Problem 87 - Chemistry The Central Science In Si Units

Question

A solution of \(105.0 \mathrm{~mL}\) of \(0.300 \mathrm{M} \mathrm{NaOH}\) is mixed with a solution of \(150.0 \mathrm{~mL}\) of \(0.060 \mathrm{M} \mathrm{AlCl}_{3} .\) (a) Write the balanced chemical equation for the reaction that occurs. (b) What precipitate forms? (c) What is the limiting reactant? (d) How many grams of this precipitate form? (e) What is the concen-

Step-by-Step Solution

Answer

(a) Balanced equation: \(3\mathrm{NaOH} + \mathrm{AlCl}_3 \rightarrow \mathrm{Al(OH)}_3 + 3\mathrm{NaCl}\). (b) Precipitate: \(\mathrm{Al(OH)}_3\). (c) Limiting reactant: \(\mathrm{AlCl}_3\). (d) 0.702 g of \(\mathrm{Al(OH)}_3\) forms.

Step 1: Writing the Balanced Chemical Equation

First, we need to identify the reaction that occurs when the solutions mix. Sodium hydroxide (\(\mathrm{NaOH}\)) reacts with aluminum chloride (\(\mathrm{AlCl}_3\)) to form aluminum hydroxide (\(\mathrm{Al(OH)}_3\)) and sodium chloride (\(\mathrm{NaCl}\)). The balanced equation is: \[ 3\mathrm{NaOH} + \mathrm{AlCl}_3 \rightarrow \mathrm{Al(OH)}_3 + 3\mathrm{NaCl} \]

Step 2: Identifying the Precipitate

Upon mixing \(\mathrm{NaOH}\) with \(\mathrm{AlCl}_3\), \(\mathrm{Al(OH)}_3\) forms as a solid, while \(\mathrm{NaCl}\) stays in solution. The precipitate that forms is aluminum hydroxide \(\mathrm{Al(OH)}_3\).

Step 3: Determining the Limiting Reactant

To determine the limiting reactant, calculate the moles of each reactant. - Moles of \(\mathrm{NaOH}\): \(105.0 \text{ mL} \times 0.300 \text{ M} = 0.0315 \text{ moles}\)- Moles of \(\mathrm{AlCl}_3\): \(150.0 \text{ mL} \times 0.060 \text{ M} = 0.0090 \text{ moles}\)From the equation, 3 moles of \(\mathrm{NaOH}\) reacts with 1 mole of \(\mathrm{AlCl}_3\). Therefore, \(\mathrm{AlCl}_3\) is the limiting reactant as 0.0090 moles requires only 0.0270 moles of \(\mathrm{NaOH}\), which is available.

Step 4: Calculating Grams of Precipitate Formed

To find the mass of \(\mathrm{Al(OH)}_3\) formed, first calculate the moles that will form based on the limiting reactant \(\mathrm{AlCl}_3\). Since 0.0090 moles of \(\mathrm{AlCl}_3\) fully react, they will produce 0.0090 moles of \(\mathrm{Al(OH)}_3\).The molar mass of \(\mathrm{Al(OH)}_3\) is approximately 78.00 g/mol.Thus, mass = moles \(\times\) molar mass = 0.0090 moles \(\times\) 78.00 \text{ g/mol} = 0.702 \text{ g}.

Key Concepts

Balanced Chemical Equation

A balanced chemical equation is a precise representation of a chemical reaction. It showcases the reactants converting into products. It is crucial that the number of atoms for each element remains consistent on both sides of the equation.
In the given exercise, sodium hydroxide (\(\mathrm{NaOH}\)) reacts with aluminum chloride (\(\mathrm{AlCl}_3\)) to form aluminum hydroxide (\(\mathrm{Al(OH)}_3\)) and sodium chloride (\(\mathrm{NaCl}\)). The balanced form of this reaction is:

\[ 3\mathrm{NaOH} + \mathrm{AlCl}_3 \rightarrow \mathrm{Al(OH)}_3 + 3\mathrm{NaCl} \]
This equation makes sure that for every 3 moles of \(\mathrm{NaOH}\), 1 mole of \(\mathrm{AlCl}_3\) is used to produce 1 mole of \(\mathrm{Al(OH)}_3\) and 3 moles of \(\mathrm{NaCl}\). Balancing equations ensures the conservation of mass, adhering to the law of conservation of matter.

Limiting Reactant

The concept of a limiting reactant is central in stoichiometry. It is the reactant in a chemical reaction that gets entirely used up first. This reactant determines the maximum amount of product that can be formed.
In our exercise, we begin by calculating the moles of each reactant:

• Moles of \(\mathrm{NaOH}\): \(105.0 \text{ mL} \times 0.300 \text{ M} = 0.0315 \text{ moles}\)
• Moles of \(\mathrm{AlCl}_3\): \(150.0 \text{ mL} \times 0.060 \text{ M} = 0.0090 \text{ moles}\)

Given the reaction \(3\mathrm{NaOH} + \mathrm{AlCl}_3 \rightarrow \mathrm{Al(OH)}_3 + 3\mathrm{NaCl}\), 3 moles of \(\mathrm{NaOH}\) react with 1 mole of \(\mathrm{AlCl}_3\). In this case, \(\mathrm{AlCl}_3\) is the limiting reactant because there is not enough \(\mathrm{AlCl}_3\) to use up all the available \(\mathrm{NaOH}\). Consequently, the amount of \(\mathrm{AlCl}_3\) determines the amount of \(\mathrm{Al(OH)}_3\) that can be produced.

Molar Mass

Molar mass plays a vital role in stoichiometry for converting between moles and mass. It is the mass of one mole of a substance, expressed in g/mol.
For the exercise, it is necessary to determine the molar mass of aluminum hydroxide (\(\mathrm{Al(OH)}_3\)) to calculate the mass of the precipitate that forms:

• Molar mass of \(\mathrm{Al(OH)}_3\) = 1 aluminum (Al) atom (≈ 26.98 g/mol) + 3 oxygen (O) atoms (≈ 16.00 g/mol each) + 3 hydrogen (H) atoms (≈ 1.01 g/mol each)
• Total molar mass = 26.98 + (3 \( imes\) 16.00) + (3 \( imes\) 1.01) = 78.00 g/mol

Using the limiting reactant \(\mathrm{AlCl}_3\), knowing that 0.0090 moles will form 0.0090 moles of \(\mathrm{Al(OH)}_3\), you can calculate that the precipitate mass is 0.0090 moles \( imes\) 78.00 g/mol = 0.702 g.

Precipitate Formation

Precipitate formation occurs when substances in solution react to form an insoluble solid. This is a key aspect of many chemical reactions.
In the given reaction of \(\mathrm{NaOH}\) with \(\mathrm{AlCl}_3\), the formation of aluminum hydroxide (\(\mathrm{Al(OH)}_3\)) marks the formation of a precipitate. As \(\mathrm{Al(OH)}_3\) precipitates out of the solution, a solid is formed, which can be observed as a cloudiness or solid particles.
This process is crucial for separating substances and is often used in techniques such as water purification and the extraction of metals in chemical industries. Precipitate formation serves as a visual cue that a reaction has occurred, which can also be used to determine whether the reactants are fully consumed.