Question

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

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

Write the balanced chemical equation

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

Calculate the moles of the reactants

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

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.

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. \[ \text{moles of }Fe(OH)_3 = 0.01875\,mol \, NaOH \times \frac{1\, mol\, Fe(OH)_3}{3\, mol\, NaOH} = 0.00625\,mol \]

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.

Key concepts

Stoichiometry

Stoichiometry is the study of the quantitative relationships in chemical reactions. It allows us to predict the amounts of products and reactants involved in a chemical reaction. By using a balanced chemical equation, we can identify the proportional relationships between different substances.
In the given problem, the balanced chemical equation for the reaction between iron(III) nitrate and sodium hydroxide is: \[ Fe(NO_3)_3 (aq) + 3 NaOH (aq) \rightarrow Fe(OH)_3 (s) + 3 NaNO_3 (aq) \] Here, the coefficients show the stoichiometric ratios: one mole of iron(III) nitrate reacts with three moles of sodium hydroxide to form one mole of iron(III) hydroxide precipitate and three moles of sodium nitrate. By employing stoichiometry, we can determine how much of each solution is needed or will be formed in this reaction.
Understanding stoichiometry is fundamental to performing reliable calculations in chemistry, making it a vital tool for predicting the outcomes of reactions.

Limiting Reactant

The limiting reactant is the substance in a chemical reaction that runs out first, thus stopping the reaction from continuing. It determines the maximum amount of product that can be formed. Identifying the limiting reactant is crucial for maximizing efficiency in chemical processes.
In this case, to find the limiting reactant, we start by calculating the moles of each reactant: iron(III) nitrate has 0.007875 mol, and sodium hydroxide has 0.01875 mol. Using the stoichiometric ratio from the balanced equation, we know one mole of iron(III) nitrate needs 3 moles of sodium hydroxide. Thus, to react completely with 0.007875 mol of iron(III) nitrate, we would need: \[ 0.007875 \text{ mol } Fe(NO_3)_3 \times \frac{3 \text{ mol NaOH}}{1 \text{ mol } Fe(NO_3)_3} = 0.023625 \text{ mol NaOH} \] Since we only have 0.01875 mol of NaOH available, sodium hydroxide is the limiting reactant. This means that NaOH will be completely used up first, dictating the amount of iron(III) hydroxide that can be formed.

Precipitate Formation

Precipitate formation in chemical reactions involves the formation of a solid from a solution. This occurs when the concentration of a compound exceeds its solubility limit. The solid formed is known as a precipitate, and the reaction is termed as a precipitation reaction.
In this particular reaction, iron(III) hydroxide, \( Fe(OH)_3 \), acts as the precipitate. It forms as a solid when aqueous iron(III) nitrate reacts with sodium hydroxide. The product's insolubility in water is the driving force behind the precipitation.

• Precipitation reactions are essential in various applications, including water purification and qualitative chemical analysis.
• They help in identifying specific ions in a solution due to their distinctive solid formation.

Observing the appearance of a solid can confirm that a chemical reaction has successfully occurred. In our case, \( Fe(OH)_3 \) precipitate is a clear indication of the chemical transformation taking place.

Molar Mass Calculation

Calculating molar mass is a fundamental skill needed to convert between moles and grams of a compound. The molar mass of a substance is the mass of one mole of its entities (atoms, molecules, ions, etc.), expressed in grams per mole.
To find the mass of the iron(III) hydroxide precipitate formed, we start by calculating its molar mass. The formula \( Fe(OH)_3 \) indicates that it contains one atom of iron and three hydroxide ions.

• The atomic mass of iron \( (Fe) \) is approximately 55.85 g/mol.
• Each hydroxide group \( (OH) \) has a mass of approximately 17.01 g/mol.

Adding these together gives: \[ \text{Molar mass of } Fe(OH)_3 = 55.85 + 3 \times 17.01 = 106.87 + 51.03 = 157.9 \, \text{g/mol} \] Given the moles of \( Fe(OH)_3 \) produced from the limiting reactant (0.00625 mol), we can convert this to grams using the calculated molar mass: \[ \text{Mass} = 0.00625 \text{ mol} \times 157.9 \text{ g/mol} = 0.987 \text{ g} \] This calculation shows how to determine the mass of a precipitate from its molar amount, linked directly to moles and molar mass.