Problem 18 - Chemistry and Chemical Reactivity

Question

A You mix 25.0 mL of 0.234 M FeC 1 , with 42.5 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 \(\left(\mathrm{FeCl}_{3} \text { or } \mathrm{NaOH}\) ) is present \right. in a stoichiometric excess. What is the molar concentration of the excess reactant remaining in solution after \(\mathrm{Fe}(\mathrm{OH})\), has been precipitated?

Step-by-Step Solution

Answer

(a) 0.625 g; (b) 0.02474 M NaOH

Step 1: Write the Balanced Equation

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

Step 2: Calculate Moles of Each Reactant

Calculate the moles of \(\text{FeCl}_3\):\[\text{Moles of FeCl}_3 = Molarity \times Volume = 0.234\,M \times 0.025\,L = 5.85 \times 10^{-3} \text{ moles}\]Calculate the moles of \(\text{NaOH}\):\[\text{Moles of NaOH} = Molarity \times Volume = 0.453\,M \times 0.0425\,L = 0.01922 \text{ moles}\]

Step 3: Determine the Limiting Reactant

The stoichiometry from the balanced equation shows that 1 mole of \(\text{FeCl}_3\) requires 3 moles of \(\text{NaOH}\). Therefore, the moles of \(\text{NaOH}\) needed to react with \(5.85 \times 10^{-3}\) moles of \(\text{FeCl}_3\) are:\[5.85 \times 10^{-3} \times 3 = 1.755 \times 10^{-2} \text{ moles of NaOH}\]Since we only have \(0.01922\) moles of \(\text{NaOH}\), and more than \(1.755 \times 10^{-2}\) moles are available, \(\text{FeCl}_3\) is the limiting reactant.

Step 4: Calculate Mass of Precipitate

Using the limiting reactant \(\text{FeCl}_3\), we find moles of \(\text{Fe(OH)}_3\):As 1 mole of \(\text{FeCl}_3\) gives 1 mole of \(\text{Fe(OH)}_3\), the moles of precipitate will be \(5.85 \times 10^{-3}\).The molar mass of \(\text{Fe(OH)}_3\) is calculated as:\[\text{Fe(OH)}_3 = 55.85 + 3(16.00 + 1.01) = 106.87\,g/mol\]Thus, the mass of \(\text{Fe(OH)}_3\) is:\[5.85 \times 10^{-3} \times 106.87 = 0.6251\,g\]

Step 5: Calculate Concentration of Excess Reactant

The excess moles of \(\text{NaOH}\) left after reaction is given by:Moles of excess \(\text{NaOH} = 0.01922 - 0.01755 = 1.67 \times 10^{-3}\).The total volume of the solution is:\[25.0 + 42.5 = 67.5\,mL = 0.0675\,L\]Thus, the concentration of \(\text{NaOH}\) remaining is:\[\frac{1.67 \times 10^{-3}}{0.0675} = 0.02474\,M\]

Key Concepts

Limiting Reactant

In any chemical reaction, the limiting reactant is the substance that is completely consumed first. It determines the amount of product that can be formed. This concept is crucial to understanding chemical reactions because, often, reactants are not present in perfect proportions. A balanced chemical equation provides the ratio of moles needed for a reaction, and from this, we can identify the limiting reactant by comparing the actual mole ratios we have.To find the limiting reactant in the reaction between \\(( ext{FeCl}_3)\) and \( ext{NaOH}\), we calculated the moles available for each reactant. Given the balanced equation: \FeCl}_3 (aq) + 3 ext{NaOH} (aq) \rightarrow \\text{Fe(OH)}_3 (s) + 3\text{NaCl} (aq)\(, \we expect \)3\( moles of NaOH for every mole of \FeCl}_3\). \After calculation, we determined that FeCl}\(_3\) was the limiting reactant since it ran out before \NaOH.

Precipitate Formation

Precipitation occurs when a solid forms from a solution during a chemical reaction. This solid is called a precipitate. In the reaction of iron (III) chloride and sodium hydroxide, \(\text{Fe(OH)}_3\) is the precipitate, which appears as a solid because it is not soluble in water.The formation of a precipitate is an important aspect of this reaction. When \(\text{FeCl}_3\) reacts with \(\text{NaOH}\), they form iron (III) hydroxide, \(\text{Fe(OH)}_3\), which precipitates out of the solution. This occurs because the solubility product of \(\text{Fe(OH)}_3\) is very low, leading to its formation as a solid. Observing a precipitate in a reaction can be evidence that a reaction has occurred.

Molarity

Molarity is a way to express the concentration of a solution. It is defined as the number of moles of a solute per liter of solution, denoted as \(\text{M}\).Understanding molarity is key to solving our exercise, as it allows us to calculate the number of moles of reactants we have. For example, with \(25.0\) mL of \(0.234 \text{M} \text{FeCl}_3\) and \(42.5\) mL of \(0.453 \text{M} \text{NaOH}\), \we calculate moles using the formula: \\(\text{Moles} = \text{Molarity} \times \text{Volume}\). \This calculation is essential to identify the limiting reactant and the amount of precipitate formed.Molarity is also used to find the concentration of the excess reactant left after the reaction. Understanding these calculations enables one to precisely measure how substances in a reaction will behave.

Balanced Chemical Equation

A balanced chemical equation provides the relationship between reactants and products in terms of moles. All chemical reactions must be balanced to comply with the law of conservation of mass, which states that mass is neither created nor destroyed in a reaction.For our given exercise, the balanced chemical equation is: \(\text{FeCl}_3 (aq) + 3\text{NaOH} (aq) \rightarrow \text{Fe(OH)}_3 (s) + 3\text{NaCl} (aq)\). This tells us for every mole of \(\text{FeCl}_3\) used, \(3\) moles of NaOH are needed to produce \(1\) mole of \(\text{Fe(OH)}_3\) and \(3\) moles of \(\text{NaCl}\). \Balancing the equation is the first step in solving stoichiometric problems, as it provides the necessary mole ratios of reactants and products, which are crucial for calculating quantities in chemical reactions.