Problem 69 - Chemistry The Molecular Science

Question

Aluminum chloride, \(\mathrm{Al}_{2} \mathrm{Cl}_{6}\), is an inexpensive reagent used in many industrial processes. It is made by treating scrap aluminum with chlorine according to the balanced equation $$2 \mathrm{Al}(\mathrm{s})+3 \mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow \mathrm{Al}_{2} \mathrm{Cl}_{6}(\mathrm{~s})$$ (a) Determine which reactant is limiting if \(2.70 \mathrm{~g} \mathrm{Al}\) and \(4.05 \mathrm{~g} \mathrm{Cl}_{2}\) are mixed. (b) Calculate what mass of \(\mathrm{Al}_{2} \mathrm{Cl}_{6}\) can be produced. (c) Calculate what mass of the excess reactant remains when the reaction is complete.

Step-by-Step Solution

Answer

(a) Cl₂ is limiting, (b) 5.067 g Al₂Cl₆, (c) 1.674 g Al remains.

Step 1: Calculate moles of Aluminum

Start by calculating the moles of Aluminum using its molar mass (27.0 g/mol). The moles of Aluminum are given by the formula:\[\text{moles of Al} = \frac{\text{mass of Al}}{\text{molar mass of Al}} = \frac{2.70 \text{ g}}{27.0 \text{ g/mol}} = 0.1 \text{ mol}\]

Step 2: Calculate moles of Chlorine gas

Calculate the moles of Chlorine gas \(\text{Cl}_2\) using its molar mass (70.9 g/mol). The moles of Chlorine are calculated as follows:\[\text{moles of } \text{Cl}_2 = \frac{\text{mass of } \text{Cl}_2}{\text{molar mass of } \text{Cl}_2} = \frac{4.05 \text{ g}}{70.9 \text{ g/mol}} \approx 0.057 \text{ mol}\]

Step 3: Determine the limiting reactant

Use the stoichiometry from the balanced equation: \(2 \text{ mol Al} : 3 \text{ mol } \text{Cl}_2\). Calculate the ratio of moles available to required:- For Al: \(0.1 \text{ mol Al} \rightarrow 0.15 \text{ mol } \text{Cl}_2\text{ required (since } \frac{3}{2} = 1.5)\)- For \(\text{Cl}_2\): \(0.057 \text{ mol } \text{Cl}_2 \rightarrow 0.038 \text{ mol Al required (since } \frac{2}{3} \approx 0.67)\)Since more \(\text{Cl}_2\) is required than available, \(\text{Cl}_2\) is the limiting reactant.

Step 4: Calculate mass of Aluminum chloride produced

According to the balanced equation, the ratio of \(\text{Cl}_2\) to \(\text{Al}_2\text{Cl}_6\) is \(3 : 1\). Therefore, the moles of \(\text{Al}_2\text{Cl}_6\) produced is:\[\text{moles of } \text{Al}_2\text{Cl}_6 = \frac{0.057 \text{ mol } \text{Cl}_2}{3} = 0.019 \text{ mol }\]Now, calculate the mass using the molar mass of \(\text{Al}_2\text{Cl}_6\) (266.68 g/mol): \[\text{mass of } \text{Al}_2\text{Cl}_6 = 0.019 \text{ mol} \times 266.68 \text{ g/mol} \approx 5.067 \text{ g}\]

Step 5: Calculate mass of excess Aluminum remaining

First, determine how much Al reacts with the limiting \(\text{Cl}_2\):- Required moles of Al for 0.057 mol \(\text{Cl}_2\): \[\text{moles of Al used} = \frac{2}{3} \times 0.057 \text{ mol } \text{Cl}_2 = 0.038 \text{ mol Al}\]Convert this to grams: \[\text{mass of Al used} = 0.038 \text{ mol} \times 27.0 \text{ g/mol} = 1.026 \text{ g}\]Finally, subtract the consumed Al from the initial amount: \[\text{excess Al} = 2.70 \text{ g (initial)} - 1.026 \text{ g} \approx 1.674 \text{ g}\]

Key Concepts

Stoichiometry

Stoichiometry is an essential concept in chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. In essence, it helps us understand how much of each substance is needed or produced in a reaction. This involves using the balanced chemical equation, which provides the molar ratio between the reactants and products.

For example, in the reaction of aluminum with chlorine gas to produce aluminum chloride, the balanced equation is:

• 2 Al (solid) + 3 Cl\(_2\) (gas) → Al\(_2\)Cl\(_6\) (solid)

This equation tells us that 2 moles of aluminum react with 3 moles of chlorine to produce 1 mole of aluminum chloride. Thus, there is a 2:3:1 molar ratio among these compounds.

Stoichiometry allows us to apply these ratios to calculate the quantities involved in the reaction. By knowing the masses of the starting materials, we can calculate:

• How much product is formed if a reaction goes to completion.
• Which reactant is limiting when not all reactants are present in the ratio prescribed by the balanced equation.
• How much of any reagent remains unreacted when the reaction stops.

Mastering stoichiometry is crucial for predicting yields and optimizing the use of reactants in laboratory or industrial settings.

Molar Mass

Molar mass is a key concept that refers to the mass of a given substance (chemical element or compound) divided by the amount of substance. It is typically expressed in units of grams per mole (g/mol). Molar mass is calculated by summing the atomic masses of all the atoms in a molecule. This measure allows chemists to convert between the mass of a substance and the amount of moles, which are central to calculations in stoichiometry.

For instance, consider aluminum (Al) and chlorine gas (Cl\(_2\)) in the reaction:

• Aluminum has a molar mass of 27.0 g/mol.
• Chlorine gas, being a diatomic molecule (Cl\(_2\)), has a molar mass of approximately 70.9 g/mol (calculated as 2 times the atomic mass of chlorine, which is about 35.45 g/mol).

These molar masses allow us to determine the number of moles from a given mass, as shown in the exercise calculations. For example, to find the moles of aluminum in 2.70 g, we use the formula: \[\text{moles of } \text{Al} = \frac{\text{mass of Al}}{\text{molar mass of Al}} = \frac{2.70 \text{ g}}{27.0 \text{ g/mol}} = 0.1 \text{ mol}\] Accurate molar masses are essential for determining the limiting reactant by comparing the computed moles of each reactant against their stoichiometric coefficients in the balanced chemical reaction.

Chemical Reactions

Chemical reactions describe the process by which substances interact and transform into different substances. This process involves the breaking and forming of chemical bonds, converting reactants into products. Each reaction can be symbolically represented by a chemical equation. A balanced chemical equation reflects the conservation of mass and matter, ensuring that the number of atoms for each element is the same on both sides of the equation.

Let's look closer at the reaction of aluminum with chlorine:

• The reactants are solid aluminum (Al) and chlorine gas (Cl\(_2\)).
• These react to form aluminum chloride (Al\(_2\)Cl\(_6\)), a compound used industrially.

In a chemical reaction like this, balancing the equation is crucial as it defines the stoichiometric coefficients needed for stoichiometric calculations. A balanced equation like:\[2 \text{Al} + 3 \text{Cl}_2 \rightarrow \text{Al}_2\text{Cl}_6\]helps ensure the conservation of atoms, reflecting what actually happens during the reaction at the molecular level.

Understanding these transformations not only aids in quantitative predictions about the outcome of reactions (such as yield and leftover reactants) but also ensures safety and efficiency when scaling up for industrial applications. By analyzing chemical reactions, chemists can develop new materials, optimize production processes, and solve relevant problems in various fields of chemistry.