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Chapter 4

Q. 4.3

GRADED

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) + 3OH^{-}(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.

fig 4.3
fig 4.2
fig 4.6

Step-by-Step

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}