Problem 118 - Chemistry and Chemical Reactivity

Question

You mix \(25.0 \mathrm{mL}\) of \(0.234 \mathrm{M} \mathrm{FeCl}_{3}\) with \(42.5 \mathrm{mL}\) of \(0.453 \mathrm{M} \mathrm{NaOH}\). (a) What mass of \(\mathrm{Fe}(\mathrm{OH})_{3}\) (in grams) will precipitate from this reaction mixture? (b) One of the reactants \(\left(\mathrm{FeCl}_{3} \text { or } \mathrm{NaOH}\right)\) is present in a stoichiometric excess. What is the molar concentration of the excess reactant remaining in solution after \(\operatorname{Fe}(\mathrm{OH})_{3}\) has been precipitated?

Step-by-Step Solution

Answer

0.625 g of \(\mathrm{Fe(OH)}_3\) will precipitate; the remaining \(\mathrm{NaOH}\) concentration is 0.0252 M.

Step 1: Determine the balanced chemical equation

The balanced chemical equation for the reaction between iron(III) chloride and sodium hydroxide is: \[\text{FeCl}_3 + 3\text{NaOH} \rightarrow \text{Fe}(\text{OH})_3 \downarrow + 3\text{NaCl}\] This indicates that one mole of \(\text{FeCl}_3\) reacts with three moles of \(\text{NaOH}\) to form one mole of \(\text{Fe}(\text{OH})_3\).

Step 2: Calculate moles of FeCl₃ and NaOH

Calculate the moles of each reactant by converting the volume to liters and multiplying by their molarity.For \(\text{FeCl}_3\):\[ \text{Moles of FeCl}_3 = 0.0250 \, \text{L} \times 0.234 \, \text{M} = 0.00585 \, \text{mol}\]For \(\text{NaOH}\):\[ \text{Moles of NaOH} = 0.0425 \, \text{L} \times 0.453 \, \text{M} = 0.0192525 \, \text{mol}\]

Step 3: Identify the limiting reactant

According to the stoichiometry of the reaction, 1 mole of \(\text{FeCl}_3\) needs 3 moles of \(\text{NaOH}\). Calculate how much \(\text{NaOH}\) is required for 0.00585 moles of \(\text{FeCl}_3\):\[\text{Required NaOH} = 0.00585 \, \text{mol of FeCl}_3 \times 3 \, \text{mol NaOH/mol FeCl}_3 = 0.01755 \, \text{mol}\]Since 0.0192525 moles of \(\text{NaOH}\) are available, and only 0.01755 moles are needed, \(\text{FeCl}_3\) is the limiting reactant.

Step 4: Calculate mass of Fe(OH)₃ precipitated

Since \(\text{FeCl}_3\) is the limiting reactant, it determines the amount of \(\text{Fe}(\text{OH})_3\) formed. According to the balanced equation, 1 mole of \(\text{FeCl}_3\) produces 1 mole of \(\text{Fe}(\text{OH})_3\).Thus, 0.00585 moles of \(\text{FeCl}_3\) will produce 0.00585 moles of \(\text{Fe}(\text{OH})_3\).To find the mass of \(\text{Fe}(\text{OH})_3\), use its molar mass: \(\text{Molar mass of Fe(OH)}_3 = 55.85 + 3 \times (16.00 + 1.01) = 106.87 \text{g/mol}\).\[ \text{Mass of Fe(OH)}_3 = 0.00585 \, \text{mol} \times 106.87 \, \text{g/mol} = 0.625 \text{g}\]

Step 5: Calculate excess NaOH remaining

First, determine how much \(\text{NaOH}\) was used by \(\text{FeCl}_3\):\[\text{NaOH used} = 0.01755 \, \text{mol}\]The initial moles of \(\text{NaOH}\) were 0.0192525, so the excess moles are:\[\text{Excess NaOH} = 0.0192525 - 0.01755 = 0.0017025 \, \text{mol}\]Calculate the concentration of remaining \(\text{NaOH}\) in the total solution volume:\[ \text{Total volume} = 25.0 \, \text{mL} + 42.5 \, \text{mL} = 67.5 \, \text{mL} = 0.0675 \, \text{L}\]\[\text{Concentration of remaining NaOH} = \frac{0.0017025 \, \text{mol}}{0.0675 \, \text{L}} = 0.0252 \, \text{M}\]

