## Subscribe \$4.99/month

Un-lock Verified Step-by-Step Experts Answers.

## Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

## Tip our Team

Our Website is free to use.
To help us grow, you can support our team with a Small Tip.

## Holooly Tables

All the data tables that you may search for.

## Holooly Help Desk

Need Help? We got you covered.

## Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Products

## Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

## Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

## Holooly Help Desk

Need Help? We got you covered.

## Q. 4.3

When aqueous solutions of sodium hydroxide and iron(III) nitrate are mixed, a red precipitate forms.

ⓐ Write a net ionic equation for the reaction.

ⓑ What volume of 0.136 M iron(III) nitrate is required to produce 0.886 g of precipitate?

ⓒ How many grams of precipitate are formed when 50.00 mL of 0.200 M NaOH and 30.00 mL of 0.125 M Fe$(NO_{3})_{3}$ are mixed?

 ANALAYSIS reactant compounds [NaOH and Fe$(NO_{3})_{3}$] Information given: net ionic equation Asked for:

STRATEGY

1. Follow the schematic diagram in Figure 4.3 to determine possible precipitates.
2. Use the precipitation diagram (Figure 4.2) to determine whether the possible precipitates are soluble or insoluble.
3. Write the net ionic equation. Start with the product

 ANALAYSIS net ionic equation from (a): [$Fe^{3+}(aq)$ + $3OH^{-}(aq)$ → $Fe(OH)_{3}(s)$] mass of precipitate (0.886 g); molarity of $Fe(NO_{3})_{3}$ (0.136 M) Information given: molarity of reacting ion, $Fe^{3+}$ molar mass of precipitate Information implied: volume of $Fe(NO_{3})_{3}$ used in the reaction Asked for:

STRATEGY

Reverse the pathway shown in Figure 4.6.

mass $Fe(OH)_{3}$ → mol ppt → mol ion : mol of parent compound → V of parent compound

 ANALYSIS net ionic equation from (a): [$Fe^{3+}(aq)$ + 3$OH^{-}(aq)$ → $Fe(OH)_{3}(s)$] volume (50.00 mL) and molarity (0.200 M) of NaOH volume (30.00 mL) and molarity (0.125 M) of $Fe(NO_{3})_{3}$ Information given: number of moles of reacting ions, $Fe^{3+}$ and $OH^{-}$ Data for moles of both reactants is given, making this a limiting reactant problem. Information implied: mass of precipitate formed Asked for:

STRATEGY

1. Follow the pathway in Figure 4.6 for both NaOH and $Fe(NO_{3})_{3}$ to obtain moles of precipitate formed.
mol NaOH (V × M) → mol $OH^{-}$ → mol ppt    $Fe(NO_{3})_{3}$(V × M) → mol $Fe^{3+}$ → mol ppt
2. Choose the smaller number of moles and convert moles to mass.

## Verified Solution

 $Na^{+}$ and $OH^{-}$ from NaOH; $Fe^{3+}$ and $NO^{-}_{3}$ from $Fe(NO_{3})_{3}$ Ions in solution $Fe(OH)_{3}$ and $NaNO_{3}$ Possible precipitates $Fe(OH)_{3}$ is insoluble and forms a precipitate. Solubility $Fe^{3+}(aq)$ + $3OH^{-}(aq)$ → $Fe(OH)_{3}(s)$ Net ionic equation

 0.886 g $Fe(OH)_{3}$ × $\frac{1 mol Fe(OH)_{3}}{106.87 g Fe(OH)_{3}}$ × $\frac{1 mol Fe^{3+}}{1 mol Fe(OH)_{3}}$ × $\frac{1 mol Fe(NO_{3})_ {3}}{1 mol Fe^{3+}}$ = 0.00829 mol $Fe(NO_{3})_{3}$ V = $\frac{mol}{M}$ = $\frac{0.00829 mol}{0.136 mol/L}$ = 0.0610 L $V_{Fe(NO_{3})_{3}}$

 0.0500 L × 0.200$\frac{mol NaOH}{L}$ × $\frac{1 mol OH^{-}}{1 mol NaOH}$ × $\frac{1 mol Fe(OH)_{3}}{3 mol OH^{-}}$ = 0.00333 mol $Fe(OH)_{3}$ mol ppt if NaOH limiting 0.0300 L × 0.125$\frac{mol Fe(NO_{3})_{3}}{L}$ × $\frac{1 mol Fe^{3+}}{1 mol Fe(NO_{3})_{3}}$ × $\frac{1 mol Fe(OH)_{3}}{1 mol Fe^{3+}}$ = 0.00375 mol $Fe(OH)_{3}$ mol ppt if $Fe(NO_{3})_{3}$ limiting 0.00333 mol < 0.00375 mol; 0.00333 mol $Fe(OH)_{3}$ is obtained Theoretical yield 0.00333 mol × $\frac{106.87 g}{1 mol}$ = 0.356 g $Fe(OH)_{3}$