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.
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} |