\text{10}}^{-3}\text{mol}\end{array}$

  $\begin{array}{l}\text{Moles}\text{of}{\text{Fe(OH)}}_{\text{3}}\text{=}\text{0}\text{.0425}\text{L}\text{of}\text{NaOH×}\frac{\text{0}\text{.453}\text{mol}\text{NaOH}}{\text{1L}\text{NaOH}}\text{×}\frac{\text{1}\text{mol}{\text{Fe(OH)}}_{\text{3}}}{\text{3}\text{mol}\text{NaOH}}\\ \text{Moles}\text{of}{\text{Fe(OH)}}_{\text{3}}\text{=}\text{6}{\text{.42×10}}^{\text{-3}}\text{mol}\end{array}$

According to the mole calculation, ${\text{FeCl}}_{\text{3}}$ is smaller than the Sodium hydroxide, therefore, ${\text{FeCl}}_{\text{3}}$ is the limiting reagent. Hence, sodium hydroxide is excess reactant.

Interpretation Introduction

Interpretation:

The concentration of excess reactant remaining in solution after the precipitation of ${\text{Fe(OH)}}_{\text{3}}$ has to be calculated.

Concept introduction:

The Molarity of the solution can be calculated by using following formula

  $\text{Molarity of the solution}\text{=}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litre}}$

Explanation of Solution

Expected not formed ${\text{Fe(OH)}}_{\text{3}}$ in the reaction = $\text{0}{\text{.00642 mol Fe(OH)}}_{\text{3}}$ expected (from $\text{NaOH}$) - $\text{0}{\text{.00585 mol Fe(OH)}}_{\text{3}}$ formed.

Expected not formed ${\text{Fe(OH)}}_{\text{3}}$ in the reaction = $\text{0}{\text{.00057 mol Fe(OH)}}_{\text{3}}$

Unreacted mole of sodium hydroxide is calculated as follows,

  $\begin{array}{l}\text{Moles}\text{of}\text{unreacted}\text{NaOH}\text{=}\text{0}\text{.00057}\text{mol}\text{of}{\text{Fe(OH)}}_{\text{3}}\text{not}\text{formed}\text{×}\frac{\text{3}\text{mol}\text{NaOH}}{\text{1}\text{mol}{\text{Fe(OH)}}_{\text{3}}}\\ \text{Moles}\text{of}\text{unreacted}\text{NaOH}\text{=}\text{1}{\text{.71×10}}^{\text{-3}}\text{mol}\end{array}$

Total volume = $\text{25}\text{ml}\text{+}\text{42}\text{.5}\text{ml}\text{=}\text{67}\text{.5}\text{ml}\text{(or) 0}\text{.0675}\text{L}$.

Molarity of the sodium hydroxide solution is calculated as follows,

    $\begin{array}{l}\text{Molarity of}\text{NaOH}\text{=}\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litre}}\\ \text{Molarity of}\text{NaOH}\text{=}\frac{\text{0}\text{.0017}\text{mol}\text{NaOH}}{\text{0}\text{.0675}\text{L}}\\ \text{Molarity of}\text{NaOH}\text{=}\text{0}\text{.0252}\text{M}\end{array}$

The concentration of excess reactant is $\text{0}\text{.0252}\text{M}$.