Key Concepts

Limiting Reactant

In any chemical reaction, the limiting reactant is the substance that is entirely consumed first, stopping the reaction because there is no more quantity to react with the other reactants. This concept is crucial because it determines the maximum amount of product that can be formed from the given reactants.
In our specific case, we are dealing with a reaction between iron(III) chloride (\( ext{FeCl}_3\)) and sodium hydroxide (\( ext{NaOH}\)). From the balanced chemical equation, we see that 1 mole of \( ext{FeCl}_3\) reacts with 3 moles of \( ext{NaOH}\).

To identify the limiting reactant:

• Calculate the moles of each reactant present.
• Consider the stoichiometry from the balanced equation to determine which reactant runs out first.

In this example, \( ext{FeCl}_3\) turned out to be the limiting reactant because we started with fewer moles compared to what is needed to completely react with the available moles of \( ext{NaOH}\). This means \( ext{FeCl}_3\) limits the formation of the product, \( ext{Fe(OH)}_3\).

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions. It is based on the conservation of mass where the total mass of reactants equals the total mass of products. This concept is fundamental when analyzing chemical equations because it allows us to predict the quantities of substances consumed and produced.
In the reaction \[\( ext{FeCl}_3 + 3 ext{NaOH} \rightarrow ext{Fe}( ext{OH})_3 + 3 ext{NaCl}\)\], stoichiometry helps us calculate the necessary \( ext{NaOH}\) needed for reacting completely with the \( ext{FeCl}_3\).

Steps in stoichiometric calculations include:

• Writing and balancing the chemical equation.
• Using the mole ratio from the balanced equation to calculate the relationship between reactants and products.
• Determining the mass or volume of reactants and products as needed.

Ultimately, stoichiometry allows us to precisely measure substances in laboratory settings ensuring all reactants and products are accounted for.

Chemical Equations

A chemical equation is shorthand for a chemical reaction, using symbols and formulas to represent the reactants and products. The equation provides important information about the proportions in which substances react and are formed.
For the precipitation reaction we explored, the balanced chemical equation is crucial:\[\( ext{FeCl}_3 + 3 ext{NaOH} \rightarrow ext{Fe}( ext{OH})_3 \, \downarrow + 3 ext{NaCl}\)\]This equation shows us that:

• One mole of \( ext{FeCl}_3\) reacts with three moles of \( ext{NaOH}\).
• Each reactant and product must maintain this ratio for the reaction to be accurate.

The arrows and other symbols in the equation, such as the precipitation arrow "\(\downarrow\)", indicate physical states and the direction of the reaction.

Understanding chemical equations is essential for predicting the outcomes of chemical reactions.

Molarity Calculations

Molarity, a vital concept in chemistry, is the measure of concentration of a solute in a solution, expressed as moles of solute per liter of solution (M or mol/L). Molarity calculations are crucial for preparing solutions with precise concentrations for reactions.
In the exercise, molarity helps us determine the amount of each reactant available. For instance, the \( ext{FeCl}_3\) solution has a molarity of 0.234 M. To find the moles of \( ext{FeCl}_3\) in 25.0 mL:

• Convert 25.0 mL to liters (\(0.0250 \, ext{L}\)).
• Calculate moles using the equation: \[\( ext{Moles of FeCl}_3 = 0.0250 \, ext{L} \times 0.234 \, ext{M}\)\]= 0.00585 moles.

Likewise, the calculation for \( ext{NaOH}\) enables us to discover its available moles using its known concentration of 0.453 M.

Mastering molarity calculations allows for precise control over chemical reactions and solution preparations.