Problem 118 - Chemistry and Chemical Reactivity

Question

A 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) On of the reactants (FeCly or NaOH) is present in a stoichiometric excess. What is the molar concentra tion of the excess reactant remaining in solution after \(\mathrm{Fe}(\mathrm{OH})\), has been precipitated?

Step-by-Step Solution

Answer

(a) 0.625 g of \(\text{Fe(OH)}_3\) will precipitate. (b) The concentration of excess \(\text{NaOH}\) is 0.0244 M.

Step 1: Write the Balanced Chemical Equation

The first step involves writing down the balanced chemical reaction equation for the reaction between iron(III) chloride (\(\text{FeCl}_3\)) and sodium hydroxide (\(\text{NaOH}\)). The balanced equation is:\[ \text{FeCl}_3 (aq) + 3 \text{NaOH} (aq) \rightarrow \text{Fe(OH)}_3 (s) + 3 \text{NaCl} (aq) \]

Step 2: Calculate the Moles of Reactants

Calculate the number of moles of each reactant.For \(\text{FeCl}_3\):\[ n_{\text{FeCl}_3} = M \times V = 0.234 \text{ M} \times 0.0250 \text{ L} = 0.00585 \text{ mol} \]For \(\text{NaOH}\):\[ n_{\text{NaOH}} = M \times V = 0.453 \text{ M} \times 0.0425 \text{ L} = 0.0192 \text{ mol} \]

Step 3: Determine the Limiting Reactant

Use the stoichiometry of the balanced equation to find out which reactant is the limiting reactant. From the equation, 1 mole of \(\text{FeCl}_3\) reacts with 3 moles of \(\text{NaOH}\).Calculate the moles of \(\text{NaOH}\) needed to react with \(\text{FeCl}_3\):\[ n_{\text{NaOH needed}} = 3 \times n_{\text{FeCl}_3} = 3 \times 0.00585 = 0.01755 \text{ mol} \]Since \(0.01755 \text{ mol of NaOH}\) are needed but \(0.0192 \text{ mol of NaOH}\) are available, \(\text{FeCl}_3\) is the limiting reactant.

Step 4: Calculate Mass of Fe(OH)3 Precipitated

Since \(\text{FeCl}_3\) is the limiting reactant, the amount of \(\text{Fe(OH)}_3\) produced will be based on \(\text{FeCl}_3\).\[ n_{\text{Fe(OH)}_3} = n_{\text{FeCl}_3} = 0.00585 \text{ mol} \]The molar mass of \(\text{Fe(OH)}_3\) is:\[ M = 55.85 + 3 \times (16.00 + 1.01) = 106.87 \text{ g/mol} \]Mass of \(\text{Fe(OH)}_3\):\[ m = n \times M = 0.00585 \times 106.87 = 0.625 \text{ g} \]

Step 5: Calculate the Concentration of Excess NaOH

Calculate the moles of \(\text{NaOH}\) in excess after the reaction:Moles of \(\text{NaOH}\) used:\[ n_{\text{NaOH used}} = 0.01755 \text{ mol} \]Excess \(\text{NaOH}:\)\[ n_{\text{NaOH excess}} = 0.0192 - 0.01755 = 0.00165 \text{ mol} \]Volume of solution after mixing:\[ V_{\text{total}} = 0.0250 \text{ L} + 0.0425 \text{ L} = 0.0675 \text{ L} \]Concentration of excess \(\text{NaOH}:\)\[ \text{[NaOH]} = \frac{0.00165 \text{ mol}}{0.0675 \text{ L}} = 0.0244 \text{ M} \]

Key Concepts

Balanced Chemical Equation

To begin any chemical reaction problem, it's important to first understand the balanced chemical equation. This helps ensure that the reaction abides by the law of conservation of mass, which states that mass cannot be created or destroyed in a chemical reaction. When iron(III) chloride (\(\text{FeCl}_3\)) reacts with sodium hydroxide (\(\text{NaOH}\)), the balanced chemical equation is:

• \(\text{FeCl}_3 (aq) + 3 \text{NaOH} (aq) \rightarrow \text{Fe(OH)}_3 (s) + 3 \text{NaCl} (aq)\)

This equation shows the stoichiometry of the reaction, meaning it describes the exact ratio of reactants to products. In this particular case, each mole of \(\text{FeCl}_3\) will react with three moles of \(\text{NaOH}\) to produce one mole of iron(III) hydroxide (\(\text{Fe(OH)}_3\)) and three moles of sodium chloride (\(\text{NaCl}\)). Knowing this ratio is crucial for determining how much of each reactant is needed and what the products will be.

Moles of Reactants

Once the balanced chemical equation is known, the next step is to calculate the moles of each reactant involved in the reaction. The number of moles can be found using the formula:

• \( n = M \times V \)

where \( M \) is the molarity (concentration) of the solution and \( V \) is the volume in liters. For the reaction of \(\text{FeCl}_3\) and \(\text{NaOH}\):

• Moles of \(\text{FeCl}_3\) = \(0.234 \text{ M} \times 0.0250 \text{ L} = 0.00585 \text{ mol}\)
• Moles of \(\text{NaOH}\) = \(0.453 \text{ M} \times 0.0425 \text{ L} = 0.0192 \text{ mol}\)

These calculations give a clearer picture of the quantities of reactants available to undergo the reaction. It's fundamental for the subsequent determination of the limiting reactant.

Precipitation Reaction

In this chemical scenario, the reaction forms a precipitate, which is an insoluble solid that emerges from a liquid solution. When \(\text{FeCl}_3\) reacts with \(\text{NaOH}\), they form iron(III) hydroxide (\(\text{Fe(OH)}_3\)), which precipitates out of the solution. This process can be visually observed as a solid forming and settling at the bottom of the container. Precipitation reactions often involve ionic compounds where two ions react to form an insoluble salt.
Precipitation is a key indicator of the reaction's progress because it signifies a chemical change. Understanding which substances form precipitates, and under what conditions, is essential when predicting the outcomes of chemical reactions. In this problem, recognizing iron(III) hydroxide as a precipitate helps us identify the completion of the reaction and calculate how much of it will form.

Excess Reactant Calculation

Calculating the excess reactant in a chemical reaction is as critical as identifying the limiting reactant. The limiting reactant is completely consumed during the reaction, limiting the amount of product formed, while the excess reactant remains after the reaction is complete.To determine the excess amount, subtract the moles used from the initial moles available. For \(\text{NaOH}\), used with our given amounts:

• \(\text{Moles used} = 0.01755 \text{ mol}\)
• \(\text{Moles excess} = 0.0192 \text{ mol} - 0.01755 \text{ mol} = 0.00165 \text{ mol}\)

Finally, to find the concentration of the excess \(\text{NaOH}\), divide the excess moles by the total volume of the solution:

• \(\text{[NaOH]} = \frac{0.00165 \text{ mol}}{0.0675 \text{ L}} = 0.0244 \text{ M}\)

Knowing how much excess reactant is left can be useful in understanding reaction efficiency and for further processing or neutralization steps.