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Chapter 4: Problem 63

What mass of iron(III) hydroxide precipitate can be produced by reacting 75.0 mL of 0.105 M iron(III) nitrate with 125 mL of 0.150 M sodium hydroxide?

Short Answer

Expert verified
The mass of iron(III) hydroxide precipitate that can be produced by reacting 75.0 mL of 0.105 M iron(III) nitrate with 125 mL of 0.150 M sodium hydroxide is \(0.987 \, grams\).

Step by step solution

01

Write the balanced chemical equation

\[ Fe(NO_3)_3 (aq) + 3 NaOH (aq) \rightarrow Fe(OH)_3 (s) + 3 NaNO_3 (aq) \]
02

Calculate the moles of the reactants

\[ \text{moles of }Fe(NO_3)_3 = 75.0\,mL \times \frac{0.105\,mol}{1000\,mL} = 0.007875\,mol \] \[ \text{moles of }NaOH = 125\,mL \times \frac{0.150\,mol}{1000\,mL} = 0.01875\,mol \]
03

Determine the limiting reactant

Using the stoichiometry of the balanced chemical equation, we can calculate the moles of NaOH required for complete reaction with the given moles of Fe(NO3)3. \[ \text{moles of }NaOH\, \text{required} = 0.007875\,mol \, Fe(NO_3)_3 \times \frac{3\, mol\, NaOH}{1\, mol\, Fe(NO_3)_3} = 0.023625\,mol \] Since we have less moles of NaOH than required (0.01875 mol < 0.023625 mol), NaOH is the limiting reactant.
04

Calculate the moles of iron(III) hydroxide produced

Now, we can use the stoichiometry of the balanced chemical equation to calculate the moles of iron(III) hydroxide produced using the moles of the limiting reactant, NaOH. \[ \text{moles of }Fe(OH)_3 = 0.01875\,mol \, NaOH \times \frac{1\, mol\, Fe(OH)_3}{3\, mol\, NaOH} = 0.00625\,mol \]
05

Convert the moles of iron(III) hydroxide to mass

Finally, we can convert the moles of iron(III) hydroxide produced to mass by using the molar mass of iron(III) hydroxide, which is 106.87 g/mol for Fe and 17.01 g/mol for each OH group (total 106.87 + 3*17.01 g/mol). \[ \text{mass of }Fe(OH)_3 = 0.00625\,mol \times (106.87 + 3\times17.01)\,g/mol = 0.00625\,mol \times 157.9\,g/mol = 0.987\,g \] Thus, the mass of iron(III) hydroxide precipitate that can be produced is 0.987 grams.

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Most popular questions from this chapter

You made 100.0 mL of a lead(II) nitrate solution for lab but forgot to cap it. The next lab session you noticed that there was only 80.0 mL left (the rest had evaporated). In addition, you forgot the initial concentration of the solution. You decide to take 2.00 mL of the solution and add an excess of a concentrated sodium chloride solution. You obtain a solid with a mass of 3.407 g. What was the concentration of the original lead(II) nitrate solution?

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